On the Lasker Noether Theorem

On the Lasker-Noether Decomposition TheoremAuthor(s): A. SeidenbergReviewed work(s):Source: American Journal of Mathematics, Vol. 106, No. 3 (Jun., 1984), pp. 611-638Published by: The Johns Hopkins University PressStable URL: http://www.jstor.org/stable/2374287 .Accessed: 27/06/2012 11:01

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ON THE LASKER-NOETHER DECOMPOSITION THEOREM

By A. SEIDENBERG

In [9], following Richman [4], we considered the question of a good definition of Noetherian ring for constructive purposes. The criterion of suitability was that the defining conditions on the ring R should be classically equivalent to the ascending chain condition on ideals and should transfer constructively to R [X], the ring of polynomials in one indeterminate over R. In [9] the following defining conditions were given:

(a) Given a chain A 1 C A2 C * of finitely generated ideals in R, one can find an i such that Ai = Ai+t; and

(b) The finitely generated ideals are finitely related, or, equivalently (by [9; Theorem 2]), the set of finitely generated ideals is closed under intersection and quotient.

This is a slight improvement on [4], where a discreteness assumption is made, i.e., given elements a, b of R one can tell whether a = b or a * b. In the present paper we assume not only discreteness but even a stronger condition, namely, that the finitely generated ideals are detachable, i.e., given a finitely generated ideal A and an a in R, one can tell whether a is in A. In this way our rings RIA will also be discrete (and Noetherian).

In these considerations the Lasker-Noether decomposition theorem, that every ideal has a normal decomposition into primary ideals, never occurred. Since this theorem is a cornerstone of Noetherian ring theory we ask: Suppose that for any given ideal of the Noetherian ring R (i.e., given via a finite number of generators) one is given a normal decomposition of it into primary ideals together with the associated primes, or can construct one, then can one construct the same for any given (finitely generated) ideal of R [X]?

Although the present work is intended as a contribution to constructive (or constructivist) mathematics, it should be realized that the problem posed is a bona fide and, we believe, formidable one, from the classical

Manuscript received February 26, 1982.

611

612 A. SEIDENBERG

point of view also. From that point of view the existence of a normal decomposition of any ideal in R [X] is trivial, or well known (see, for example, [11: p. 208f]), and follows simply from the Noetherian character of R [X], but the problem of constructing a normal decomposition (on the basis of hypotheses on R) remains.

The essential difference between the classical and constructive points of view is that the classical makes a distinction between existence and construction whereas the constructive does not. Thus classical mathematics often obtains the existence of some object by means of convictions very widely, though not universally, held, whereas in constructive mathematics the existence can only be established via a construction. Thus the construction problem, which is central from the constructivist point of view, may become peripheral from the classical.

The point of view also bears on the difficulty of the problem. Thus classically we may refer to the structure theorem itself in the course of the argument and, for example, to the associated primes of an ideal A, which are usually defined in terms of a normal decomposition, whereas constructively we have to circumvent the structure theorem -and cannot even refer to the associated primes of A, unless a different definition can be found for them. (In fact, this can be done: classically it is known that a prime ideal P is an associated prime of A if and only if there is an a in R such that A: (a) = P (cf. [3, (8.8), p. 23]), and this property could be used to give a definition of associatedprime which might be useful.) Classically, too, there are methods available, especially the probing methods (cf. the remarks on p. 685f of [10],' which we do not know how to exclude, classically, but if they had suggested themselves, we would have rejected them, as we would not know how to find a constructivist counterpart for them; actually, no such suggestions occurred. In any event, we believe that the problem here posed would meet with the same order of difficulty regardless of the point of view taken. In the following we first solve the problem from a classical point of view and then add some remarks, brief as it turns out, for putting the solution on a good constructivist basis.

For a reason that will appear we prefer to pose the problem for R [X 1 .. ., Xn ] rather than for R [X], i.e., for several variables rather than just for n = 1.

Classical. Let R be a commutative ring with identity. We assume we can compute in R, i.e., add, subtract, and multiply, and that we are given the 0 and the 1. We assume that R is discrete, and even that its fi-

ON THE LASKER-NOETHER THEOREM 613

nitely generated ideals are detachable, so we can tell whether a given element = 0, and in particular can tell whether 1 = 0. Dismissing the case 1 = 0 as trivial, we suppose 1 ? 0. We are not given the characteristic: in our computations we proceed as though the characteristic were 0 unless we come upon a non-zero integer multiple n *1 of the identity; as soon as we do we test whether n *1 = 0; if n *1 0 we proceed with out computations; if n 1 = 0, then we can soon find the characteristic, and then start over.2

We assume conditions (a) and (b) for R. Classically, of course, it already follows from (a) that all ideals of R, and in particular A n B and A :B for any given ideals A and B, are finitely generated; but by (b) we mean that given two finitely generated ideals (given, it will be understood, via finite bases), then we are also given, or can construct, finite bases for their intersection and quotient.

We assume that for any given ideal A = (a1, ..., a,,) of R we are also given a normal decomposition of A into primary ideals, together with the associated primes. One can then tell whether a given ideal A of R is prime: A is primary if and only if in a normal decomposition of it, which we are given, only one primary component occurs; and it is prime if and only if it is primary and equal to its associated prime.

There are further necessary conditions on R for our problem to have a positive resolution. Assume for a moment that R is an integral domain, and let K be its quotient field. Then any polynomial in K[X] (* 0) can be effectively factored into irreducible factors. In fact, let f be a polynomial in K[X], say of positive degree. Since the coefficients of f are written as quotients of elements of R, we may assume f E R [X]. Let (f) = Q1 n ... n Q be a normal decomposition of (f) into primary ideals (which we would have if our problem has a positive resolution). By [9; p. 57, Lemma 4, Cor. 1] for any given ideal A of R [X] we can construct R n A, and thus tell which Qi meet R - 0. Hence we can construct a normal decomposition of K[X](f) and so get a complete factorization of f over K. Thus the quotient field K of R must satisfy what in [8] we have called condition (F).

The field K is discrete. Let k be any discrete field, in effect what in [8] we called an explicitly given field, and let its characteristic be p. Since the condition (P) to be mentioned in a moment is vacuous if p = 0, we may suppose p > 0. In [8; Section 45, p. 293] it was shown that one can construct the associated prime of any given primary ideal in k[X1, ..., Xn ] if and only if k satisfies what we call there condition (P). Condition (P) for k

614 A. SEIDENBERG

is defined in Section 39 and in Section 43 it is shown that k satisfies (P) if and only if any elements zI, . . ., zs can be checked for p-independence (i.e., one can check whether [kP(zI, ..., zs):kP] = pS); and in the case they are p-dependent, one can construct an equation exhibiting this. Thus if our problem for R [XI , ..., X,, ] is to have a positive resolution, then K must satisfy (P). Since the characteristic of R is not given, this condition will not be invoked unless in the course of our constructions we find the characteristic.

If our problem has a positive resolution for R, then it also, as one easily checks, has a positive resolution for R/A for any given ideal A, and in particular for any prime ideal A = P of R. Hence the quotient field of R/P would also have to satisfy (F) and (P). Thus a necessary condition for our problem to have a positive resolution is that Rp /RpP satisfy (F) and (P) for any given prime ideal P of R.

The same can be said in the general case that R contains zero-divisors 0. As before, the quotient field of any ring R/P for P a given prime ideal

of R would have to satisfy (F) and (P); and this field can still be written as Rp/R pP. Recall that Rp may be obtained as follows: Let S = R -P and letn betheS-componentof (0); i.e., n = {x eR lsx = Ofor somes eS}. Let (0) = QI n ... Q, be a normal decomposition of (0). Then n can be obtained from QI n ... Qt by dropping the Qi that meet S or, alterna- tively said, the Qi whose associated primes Pi are not contained in P. Thus we can construct n and R = R/n. LetP = P/n, S = R - P. The elements of S are not zero-divisors in R, and R - is obtained in the expected way as a subring of the total quotient ring of R :R - {x/3js I, s E R, s 3 P} . By R one means Rp and by RpP one means RpP; and Rp /RpP is the quotient field of R/P = R/P. So our necessary condition is again expressed in terms of the RpP. This necessary condition, along with the other assumptions on R, will also prove sufficient.

The ring R [XI , ... , Xn I was brought in to get the necessary condition about (F) and (P), but in what follows we will work with R [X] simply and the result will follow for R [XI, ..., Xn"] by induction, provided that we can show that conditions (F) and (P) transfer from R to R [X]. That is so by [8; Section 41, p. 291], which says in a special case that if (P) and (F) hold for a field k, then they hold for a simple field extension k(x) of k; here one will have to know the defining equation for x over k. If P* is a prime ideal in R[X], then as already mentioned once we can construct Rfn P* = P; and R [X]/P* = (R/P)[x], where x = X mod P*. If P* = (f I 9 . . .fs) and

f1, . . ., fs are the residues of the fi mod R [X]P, then also R [X]/P* =


(R/P)[X]/(f1, ... f), from which one soon gets the defining equation (possibly 0 = 0) for x over the quotient field of R/P.

LEMMA. Let F = R (E ... (E R (g summands) be a free R-module; we write F = R I (i *... * Rg with Ri = R , i = 1, . .. g; and let M be a given finitely generated submodule of F. Let P be a given prime ideal of R and assume S = R - P consists of non-zero divisors. Then one can construct RsM fl F.

Proof. Let (a1i, . .., ag),i 1, .. ., r, be a given basis of M. As a = a1 + * * * + ag, ai eRi, runs overM, ai sweeps out an idealAi, which we can construct. LetN = RRsM nF. As b = b1 + ? * * + bg, bi eRi, runs overN, bi sweeps out an ideal Bi; and one will haveBi C RsAi n R. Since we have a normal decomposition of Ai into primary ideals and the finitely generated ideals are detachable, we can write down a basis

(bil * *. *, bik,.) of RsA1 f R. Every element of N can be written in the form (E u 1jb Ij, .. ., E ugjbgj) for some u ij E R , and we ask for which u ij E R we will have (E u Ijb Ij, ..., ugjbgj) E N; or for which uij E R there is an s e S, s = s(uij), such thats(E u Ijb Ij, . . ., E ugjbgj ) eM; and we will find a finite R-module basis for the u ij. Now multiplying the bij by a specified element t E S does not change the set of solutions. We take, i.e., construct, a t such that tb Ij E A1 for every j, j= 1, ..., k 1. We write tb =

Er C (Ja lk . Subtracting Ej E c (J)(a k, agk )U ij from t(E u Ijb j, , ugjbgj) we get an element

X = (0, E tu2jb2j - S c()a2ju1j, .. ., 9E tugjbgj- c(kagku1j),

which is in Nn(R2(9 ... (? Rg) if and only if (E u Ijb Ij, .. .,ugjbgj) is inN. LetN' = Nn (RJ2 * * *Rg), M' = M n(R2(9 ... (3Rg). Then N' =RsM' n (R 2 ?) * * Rg). We can construct M' (since A 1 is finitely related) and then N' by induction on g. Then write X as a linear combination with unknown coefficients of a basis of N'. Then we have a system of homogeneous linear equations over R and can construct (or are given) a finite R-basis for its set of solutions. This yields a basis for the u ij and, via this basis, a basis for N.

COROLLARY. One can find an s E S such that sN C M. The proof is by induction on g.

THEOREM1. Let A = (f 1, . . ., fq) be a given ideal in R[X], P a given prime ideal in R. Then one can construct R, [X]A n R [X].'

616 A. SEIDENBERG

Proof. Let S = R -P and let n, which we can construct, be the S-component of (0). Let R = R/n, P = P/n, and S = S/n, so S consists of non-zero divisors in R; and let A = A mod R [X]n. Then Rp [X]A n R [X]

is the inverse image under R [X] -- R [X]/R [X]n = R [X] of Rp [X]A n

R [X]. So with a change of notation, we may suppose that S consists of non- zero divisors.

ByA:fP for large p we meanA:fP if A:fP = A:fP+1; by [9], or (a), one can construct such a p. We use the well-known identity A = (A :fP) n (A, fP) for large p. Write A = (A :Xp) n (A, XP) for large p and thus reduce the problem to the cases

(i) no associated prime of A contains X, and (ii) every associated prime of A contains X.

First suppose no prime of A containsX. Letfi = ajo + a1iX + * and let A0 be the ideal of constant terms of elements of A; thus A0 = (a 10, . . ., a qO). Now let B = Rp [X]A f R[X]. The constant terms of elements of B are contained inRpAo n R = (c1, . . ., cm). Let tici eAo, ti E

R -P, and t = H t1, so tci eA0 for everyi. If b eB, then tb has a constant term in Ao. Let N be the R-module generated by fl, ..., fq. Then tb - n 0(X) for some n E N, deg n c max {deg f}i , so subtracting n from tb does not touch the terms of degree > 1 + max {deg fi } in tb. If deg b ? 1 + max{deg fi}, then deg(tb - n)/X = deg b - 1 and (tb - n)/X E B, since the primes of B are amongst those of A, and so B:X = B. If deg(tb - n)/X > max{degfi}, repeat the process: (t(tb -

n)/X -n 1 )/X E B, i.e., (t2b - a)/X2 B, where a E A and deg(t2b -

a)IX2 = deg b - 2. Repeating if necessary, we find a p such that (tPb - a1)/XP e B and has degree c max {degfi}; a1 eA. LetMjbe the R-module of elements of A of degree cj and let MJ be the R-module of elements of B of degree 1j. Then MJ = RpMj n (R + RX + * * * + RXj). By [9; Lemma 4, Corollary 1] one can construct M1 and then by the

lemmaMJ, for anyj, in particular forj = max{degfi} . Then (tPb - al)! XP is in MJ' and tPb is in the R [X]-ideal C generated by A and MJ. So b e C: tP for large p. Conversely, if b E C: tP, then tPb E B, whence b e B. Thus Rp [X]A n R [X] = C: tP for large p, and this we can construct.

If A contains XP, the matter is even simpler, since in that case the problem will be solved upon construction of M'_1.

THEOREM 2. One can construct the minimal primes of any given R[X]-idealA = (f, * * *, fq)


Proof. A = (1) has no minimal primes, so we suppose A * (1). Let PI, ..., P be the minimal primes of A, and let Pi = Pi n R. The minimal amongst the primes Pi are the minimal primes of A n R, as one may see by considering a Lasker-Noether decomposition of A and contracting it toR, so we can find these. LetP' be one of them (and letP1 n R = P'). If R [X]Pj D A, then R [X]Pf is a minimal prime of A (and P1 = R [X]P1'). If R [X]P' ;$ A, then some element of A, one of the fi for example, is not in R [X]P1'. Passing to Rp [X], one will have an element f in Rp [X]A - Rp [X]P . Look at f mod P' = f in (Rp, /Rp;Pf')[X] and factor f completely, say f = I ... fh; take counterimages f, .. , f off fI, ... fs. Since Rp1[X]PI /Rp;[X]Pj' = (fi) for some i, we get candidates RP1 [X](P1, fi) for Rp1 [X]P1; at least one of these contains Rp1 [X]A, and any one containing Rp1 [X]A will be a minimal prime of it. Contracting to R [X] we get at least one minimal prime P of A.

The ideal A: PP for large p, which we can construct, has P2, ..., P. as its miniumal primes, as one sees from a simple consideration on a normal decomposition of A. If A: PP = (1), then P is the sole minimal prime of A; otherwise we construct a minimal prime of A :PP. This will be a second minimal prime of A. Etc.

LEMMA. If one can write any given R [X]-ideal A * (1) as a finite intersection of primary ideals intersected with a strictly larger ideal A 1, then one can write A as a finite intersection of primary ideals (and hence find a normal decomposition for it).4

Proof IfA1 = (1), we already have A written as a finite intersection of primaries. If A 1 * (1), then we write A 1 as the intersection of a finite number of primary ideals and an ideal A2 strictly greater than A 1, and we repeat the argument if A2 * (1), etc. This process must terminate by property (a).

THEOREM 3. If for any given R [X]-ideal A, having P1, ..., P., as its minimal primes, one can construct the associated primaries Ql ... ., Q. of a Lasker-Noether decomposition of A, then one can construct a Lasker-Noether decomposition of A.

Proof. If A = Q, n .. Q,, then one already has the required decomposition. Otherwise A: (Ql n ... n Q,) * (1). By the structure theorem one sees that the associated primes of A: (Ql n fl. Q. ) are amongst those of A and that none of them is a Pi. Hence A: (Ql n ... n Q,) is in no Pi and one can find (by a familiar construction) an f e

618 A. SEIDENBERG

A:(Q1 n nQ) - UPi. WehaveA = (A:fp) n(A, fP) =A, n cl for large p, where A:fP = A1 and (A, fP) = C1. Here A < A 1 since f is in some associated prime of A, and A < C1 since f 0 P1. A 1 has the same minimal primes as A and in a normal decomposition the same associated primaries Q1 , . . . , Q,. If A * Q1 n ... n Qs, we repeat the process with a g eA1:(Ql n fl n Q) - UPi. We getA = (A:fP):gf n [(A:pf, g) n (A, fP)] =A2 f C2 for large a, where A 2 = (A:fP): g and

C2 is the ideal in the brackets. Here A < A 1 < A 2 (for a reason given) and A < C2 since fpg' E C2 - A. We repeat the process over and over, getting sequences A, A1, A2, . . . and C1, C2, . .. with A < A1 < A2

A < Ci, andA =Ai n Ci. The sequence A < A 1 < ... must terminate by property (a), and when it does, say with A ,, we will have A =

(Q1 n fl. n Qs) n Ci, withA < Ci. By the lemma the proof is complete. So our problem comes to finding the primaries Q 1, . . ., Qs associated

with the minimal primes PI, . . ., Ps in a normal decomposition of A. Let P be one of the minimal primes. Dividing A by a large power of the product of the other minimal primes, we get an ideal A 1 with only one minimal prime, namely P, and with the same associated primary in a normal decomposition. Thus we may assume A has only one minimal prime P. Let P= P n R. Passing to Rp,[X], there are two cases:

(i) Rp,[XJA ? Rp,[X]P', and (ii) Rp,[XJA C Rp,[X]P'.

In case (i) all the minimal primes of Rp, [X]A are maximal, and there could not be more than one of them, as otherwise A too would have more than one minimal prime. In this case Rp, [X]A is primary and Rp, [X]A n R [X]

is the sought primary ideal. In case (ii) Rp, [X]A has Rp, [X]P' as sole minimal prime, but Rp,[X]A may have embedded primes. A main difficulty is to slough off these embedded primes (without changing the primary belonging to Rp,[X]P').

Thus we come to the crucial case that R is local and A has the extension of the maximal ideal of R to R [XI as sole minimal prime. With a change in notation, we call P the maximal ideal of R. We can construct an n such that pn C A, and passing to R/Pn, we may suppose pn = (0). In that case R is complete. We wish to apply Cohen's structure theorem for complete local rings (cf. [1]). First we deal with the equicharacteristic case, in which case, classically, R contains a field (cf. [1; p. 72]).


The equicharacteristic case with p = 0. R is a local ring with maximal ideal P = (x1, ..., xm); pn = (0); A = (fi, . . ., f,) is an ideal in R [X] with A C R [X]P. We do not know the characteristic of R or of R /P, much less that they are equal, but we proceed as though the characteristics of both R and R /P were 0. This provisional supposition only means that we suppose we do not come upon a relation m * 1 = 0 or m 1 0= (P) for a non- zero integer m in the course of our computations; if we do, then we will soon have the characteristic pn, p a positive prime, of R; if n > 1, we go to the unequal characteristic case and if n = 1 we are definitely in the equicharacteristic case with ch. p > 0.

We write the elements of R in the form E ciI ... imxii ... xI'm and will be adding and multiplying them. Our computations will involve only a finite number of such elements and thus only a finite number of coefficient c's. Let k be the quotient-ring of the prime ring of R with respect to the integers 0 P that occur in our computations. Thus we will be working in k[..., cil . , ...]. In addition, if we come upon an h E k[....

Xi,.. im, .1 such that h(c) E O(P), then we may want to consider 1/h(c).

Let the c's in some order be c1, . . ., cu. We would like to arrange matters so that if we come upon a relation h(c) = O(P), then we will have h(c) = 0. If we had a field of representatives I (of R/P), this could easily be done, as we would choose the c's from L It is not to be expected, however, that we can construct I, as I is usually obtained by trans-finite arguments. Nevertheless, because our computations are finite in number, we will be able to arrange the desired situation.

LetF = Ecil ... imXi X m andG = Edi ... i X" .. X'm ,where

the ci1. im and di,1 . i m are in k [c1, .. ., c . Then the coefficients of FG are in k[c1, ..., cu] and some of these may be =0(P). We will be concerned with such relations and similar ones obtained from F + G.

Let c-1, . . ., c-u be the residues of the ci mod P. We arrange these in some order so that there is no relation over k = k mod P (amongst the relations we have) of c I, . . ., c r whereas c-r+i is algebraic over k [c-, . ... &r+i- 1] for i ? 1. We may have several relations for c-r+1 over k [RU, ..., &cr]. If so, then using the division algorithm, we may reduce these to one relation Fr+I (c-I, .. ., cr+) = 0, Fr+j E k[X1, * .* , Xr+1I. In applying the algorithm, we will at times be testing whether a leading coefficient is =0. If it is, we get a relation between c-l . . .. c- In that case we start over, with a reduction in r. Since this kind of reduction can take

620 A. SEIDENBERG

place only a finite number of times, we may suppose no relation in c1, ..., Zr arises. A similar argument shows that we may suppose Fr+1 (c-1, ... ., c-r, X) has no multiple roots. Here we will be taking a deriva- tive of Fr+1 (c-1, . . ., cr, X) and will have to examine certain integral mul- tiples of 1 1 mod P. In doing so we may find that the characteristic of R/P (and hence also of R) is * 0. For a reason given, we may suppose that doesn't happen; similarly at further points in the argument.

Using Hensel's Lemma we can construct an element c /1 such that r+1 -Cr+1 modP and Fr+j(c, . . ., Cr, cr1) = 0 (cf. [1, Theorem 4,

p. 63]); in general the proof of Hensel's Lemma involves an infinite construction, but in our case it is finite since P' = 0. Then we construct an element c?r+2 C cr+2 mod P such that Fr+2(c 1 r ** Cr, c,41, cr+2) = 0; etc. for cr?3, ..., cu. We write c' = ci for i = 1, .. ., r. Thus we have elements satisfying Fr+j = 0, ..., Fu = 0.

We have a homomorphism k[c, ..., c ] -+ k [c1, 9 ..., C u] with c/ 4 C-i. We cannot say this homomorphism is an isomorphism because we do not know all the relations between the c/; but as far as the relations that occur in our computations are concerned the cl' and the c-i satisfy the same relations. Thus we could say that the homomorphism is isomorphic- like. Similarly, though k [cl, . .. , cU is an integral domain, we cannot say that k[c , ..., c ] is, though we can say that it is domain-like. If g, h E

.C[Xl . ... . Xuj, h arises in our computations, and h(ci, ..., c') * 0, hence h(c, . . ., c') # 0(P), we can then form the fraction g(c')/h(c'); and forming all such quotients we could speak of a field-like ring, which we will denote by "k(c', ..., cr),'" with quotation marks to indicate that we are not claiming that we really have a field. k itself is a field-like ring.

Our data is given in the form of several polynomials h = i i E ci1 ... X m. The cil . im in some order we have called

... , cu, and we have computed c, ..., c I with c c ci(P). In 2: cil .. mx'l .. **xmm we write E (c!' -i + di, . .. im)xt11** x'm, where if Cil ..., im =Ck 9 then cl im Ck; and di, ... iE P. Write the di, .. im 9 and especially do ... 0, as linear combinations of x1, .. ., xm with coefficients from R. Then h(x) takes the form c' ... 0 + E c* ... jiX1 * m.

(i1 + .. + im > 0), and we have the polynomial h* = co' o +

Ic1 . . Xl ... Xmm. With these new "data" h * we go over our computations with the h. Unless new relations in the c1, ..., c arise, which we can suppose not to happen, we will not compute new "primed" values for the cI, . . ., c9; butwewill compute "primed" valuescz'...i fortheci*1 ... im We write thecz .c ., and especially the c for i1 + + im = 1, in


the form ct'... + a linear combination of x1, ... xm with coefficients in R, so that h*(x) takes the form c0 ... o ? S c/1. iX11 * xim +

I2c *z ... imx1* x m, where in the middle sum i1 + + im = land in the last il + * + im > 1; and we have the corresponding polynomials h **. We repeat a similar process for the h **, etc. The process terminates since i + + + im < n; and when it does we will have rewritten our elements h(x) as polynomial expressions in x1, ..., x m with coefficients from a field-like ring.

Thus our eventual field-like ring "k(c , ..., c"')" is analogous to the coefficient field in Cohen's structure theorem for complete local rings. We can think of the subring of R in (or over) which our computations take place, as having the form I[x 1, ..., m ], where I is a field-like coefficient ring.

Now we want to carry out computations back to I[X1, . .. , Xm ] and to do this will need to know the relations satisfied by the xi, ... * *m over I. We can decide whether one of the xi is in the R-ideal generated by the others, or alternately, whether x 1, . . ., x m is a minimal basis of P, and we may as well suppose that it is; equivalently x1, . .. -xm mod P2 is an RIP- basis of p/p2. Consider now the power products a1, . .., aq of the xi of degree 2. We can get an R-module basis for the relations of a1, . . ., aq

over R, and if in any of these a coefficient OP, we can write some ai in terms of the others. Continuing, we can get an R/P basis, say o1, a2, ... a Ur

mod P3, for p2/P3; and we have relations expressing Ur+?l .. ., Uq in terms of o1, .. ., aOr and power-products of the xi of degree > 3. We proceed similarly with p3/p4, ..., pn-l/pn. Eventually we get elements to = 1, tt, ... . tq, power-products of the xi, such that in any linear combination of the ti of degree d that is o0(pd+l) the coefficients are inP; and we have a set of relations S giving the other power-products as linear combinations of t1, . . ., tq. With a possibly increased I we may suppose the coefficients in the relations in S to be in I. The construction allowing this arrangement goes forward, since in a relation between the ti if the coefficient ci of ti is replaced by ci + d with d e P, then dti can be written as a linear combination of tj of greater degree. Then S will be , as far as our computations are concerned, an I-module basis for the relations between the xi, ... X'm over I. In fact, in any such relation we may replace any power-product not a ti by a linear combination of the ti with coefficients in I to get an equivalent relation E aiti = 0 with the ai E L Then all the a= 0, for if not, write the relation as b 1t 1 + * + bst i = O with all the bi * 0. Let, say, ti1, t, be the tij of lowest degree, sayd. Then b 1til +

622 A. SEIDENBERG

* + brtir = (pd+1), so bi 0(P); and since bi e I, bi = 0, a contradiction.

We have the ideal A = (fi, ... , fs) in R [X], where we suppose the fi to be written with coefficients in I[x 1, . . ., x] . In the fi we replace the xj with Xj to get polynomials f*, ..., f* in I[X1, ..., XmJ. Let g1= 0 . **, gt= 0, where gi = c . xXl * * Xmm, be our I-module basis S; and consider the polynomials gr = c31 ... jmX1i1 ... Xlm; these together with the power-products 'n-1, ..., vu of the Xi of degree = n, gen- erate the known part of the kernel of the homomorphism

[X1 I *

X *j[X] [X] I[X1, *, X m I [X], where Xi - xi . Let A* = (f . *.. f I g 1 **.* gIC, *1r *1 * *, u). Next we are going to construct I[X]s[Xl, *.., Xm]A n fi[X, X I *.. *, Xm, where S = the multiplicative system of elements in I[X] not all of whose coefficients are in P. Note that S is detachable from I[X].

LEMMA. Letkbeadiscretefield,R =k[X,Xl. .,Xn ], n2O,let RZ1 + * * * + RZt be a free R-module having Z1, . . ., Zt as free genera-

tors, letM = (Q1, , * Ss)I Qi = fi Z1 + * * * + fit,Z, be a submodule of it, and let S = k [X]- 0. Then one can construct k [XI s [X 1, . . . I XnjM n E RZi; and moreover one can find an s e k [X] - 0 such that this intersection is M: s.

For the proof, which is an induction on n, first assume n = 0. In that case one can find, as is well-known, free generators Z, ..., Z/ of E RZ, and a new basis (, , Q' of M such that (/ = gi(X)Zi' with gi e

k[X] -0 (cf. [8; Section 8, p. 278]). Then clearly RsM n E RZi = (Z1', ...,Zr') ands =g1 gris the desired s. In the case n > 0, in order to meet the regularity condition on D in [8; Section 9, p. 278], one makes a homogeneous non-singular linear transformation on X1, . . ., Xn as in [10;

Theorem 8, p. 698].6 One computes as there (with a slight change in the notation) an element t e k [X] - 0 and works temporarily over k [X]T,

where T is the multiplicative system {1, t, t2, . .. }; this will introduce a power of t as a factor into our eventual s. One establishes

RsM n E2RTZi = R TM + (RIM' nf R 'ZiXXn),

where R' = k[X, X1 X R] ZiXj, is a finitely generated free RT-module with the ZiXj as free generators, and M' is the finite R '-module generated by certain (iXI. Applying the inductive assumption to the second summand on the right, one soon has a desired s, at least relative to


RT. After constructing RsM nf RTZi = N, one still has to contract to E RZi. If N* is a finite submodule of E RZi extending to N, then the desired contraction is N*: tP for large p, and this we can find.

In applying the idea of this lemma to I[X, X1, . .. , X,, I and M = A *,

we take S = multiplicative system of elements in I[X] not all of whose coefficients are in P. Although any constructed element of I in P is =0, we still do not know that P n I = (o); and although I is field-like, we do not know that it is an integral domain. We come to the assertion that RsM n S RZi = (ZI, *., Zr). To get this we have to show that sQ e RZ + +RZ' implies(eRZ1' + +RZr';hereseS,(e2ERZi. In the lemma this follows because k[X] has no zero divisors (except 0). In the application it follows because (0) is a primary ideal in R [X] having R [X]P as associated prime. Then towards the end of the proof, we come to constructing N*: tP for large p; in the application we come to a constructed ideal B * in I[X, X1, ... , Xm ] and would need to construct B*: tP for large p. In the lemma, k and k [X, X1, ... , Xj ] are Noetherian, and this allows the construction of N*: t P for large p. In the application, we do not have that I is Noetherian, so another modification in the proof of the lemma is needed. We give two proofs, the first depending simply on the field-like character of L In [8; p. 306] we showed that if f (i) i = 0, 1 2, . . ., is a non-negative integer andA i is an ideal in k[X1, ... , X], k a field, having a basis of elements of degree f (i), then one can compute a bound g", depending only on f and n, on the length of any strictly ascending chain

AO < A1 < ... < As; cf. [5] and [6].7 In [8; Section 56, p. 296] we showed that if A = (f1, . . ., fr) B = (fr+l ? ... . fs) are two ideals in

k[Xl, ... , Xn J and d 2 max{degfi }, thenA :B has a basis of elements of degree ?n((d + n)'d)2 . These assertions also obviously hold for a field-like ring k. Hence we can put a bound on the length of a strictly in- creasing chain B* < B*: t < B*: t2 < ... , in terms of m and a bound on the degrees of a basis of B *, and so can construct B *: tP for large p.

Here is another construction of I[X]s [X1 .. ., Xm ]A * n I[X, X, .. .,Xm ], taking into account that (X1, . .., Xm)n C A*. Let ui, j,] = 0, 1, ..., m be (m + 1)2 indeterminates and introduce Xi, i = 0, 1, ... ., m, via the formulae Xi = E u ij', writing X0 for X; in matrix notationX = UX'. Let d = det(u ij) and let H be the multiplicative system generated by the elements in I[..., ui,j ... ] not all of whose coefficients are in P arising in our computations, so that I[..., uil, ... ]H is a field- like ring. Let Su be the multiplicative system of elements in I[..., ui1, .. .][X, X1, ..., Xm] not all of whose coefficients are in P;

624 A. SEIDENBERG

note, as once before in a similar situation, that the elements of Su are not 0-divisors, since (0) is primary in R with P as associated prime. Since d E H, we have that I[ ..., uij, ... *1H4X, X1, ... , Xm] =

I[... * Uij, ... *]H[XO' Xl,.., X,]; call this ring R 1. Note that X0, X1', ..., X' are algebraically independent over I[ .. ., ui, ... ] (and over I[...., uij, ...IH), for if G(U, X') = 0 for a polynomial G E

I[ .. ., ui, . . ][X, X1, ... .,Xm ], then G(U, U 1X) = 0, and now replac- ingXby UX, we get G(U, U-1UX) = G(U, X) = 0, i.e., G = 0. We will consider a situation like in the lemma, i.e., we will consider a free R 1 -module E R lZj, etc. This time the regularity condition on D relative to

X0o', ... . X, is automatically fulfilled: the coefficient of X Ideg D is an element h E H. One will then obtain

RIsuM nf RZj = R1M + (RsuMm n E2R 'Z X,i)

where R = I[..., uij, ... .IH[XO Xl, ..., Xm-] etc. To get the regularity condition for M', we could introduce further indeterminates and another linear transformation, as was explicitly done in [8; Section 11, p. 280], probably without creating any difficulty, but we may note, following [6; p. 309f], that M' (and the other modules appearing inductively) already meet the regularity conditions. Hence we can construct R 1suM n S R 1Zj; and in particular R IsuA* n R 1 = BU*; and we have an element s(u, X0) E I[ ..., ui, **... ][X, X1, ..., Xm ] such that Bu* = R 1A *: (s (u, Xo)). It is now a matter of unloading the u ij. We first observe that B * has a basis in I[X, X1, ... , Xm J: We may suppose it has a basis in I[.. . , ui, ... ][X, X1, . .,X, ]; if b is one such basis element, we regard it as a polynomial in the uij with coefficients in I[X, X1, ..., Xm], and taking the appropriate partial derivatives of b with respect to the u ij, we get that the coefficients are in B *. Thus B* has a basis in I[X, X1,

. Xm ]. Let B* be the I[X1, ..., Xm ]-ideal generated by this basis. We claim that I[X, X1, ..., Xm]sA* n I[X, xl, ..., XmI = B*. For let b E B*. Then s(u, X0')b E R 1A *. The coefficients of s (u, X0) are not all in P, and since s(u, X') has been constructed, we may suppose that any coefficient in P is =0. Note, too, that s(u, X6)1Xl=...=xm=o s 1 (u, X0) * 0. Write s(u, X0') = s 1 (u, X0) - R, where every term of R is of positive degree in X1, . . .,Xm. Multiplying by Sn-1 + Sn-2R + ... +

n- , one sees that snb E R 1A *. Writings .nb as in R 1A *, multiplying by a denominator in H, and comparing coefficients of the power-products in

theuij, one finds that b EI[X,X1, . . .,Xm]sA* n I[X,X , ... 9,XmI.


ThusB* C I[X, X1, ...,Xm]sA* n I[X, Xi ...,Xm]. The opposite inclusion is proved similarly, again taking into account that (X1, ..., Xm)' C A *. One can also find a desired s(X) E I[X, X1, ... , X,,, I, as one easily sees. The construction is complete (in characteristic 0).

Our problem is to decide whether A is primary, but (as will be seen) we cannot do this confining ourselves to the characteristic 0 case, but will have need of the construction of I[X, X1, ..., Xm]sA* n I[X, X1, . . . I, ] also in the equicharacteristic case with p > 0 and for the unequal characteristic case. So we continue with this subsidiary problem.

The equicharacteristic case with p > 0. With the same notation as before, we haveP' = (0), and we take an e such that pe > n (so alsoPP = (0)). Then Rp = {xPe x E R } is a ring, because of the characteristic, and it is even a field, since if xP * 0, then x 0 P, x has an inverse y ER, and yP

is the inverse of xp . In ch. p > 0, we take this field to be our field k. Note that k is isomorphic to (R/P)Pe (via R -* RIP) and to RIP, hence satisfies conditions (F) and (P). We write (R/P)Pe = k. As before we have the elements ct, *..., cU, but this time they are purely inseparable over (R/P)Pe = k and we can get k(zF, . . ., cu) by a succession of adjunctions of irreducible pth roots. After adjoining a pth root to k, the resulting field continues to satisfy (F) and (P); and similarly for the adjunction of the other pth roots. Stepwise we get an isomorphic copy I of that field in R. This time I actually is a field, and not merely a field-like ring. The rest of the considerations are just the same as before in ch. 0, except for a minor difficulty that may arise when one takes partials with respect to indeterminates uij (cf. [8; p. 282] or [6; p. 310]). The first proof becomes simplified, since here I is a field and hence Noetherian.

The unequal characteristic case. Here we suppose that we have found RIP to have positive characteristic p; and R will have ch. ps , s' > 1, in view of Pn = (0). Now we follow Nagata's "Local Rings," pp. 106- 109, to get a coefficient ring I of a certain simple kind: I will be a complete local ring (contained in R) with P nfI as maximal ideal, andP n I = (p); moreover, I maps onto RIP in the homomorphism R -+ RIP; and any element of R can be expanded into a power series (in our case a polynomial) in x 1, .. ., xm, generators of P, with coefficients in I. Nagata deals with an arbitrary complete local ring, thus does not have pn = (0); he first constructs for every n a coefficient ring Jn in RIPn and then puts the Jn together to get the coefficient ring L This last part will not have to be done

626 A. SEIDENBERG

by us, since R/P' = R and Jn will be the desired coefficient ring. Nagata first observes (p. 107, (31.3)) that if a - b E P, then a Pn -bp E pn+l; the proof is a simple computation involving an elementary property of the bi- nomial coefficient (f), where q = pn . Thus one gets a 1-1 map an ofK = RIP into R/Pn via the map :R/P -+ R/P' defined by 0n(a) = a

(Nagata defines a 4n by 0n(a) = aP , apparently taking the exponent 2n large enough for (31.3) to apply, but n, and even n - 1, is large enough for that; we will write n where Nagata has written 2n. ) Let An be the image of RIP under an. Since 4 n induces an isomorphism from K = RIP to K

there is a 1 - 1 map between Kp and An such that (a mod P)Pn corre- sponds to (a mod P))Pn . Nagata next takes a p-basis B* of K and a set B of representatives of the elements of B*, one representative b for each b* E B*, and considers the set Sn of polynomials in elements of B taken modulo pn with coefficients An such that the degree of the polynomial in each element of B is less than pnf. Since Kp (B*) = K and the monomials in the elements of B* of degree less than pn in each element are linearly independent over Kp (by (31.2)), one finds that K and Sn are in 1 - 1 correspondence and Sn is a set of representatives for K. Placing Jn = Sn + Snp + * * * + Snpn-t , Nagata proves that Jn is a ring. For the proof, note that a product of monomials in the elements of B mod pn is in Sn, so that if we knew that Jn were closed under addition, the sum and product of elements in Sn would be in Jn and Jn would be a ring. Thus it suffices to prove that the sum of two elements in Sn is in Jn . Let E aMM, 12 bMM be elements of Sn , where aM, bM EAAn, M = b ... b Ir with bi E B (mod Pn) and 0 c ei < pn. The proof takes the form of proving pi (E, aMM + E bMM) is in P 0Jn O c i c n, by an induction on n - i. The assertion is

obvious for n - i = O, so assume that 0 c i < n. Let q = pnf. Each aM and bM is a qth power, so we have elements CM, dM in R/Pn with aM = cm, bM = dM. Since CM, dM may be replaced by elements congruent to them modP/P, we may assume cM, dMESSn. Now aM + bM = (CM + dm)

q1 (q)cqr dr, and since each (q) is -O(p), one sees that pi

(q)C qrdrM is in Pi+lJn by induction on n -i (and keeping in mind that CM, dM E SO.) Hence E p1(aM + bM)M = E p1(cM + dM -

pi (q)Cq rdrM is in piJn 9 and the assertion is proved. One checks easily that the ring Jn has the desired properties (P/lp n Jn = PJn 9 etc.).

Thus except for the existence of the p-basis B*, which Nagata gets by an application of Zorn's lemma, Nagata's construction would offer us no difficulty. To get around the use of Zorn's lemma, we can proceed as follows. We shall construct a sequence b 1, b2, ... in K having certain de-


sired properties, in particularly, they will be p-independent and the P-residues of all the elements c 1, c2, ... appearing in our computations will lie in KP( b 1b, ...). Let c 1, .. .,c be, as above, the elements that appear in our computations. From cj . .,u we pick out a p-basis (i.e., a maximal p-independent subset), say c , .* r . which will be the first of our bj: we can do this since K satisfies (P). Let y(j) = .. **r, 0 <

1k < p; here (j) = (jl, I . ., jr) is a multi-index. Then we can write i = E djpj)y(j), i > r, and complete c * - . . . r (=b .. bIr) to a p-basis b I , * , b5 of c1, *I r . di(j), . . .. This operation is repeated over and over (i.e., we write the d i(j) as a linear combination of the power-products b 1, ... b's, 0 < ik < p, with coefficients from KP, etc.). This proce- dure will occasionally be interrupted as follows.

In the proof that J,n is a ring, we come upon an element CM and we had to replace CM by an element in Sn; in effect, we had to write CM, the residue of CM mod P, as a polynomial in the elements of B* with coefficients in K . To obtain the corresponding result, having constructed b1, ...,b

say, we test the p-independence of_cM on b 1, . . ., bt. If_ M is p-independent, we simply adjoin cM to b 1I, . . ., b t, placing c M = b t + 1; if CM is p-dependent, we write CM as a linear combination of the power products b 1 b. t , O 0 ik < p, with coefficients from KP, etc. Thus we are getting involved with the construction of a second sequence, but we can combine the two sequences into one by working on one of them for a while, then going to the other for a while, and working back and forth between the two sequences. Thus we construct our sequence b I, b 2, . Note that any element occurring in our computation lies in KP (br, b2, ...). For example, we write cu = E dp(j),y(j) then displace some of the dk(j), getting C-U as a linear combination of power-products of some of the bi with coeffi-

,F2 p2 cients e. in K , etc., and eventually get c-u as a polynomial in some of the bi with coefficients in KP . Thus here, too, we can get a desired ring J. J is not quite a coefficient ring, but it encompasses all the coefficients that occur in our computations. We do not construct all of J, but only that part of it needed for the construction of J[X] s [X1, . . ., Xm ]A* n J[X, X1, .. ., Xm ] for a certain ideal A * and S = multiplicative system of elements in J[X] not all of whose coefficients are -O(p), but J is a true ring in which we can work constructively much as one works classically in I.

Let us check some of the desired properties of J, in particular, let us find the units in J. For notational reasons, consider first the classical case with Jn = Sn + Snp + * * + Snpn - . Let co E Sn = O. Then co

0 plpn

628 A. SEIDENBERG

and using K = KP (B*), there exists (and in the case of J one can find) a c0 E S, with coc6 = 1 (P/P'). Then (coc )Pn = 1 and cV ncjPn E Jn is the inverse of c0; similarly for any element c0 + c1p + * * * + cnIPn1

with ci e S,, and c0 * 0. So we have the units in Jn. Similar, or the same, considerations hold for J. We also see that J contains precisely n + 1 ideals, namely, J, Jp, . . ., Jpn-1, Jpn . Discreteness holds in J and we can obviously construct the quotient and intersection of any two given (finitely generated) ideals of J. So J is Noetherian. On the other hand, J/Jp * K, but merely = Kpn(b1, b2, ...); but we need not concern ourselves with whether J/Jp satisfies (P) and (F). If we want to factor a polynomial over J/Jp, we factor it completely over K, and introducing the coefficients into our computation of the sequence b 1, b2, . . ., we may suppose we have a complete factorization over J/Jp. Thus in effect, as far as our computations are concerned, we may assume J/Jp satisfies (P) and (F).

As before we carried the considerations back to k[X1, . .., X,, XI with k a field or field-like ring, so now we carry them back to J[XI, . . .. Xm, X]. Every prime ideal p in J[X1, ..., XM, X], contains p, since pn= 0, and we give p a dimension according to the dimension of its image in J[X, XI, .. , Xm ]/(p) = J/(p)[X,X, . ., Xm I. We have an ideal A * in J[X, X1, ..., Xm ], like our previous A*. A* contains a power of each Xi, so the construction of J[XIs[Xi, ..., XmIA* n J[X, X1, ..., Xm I amounts to finding an element s e S that is in all the 0-dimensional primes of A *, for the desired intersection is A *: sP for large p. A similar situation, with J replaced by the ring of integers Z, occurred in our paper "Construc- tions in a polynomial ring over the ring of integers," and our present problem yields to similar considerations.

Theorem 4 of that paper reads: Let R = Z[X, X1, .. ., X, ], let RZ1 + + RZt be a free R-module

with Z1, ..., Zt as free generators, and let M = (fI, . . ., ES ) with fi =

f1iZ1 + ... + fitZt be a finite submodule such that pbZi E M, i = 1, . .., t, for some given prime number p > 0 (one can then construct a suitable b, or may suppose this given). Let S be the multiplicative set in Z[X] of elements +0(p). Then one can construct RsM nf RZi.

Now we make a similar statement for R I= J[X, X1, . . ., Xm J. Here, of course, p is a generator of the maximal ideal of J, and we need say nothing about b, since pn = 0.

The proof follows closely the proof of Theorem 4, though there one could cancel a p, whereas now p is a 0-divisor, so we can expect that some changes will have to be made. The proof is an induction on n and starts


with the case n = O and t = 1. Thus M is an ideal A* = (f l, f ) in R1 = J[X]. Letfi = prif; withf/ 0 0(p). ThenRIsA* n R1 = (pr), where r = min{ri} (obviously). Moreover if, say, f1 = prfl (with ft +

0(p)), then A*: = (pr), and f is a desired s (here we varied from the proof of Theorem 4). Now, for n = 0, there is an induction on t. Let, then, e =f1Zl+ + +ftZt E ERlZi ands eS such thatsQ E M =

(e1, * * * Q), Qi = f ijZj Then Sft e (Ftl, . .9, f ), so one can find an s with s ift E (ft, .1 . ., ft1 ), and correspondingly an element rtet + * * * + r e. such that s 1 Q- ri f i e R 1Z2 + * * * + R l Z t *Then s(s 1 Q -E rifi) E M n (R tZ2 + * * * + R lZt). Since we can construct M n (R tZ2 + * * * + R ZtZ), we can construct an s 2 E S such that s 2 (s 1 -

E rifi) EM n (R tZ2 + + R lZt). Hence sls2f E M and sls2 is the desired element s.

The rest of the proof of Theorem 4 holds without changes (except for the notational ones) for J[X, X1, ..., X1 ]. At one point in the proof we say that we may as well suppose M 0 0(p), as if M = 0(p), we can simply cancel a p; but we added that this was not necessary. Here, of course, we would not attempt to cancel p.

Denote an element s obtained by one of the above algorithms (such that I[X]s[Xt . ... , Xm]A* n I[X, X , .. ., XmI =A*:s) for a coefficient ring I by a(I).

We go to the question of deciding whether A = R [XI(f, .. ., fs) is mixed. At the start we do not know the characteristic of R or even whether we are in the equicharacteristic case. However, we start as though the characteristics were 0; and we can suppose that any set program, for example, constructing I[X]s[X, *. ... Xm]A* n I[X, Xi, ... ., XmI, can be com- pleted without our coming upon a non-zero characteristic, as otherwise we start over with the increased knowledge and the argument will proceed in a similar (and even simpler) way. Suppose, then, that we have constructed a field-like coefficient ring I. Suppose now, too, that for a reason that will appear, we want to continue our computations, bringing in further coefficients ci, and that we construct a new coefficient ring I'. In constructing I', we may have come upon a positive characteristic, so the construction of I' is not just a continuation of that for I. What is the relation of I to I'? In constructing I we have the computations yielding the coefficients

. c. In constructing I', these c/ are themselves conceivably replaced by "primed" values ci". However, all relations holding for the c! hold for the cj", i.e., h(c1, ..., cu) = 0 implies h(ct", ..., ci') = 0, as h(ctf, ...,cu') = Oeven follows from h(c, ,) 0(p).8 There may

630 A. SEIDENBERG

be more relations between the cl'. However, if in the construction of I we have constructed an h with h(cl, ..., cu) * 0, then also h(cf, ..., cs') * 0, for c!' = c!" mod P, so h(c', ..., c') - h(cl. .., c ') modP, so if h(cf, ., ci') = 0, then h(c', ., cu)- O(P), whence h(c, ..., c ) = 0, a contradiction. In effect, then, we may suppose I C I'; and the computation for I to be amongst those for I'.

Let, then, I' be a further coefficient ring, with I C I' (as explained). For I' we also compute a a = a(I'). We say that according to the algorithms indicated one can find positive integers X, ,u such that a(I' )X = a(I)". (We surmize that X = ,u = 1 suffices, but do not need this.) If I' should also be in the characteristic 0 case, this is obvious, for the directions for computing a(I') are identical with those for computing a(I). 9 Even in the equicharacteristic case with p > 0 this is so, for the directions for computing a in the case of a field-like coefficient ring in no way depend on the characteristic, except for the minor difficulty relative to taking partials with respect to the u i>, a difficulty which does not even occur in the present situation, since in the computations relative to I the exponents of the power-products of the uij will have been examined and none will have yielded a positive characteristic; the characteristic merely comes in the directions for computing the coefficient ring itself. Thus it is essentially a matter of showing in the unequal characteristic case, with I' = J, that the directions for computing a(J) coincide (or can be made to coincide) suffi- ciently with those for computing a(I). We have to look at the details of Theorem 4 of [10], already cited, or rather of that theorem with Z replaced by J. The theorem deals with a module (Q1 9 ..., * *Q fi = f1Z1 + *.. +

fitZt submodule of a free J[X, X1, . .., XmI-module having Z1, ..,t

as free generators. In the application, the fij come from or via f .I . , f the generators of A, and from the elements of S. Assume then that the fii are in I[X, X1, . .., Xm ]. The proof of Theorem 4 calls for picking out a maximal subset of Q1, ... Qs linearly independent mod p over J[X, X1, * * Xm ]; the instructions relative to I call for picking out a maximal subset linearly independent over I[X, X1, ..., Xm]. In our case, this is the same; or, otherwise put, the rank of jj fj ii = the rank of II f j II mod p. In fact, let say rank jj fi; II=r and that

fll. fIr *0.

frl ... frr


Any (r + 1) X (r + 1) subdeterminant of 1j f ij is = 0 and a fortiori 0(p). On the other hand, one could not have A - 0(p) as then we would have A O(R[X, X1, ..., XmJP), whence A = 0, a contradiction. In Theorem 4 one writes ACr+i e (C1, . . ., Cr) + p(M:p) and then has to deal with the terms from p (M:p). Here we have Afr+i E (f1, . . ., fr) and the terms from p (M:p) sum to zero. Using Afr+i E (C1, . . ., fr) and the computations over I (which include making a linear transformation on X, . . ., X,, such that A becomes regular in the new variables) we can go directly to the conclusion

(*) R 1sM fl n R lTZi = RITM + (R [sM' n f R TZiXJ ),

where S = multiplicative system of elements in J[X] not =O(p). In (*) there is no further computation and the argument establishing the equality for S =multiplicative system of elements of I[X] not all of whose coefficients are in P holds equally well for S =multiplicative system of elements of J[X] not =O(p). In our instructions for the unequal characteristic case we can say that if it should turn out that rank 11 fij 11 - = rank 11 fij 11 mod p, then one uses the simpler argumentation to get (*). Thus the two programs agree up to getting (*). Of course, we still have to concern ourselves with R jsM' n l R ITZiXj , but the usual type of induction argument holds.

Thus the programs for I and I' = J coincide up to getting MT

RisM nf R1lTZi for a certain T = {1, t, t2, . . .} with t E S. It then remains to contract MTto E R 1Zj. MT has a basis in EYR lZj, and if MT is the E R 1Zi module generated by such a basis, then in each case the contraction will be M*: tP for large p; but the algorithms here differ, and in the one case may lead to M*:tX = MT:tX+l and in the other to MT:it =

MT:tI+l. In the application we will find u(I) = tx, u(J) = tl', and so (I)l = a(D)\. Perhaps by going over the considerations of [9] with a coef-

ficient ring of finite length, like our R, in mind, one could establish a result like that of [51, and so put a uniform bound on the p needed to get m*:tp = M*:tP+l, in which case one could take X = ,u = 1, but as already remarked, we do not need this and the equation a(I)' = u(J)x will suffice.

Now, finally, we come to the question of deciding whether A =

R [X](f , . . ., f,) is mixed and of finding the primary component of A belonging to R [X]P. The ideal A: u(I), which we can construct, has the same primary component belonging to R [X]P as A does. If A: u(I) > A,

632 A. SEIDENBERG

the construction has been advanced. Suppose, then, that A: u(I) = A. In that case A is primary. (Note that in going from I to J we may have replaced u(I) by a a congruent to it mod R [X]P, but the equality A: a = A remains.) Let, then s1 E R [X] -R [X]P and b E R [X] be such that s 1 b e A. Let I' be a coefficient ring containing enough coefficients to express that

sIb eA; and letI C '. Then inI'[X, X1, . . ., Xm] we will get a corresponding relation s*b* E I'[X, X1, ..., Xm]A*.lO Then u(I')b* E I'[X, X1, I.., Xm]A*; and u(I)Ib* E I'[X, X1, ... Xm]A*. Hence u(I) tb eA and b eA. Q. E.D.

Constructivist. From a constructivist point of view the main difficulty with the above is that it assumes the existence of a normal decomposition of a given ideal A of R [X] in the course of the construction of a normal decomposition. In the writing, this difficulty was loosely kept in mind: from the present point of view the classical part is a kind of analysis of the problem. It is a matter, then, of going over the above considerations and circumventing the use of the structure theorem (and, of course, meeting any other difficulty which may arise). We go over our results seriatim.

There is no occasion for comment until we come to the equation in the proof of Theorem 1, A = (A :Xp) n (A, XP) for large p, and the statement that the problem (of constructing Rp[X]A n R[X]) reduces to the cases

(i) no associated prime of A contains X, and (ii) every associated prime of A contains X.

But we could equally as well have reduced to the cases

(i) A:X = A (ii) XP eA for some p.

(Of course one will check that Rp[X](A1 f A2) n R [X] = (Rp [X]A,1 n R [X]) n Rp[X]A2 n R [XI), which depends simply on the multiplicative property of S = R - P, and not on the structure theorem.) Later in the proof we have written B = Rp [X]A n R [X]: if this is regarded as a definition, which it is, and not as a declaration of existence, there is no reason to take exception to it. Then, however, we say B :X = B, giving as a reason some structure theorem about A and B, so we will have to find a different reason. By B :X = B we mean simply that if one has a b E R [X] and an s E

R - P such that sXb E A, then one can find an s' E R - P such that s'b EA. Since A :X = A, this is immediate with s' = s. The rest of the proof holds without change. Thus we see that the use of the structure theo-


rem in the proof of Theorem I is more a matter of a customary way of looking at things rather than one of need or even convenience.

We come next to Theorem 2, which speaks of the minimal primes of a given R [X]-ideal of A. By a prime ideal P we mean, as classically, an ideal ? (1) such that ab e P and a 0 P implies b E P; since our ideals wvill always be detachable, this definition is subject to the familiar logical transforma- tions. By a minimal prime of A we mean, as classically, a prime ideal P containing A and such that no other prime ideal contained in P contains A, though throughout we speak only of finitely generated ideals. Let A be a given ideal ? (1). We show that A has at least one minimal prime, and to do this we first show that A is contained in a prime ideal. Let A' = R n A, and let P' be a minimal prime of A' (obtained from a normal decomposition of A'). Then A':P' > A' and hence also A:R[X]P' > A. If A :R[X]P' ? (1), the problem has been advanced (by condition a), so we may assume R [X]P' C A. If A C R [X]P', then A = R [X]P' and A is prime. So assume A ? R [XJP'. Pass to Rp [XJ. By elementary properties of localization, Rp [XIA Z Rp [X]P'. Now Rp [XIA ? (1), as otherwise we find an s e R - P' with s E A' C P', a contradiction. Since R [X]P' C A, we can write Rp([X](P', A) * (1). Let A = (fI, . . ., f5) and consider the ideal A = (fi, ..., fi) in Rp [X]/Rp [XIA = (Rp,/R pP')[X], where fi = fi mod R p[X]P'. Let f be the GCD of

f l, * * *, f5 in (Rp,/Rp,P')[XI, which we can find using the Euclidean algorithm. deg,f > 0. Let g be an irreducible non-unit factor of f and let g be a counterimage of g in Rp [XI. Then Rp [XI(P', g) is the counterimage of (g) and hence prime; it contains Rp,[X](P', A) since (f) C (g). The ideal (P', g), as any ideal in Rp [XI, is the extension of an R[X]-ideal, so we can construct (P', g) n R [XI = P, and this is a prime ideal containing A. By going over the argument in a different way we now get a minimal prime of A. Let A' and P' be defined and obtained as before. If A C R [X]P', then obviously R [X]P' is a minimal prime of A. Note in this case that A:R[X]P' > A, since A':P' > A'. Consider now the case A Z R[X]P'. We first seek a minimal prime of Rp/[X]A. As before, we have Rp,[X4A ? (1), so there is a prime ideal P" in Rpf [XJ containing Rp [XIA (by the previous argument applied to Rp/). Then P" contracts in Rp, to a prime contained in Rp,P' and contracts in R to a prime contained in P'. It must contract to P', otherwise the contraction will be a prime properly contained in P' and containing A', a contradiction. So P" D Rpt[X]P'. Hence Rp'[X](P', A) * (1), as before. We now get the prime

634 A. SEIDENBERG

Rp, [X](P', g) and its contraction P to R [X] as before. One easily sees that Rp,[X](P', g) is a minimal prime of Rp[X]A. Moreover, (P', g) n Rp, = Rp,P', so P n R = P'. From this it follows that P is a minimal prime of A. Moreover, here too A :P > A, at least for some g. First we will prove Rp, [X]A :Rp, [X](P', g) > Rp, [X]A; A :P > A follows directly. The proof will be by reductio ad adsurdum, starting from Rp,[X]A :Rp,[X] (P', g) = Rp,[X]A. Since (P', g)P C (P'P, g) C (P', g) we get Rp,[X]A:Rp,[X](P'P, g) = Rp,[X]A. Let p be large enough to get A':P'P =A':P'P+l. Then one will find an s e R' - P' such that sP'P C A', so Rp,[X]P'P C Rp,[X]A. One will get a contradiction, then, if one finds an h 0 Rp,[X]A such that hg E Rp,[X]A. Take p as small as possible with Rp,[X]P'P C Rp,[X]A; p - 1. Then Rp,[X] >

Rp,[X]A:P'P-1 > Rp,[X]P'. LetRp,[X]A :P'P-/Rp,[X]P' = (ft); then (f) C (f 1). Take k to be a factor of f1 and let f1 = h. Then hP'P-I ? Rp,[X]A, since h 0 (f1); but ghP'P-1 C Rp,[X]A, so we have a contradiction.

Thus in every case (with A ? (1)) we have constructed a minimal primePofA such thatA:P > A.

If P1 is another minimal prime of A, then, of course, P1 D P, so P1 :P = P1 andA :P C P1. ThenA :PPC P1 for large p andP1 is afortiori a minimal prime of A :PP. Thus the minimal primes of A, other than P, are amongst the minimal primes of A :PP for large p. In particular, if A:PP = (1) for large p, then P is the sole minimal prime of A. Assume A :PP ? (1) for large p. Let P1 be a minimal prime of A :PP such that (A :PP) :P1 > A :PP; and considerA :PPP' for large a. Then except perhaps for P and P every minimal prime of A is a minimal prime of A :PPP'. We have A < A :PP <A :PPP'. If A :PPP' * (1), we repeat the process, etc. This process must terminate, and if it terminates with, say, A :PPP* P. = (1), then the minimal primes of A will be amongst P, P1, . Pd. These primes contain A and any prime containing A contains one of P, P1,

. Pd. Hence the minimal primes of A will be the minimal primes amongst P, P1, * * . Pd

Thus we have constructed afinite set S of prime ideals such that every element of S is a minimal prime of A and every minimal prime of A is an element of S.

COROLLARY A. Let P1, ... P' be the minimal primes of A, then A: (PI, ...., -P ) = (1) for large X. Every prime ideal containing A contains a minimalprime of A. If P is a minimalprime of A, then A P > A. If


P is a prime ideal containing A, then the minimal primes of A: PPfor large p are, with the possible exception of P, precisely the minimal primes of A. If P is a minimal prime of A, then the minimal primes of A: PPfor large p are precisely the minimal primes of A with the exception of P.

Proof. We constructed prime ideals P, PI, . . ., Pd and integers p, a, . such that A: PPP' - * * P' = (1); and each Pi contains a PJ . A for-

tioriA: (P, ..., PeV' = (1) for large X. This proves the first sentence. As for the second, if P is a prime ideal containing A, then (P1', ..., P ) C P for large X, so some P' C P. As for the third, if A: P = A, then A: PP = A for large p. If P is the sole minimal prime of A, we get A = (1), a contradiction. If P = P' and there are other minimal primes P', ..., P , then (P2, ..., P')X C A C P for large X, again a contradiction. SoA :P > A. In the fourth sentence, if P is not a minimal prime of A, then P ? P!, every i, soA:PP C P':PP = P', every i, so every P is a minimal prime of A: PP; conversely, if P' is a minimal prime of A: PP, then P' C P!', and therefore P' = P!, for some i. If P is a minimal prime of A, say P =P

then first we observe that P', ..., P' are minimal primes of A PP; and second if P' is a minimal prime of A: P P for large p, then (P' * P' )X C A:P'P C P' forlargeX, soP! C P', andhenceP! = P', forsomei = 2,

e. In the fifth sentence it remains to see that P, a minimal prime of A, is not a minimal prime of A :PP for large p. If P were a minimal prime of A :PP for large p, then by the third sentence (A :PP) :P > A :PP, whereas (A :PP) :P = A :PP.

The lemma to Theorem 3 speaks of primary ideals. By a primary ideal Q we mean, as classically, an ideal *(1) such that if ab E Q and a 0 Q, then Q: bP = (1) for large p. The lemma and its proof do not otherwise call for comment. We may note, though, that if Q is a given primary ideal, then we can construct its radical, i.e., find a finite basis for it. For one will note that Q cannot have more than one minimal prime, and if P is the minimal prime of Q, then PP C Q C P, by Corollary A, so P is the desired radical.

Coming to Theorem 3, we speak of "the primary ideal Q I in a Lasker- Noether (or normal) decomposition of an R [X]-ideal A belonging to a minimal prime P1 of A. " By this for the present we shall mean R [X] pA fn R [X]. Later we shall construct this contraction, but for the moment suppose this has already been done, so that we have the existence of QI = R [X]P1A n R [X]. There is no difficulty in speaking about R [X]P1, since we have a normal decomposition of (0) in R and this extends to a normal decomposition of (0) in R [X]. Note that Q I is primary. For the proof, ob-

636 A. SEIDENBERG

serve that R[X] pA C R[X]p1P and (R [X] P1)P C R[X]p,A for large p by Corollary A, so if ab E R[X]p1A and a 0 R[X] pA, then surely b E R[X] pPI (as otherwise a E R[X] pA), whence bP E R[X]p1A. Then Q1, since it is the contraction of a primary ideal, is itself primary.

Now, as at the beginning of the proof of Theorem 3, we can say that if A = Qi n ... Q5, then one already has the required decomposition, and that otherwise A:(Q1 n ln Q5) * (1). Then we assert that A:(Q1 n .nQ5 ) is in no Pi. Since A: Qi C A: (Q 1 n .n Q,) it will suffice to show that A: Qi ? Pi. By elementary properties of localization, onehasR[X]p.Qi = R[X]pA andsQi C A for somes eR[X] -Pi, so A: Qi ? Pi. As before, one finds an f E A:(Q1 n ln Q) - UP1i; and one has A = (A:fp) n (A, fP) = A1 n C1 for large p, where A:fP = A1 and (A, fP) = C1. Sincef (Q1 n fl. Q5) C A, we have Q, n ... Q5 C A:f, and sinceA <Qin fl f Q5, we have again, as in the proof of Theorem 3, A < A:fP = A 1; and A < C1, since f 0 P1. The rest of the proof of Theorem 3 now holds word for word as before, though with a different meaning and different reasoning. Thus A and A 1 = A :f have the same minimal primes, since f is in no Pi; and "in a normal decomposition A1 has the same associated primaries Q,, * -, Q5," since the image of f in R[X]Pi is a unit and so R[X]pi(A:f) = R[X]piA.

Next comes the reduction to the case of a complete local ring, etc. By a local ring we mean a Noetherian ring with a sole maximal ideal. The reduction is easy, following the lines already laid down, and further comment is hardly necessary. In case (ii) there was a remark on the embedded primes of Rp, [X]A, but this wasn't necessary and can simply be omitted.

Once the reduction to the case of a complete local ring with Pn = 0, etc., has been made, there were no further references to the normal decomposition theorem, except for a couple of unnecessary ones: At one point we spoke of the 0-dimensional primes of A * in J[X, X1, . . ., Xm I, but this was done for the sake of orientation and the statement did not enter logi- cally into the argument. Later we referred to "whether A = R [X](f1, . L. , fs) is mixed," but we could equally well have asked whether A is primary. No other difficulties arise. The construction is complete.

UNIVERSITY OF CALIFORNIA, BERKELEY


REFERENCES

[1] I. S. Cohen, On the structure and ideal theory of complete local rings, Transactions Amer. Math. Soc., 59 (1946), 54-106.

[2] W. Krull, Parameterspezialisierung in Polynomringen, I, II, Arch. Math., 1 (1948), 56- 64, 129-137.

[3] M. Nagata, Local Rings, New York (1962). [4] F. Richman, Constructive aspects of Noetherian rings, Proceedings Amer. Math. Soc.,

44 (1974), 436-441. [5] A. Seidenberg, On the length of a Hilbert ascending chain, Proceedings Amer. Math.

Soc., 29 (1971), 443-450. [6] , Constructive proof of Hilbert's theorem on ascending chains, Trans. Amer.

Math. Soc., 174 (1972), 305-312. [7] , The prime ideals of a polynomial ideal under extension of the base field, Annali

di Matematica Pura ed Applicata, Series IV, 102 (1975), 57-59. [8] , Constructions in algebra, Trans. Amer. Math. Soc., 197 (1974), 273-313. [9] , What is Noetherian?, Rendiconti del Seminario Matematico e Fisico di Milano,

44 (1974), 55-61. [10] ' Constructions in a polynomial ring over the ring of integers, American Jour. of

Math., 100 (1978), 685-703. [11] 0. Zariski and P. Samuel, Commutative Algebra, Vol. I, Princeton (1958).

NOTES

'For another example, see [5; p. 449f] and [6; p. 312]. 2 Making the assumption that the characteristic of R is known would not disturb us, but

if it is 0, we would not be able to tell whether a given ideal A contains an integer n * 0, and so we would not know the characteristic of R/A. A part of our difficulties could be met by as- suming that R contains a field, but we are unwilling to limit the generality of our result in this way.

3 If S is a multiplicative system in R, one denotes the quotient ring of R relative to S by Rs. If P is a prime ideal in R and S = R - P, then it is standard to denote Rs by Rp. If X is an indeterminate and P is a prime ideal in R, then R[X]P is prime, and one frequently denotes R [X]P by P, simply (usually without confusion, since R [X]P n R = P). If this is done and R [X]R1XIP is written as R [X] p one should be careful not to confuse R [X] p with R p[X]. Indeed, as we shall see, for a given ideal A of R it is relatively easy to construct Rp[X]A n R, whereas to construct R[X]pA n R[X] is the main difficulty. On the other hand, if S = R - P, one may write Rs[X] or R [X]s indifferently, as the two are equal.

4Given a primary ideal Q, one can construct its associated prime P, i.e., find a finite basis for it, since P is the sole minimal prime of Q, and we have just seen that the minimal primes of a given ideal can be constructed.

5Fr+i is the polynomial arising from Fri+ in replacing the coefficients of Fr?+, which are in k, by their correspondents in k.

60n p. 698 of [10] one should write Xi = u+2X? ? * ? uinX' instead of Xi' Ui2X2 + ? '' + u inX7

70n p. 306 of [8], instead of 79 write 77. The reference is to the first 15 lines of Section 77. In Section 77 one has k(u)[X]A = k(u)[X](A, El, . . ., En) n B. Here B:X1 = B and

638 A. SEIDENBERG

k(u)[X]/k(u)[X](A, E,.E. ., En) is a k(u)-linear space of dimension < 7r deg Ei, so for p = H deg Ei, A:Xf = A:X'+1 . On p. 307 it would have been clearer to write i and il < gn_ . .) rather than < 1 + g ( . . ). Later i + l is a misprint for ij-1.

8 Whenever we construct a polynomial h we can test whether h (c1. c,) O(P) and so for constructed polynomials h will have h(c c . ) = 0 if and only if h(c1. cu) O(P).

9 For an application of this idea, or a similar one, in a considerably simpler situation, see [7], where it is used to solve some problems left open by Krull in [2].

10We may assume that s* E I'[X] - 0 and that the coefficients *0 are 30(P).

Documents

On the Lasker Noether Theorem