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4 4.1 Ideals in commutative rings Noetherian rings In this section we don’t assume that the ring R is commutative, because we don’t need to. An ideal I of R is finitely generated if there is a finite subset X of I such that I = hXi - i.e. such that I can be generated, as an ideal, by a finite set of elements. The ring R is noetherian if every ideal of R is finitely generated. If I and J are ideals of R then both notations I ⊆ J and I ≤ J will be used for “I is contained in J” and both notations I ⊂ J and I < J will be used to denote that I is a proper subset of J. Theorem 4.1. For a ring R the following conditions are equivalent. (i) R is noetherian. (ii) R satisfies the ascending chain condition (acc) on ideals - that is, every increasing sequence I0 ⊆ I1 ⊆ · · · ⊆ In ⊆ . . . of ideals of R stabilises, in the sense that there exists n0 such that In = In0 for all n ≥ n0 . (iii) R has the maximal condition on ideals: if I is any non-empty set of ideals of R then there is a maximal element of I, that is, there is I ∈ I such that there is no J ∈ I with I < J.1 S Proof. (i)⇒(ii) Given such a sequence of ideals let I = n In be their union. We check that I is an ideal. First, 0 ∈ I0 (I0 being an ideal) so certainly 0 ∈ I. Also, if a, b ∈ I then there are m, n such that a ∈ Im and b ∈ In . Let l be the larger of m and n; then both a, b ∈ Il so (Il being an ideal) a + b ∈ Il ⊆ I. Finally, if a ∈ I, say a ∈ Im , and r ∈ R then, since Il is an ideal, ra, ar ∈ Il , hence ra, ar ∈ I. Then, by (i), there is a finite set X ⊆ I such that I = hXi. Since there are only finitely many elements in X there is n0 such that X ⊆ In0 . Therefore hXi ⊆ hIn0 i = In0 : but also In0 ⊆ I = hXi. Therefore I = In0 and so In = In0 for every n ≥ n0 (since such an In is sandwiched between In0 = I and I). (ii)⇒(i) Suppose we have (ii) and let I be an ideal of R. Suppose, for a contradiction, that I is not finitely generated. Choose a sequence rn of elements of I as follows. Let r0 be any element of I and set I0 = hr0 i (note that I0 ≤ I). Since I is not finitely generated, we have I 6= I0 so choose r1 ∈ I \ I0 . Let I1 = hr0 , r1 i. Note that I0 < I1 and I1 < I. Continue inductively, setting In = hx0 , . . . , xn i and then choosing xn+1 ∈ I \ In . In this way we get a strictly increasing sequence I0 < I1 < · · · < In < . . . of ideals of R, contradicting our assumption (ii). We conclude that I must be finitely generated, as required. (iii)⇒(i) Let I be an ideal of R. Set I to be the set of finitely generated ideals contained in I; then I is not empty since it contains, for instance, the zero ideal {0}. By assumption there is I ′ ∈ I which is maximal in I. If we had I ′ < I, say s ∈ I \ I ′ , then the ideal generated by I ′ together with s would be finitely generated (since I ′ is) hence in I and would properly contain I ′ , contrary to I ′ being chosen maximal in I. Therefore I = I ′ is finitely generated. (ii)⇒(iii) If I is a non-empty set of ideals of R then choose I0 ∈ I. If I0 is not maximal within I then choose I1 ∈ I with I1 > I0 . Continue. In this way we generate a strictly increasing sequence, I0 < I1 < . . . , of ideals which, 1 This is weaker than saying that J ≤ I for every I ∈ I: we’re saying just that there is no ideal in I bigger than I, not necessarily that I contains every ideal in I. 22 by assumption, must stop at some In . But the process can stop only if In is maximal in I, as required. [By the way, the proof of (ii)⇒(i) is redundant, given the proofs of the other implications, but it is instructive, hence included.] Examples 4.2. (1) Any field (e.g. Q, R, Zp p prime) is noetherian since it has only two ideals. (2) Any finite ring (e.g. Zn ) is noetherian because it can have only finitely many ideals and hence any increasing sequence of ideals must stabilise. (3) The ring Z of integers in noetherian, as is every principal ideal domain (immediately from the definitions). We will show in the next section that any polynomial ring K[X1 , . . . , Xt ] over a field in finitely many indeterminates is noetherian. Corollary 4.3. Suppose that R is noetherian and that I is an ideal of R. Then R/I is noetherian. Proof. Denote the canonical projection from R to R/I by π. Suppose that J0 ≤ J1 ≤ · · · ≤ Jn ≤ . . . is an increasing sequence of ideals of R/I. Define In = π −1 Jn = {r ∈ R : π(r) ∈ Jn }. Since π(r) ∈ Jn implies π(r) ∈ Jn+1 we have r ∈ In implies r ∈ In+1 - that is, In ≤ In+1 for each n. Recall (from Examples for Section 1) that each In is an ideal of R. Since R is noetherian the increasing sequence I0 ≤ I1 ≤ · · · ≤ In ≤ . . . of ideals of R must stabilise: say In0 = In0 +1 = . . . Then we have π(In0 ) = π(In0 +1 ) = . . . , that is Jn0 = Jn0 +1 = . . . and so the sequence J0 ≤ J1 ≤ · · · ≤ Jn ≤ . . . stabilises, as required. The following lemma will be used later. Lemma 4.4. Let T be a subset of the ring R. If the ideal, hT i, generated by T is finitely generated then there is a finite subset U ⊆ T such that hT i = hU i. Proof. Say a1 , . . . , an is a finite set of elements which generates hU i. Each aj is in thePideal generated by T and hence can be written in the form (a finite sum) i rij tij sij for some elements rij , sij ∈ R and tij ∈ T . Then the ideal hV i generated by the finite subset V = {tij }ij of T contains the aj and hence contains the ideal, hT i, that they generate. On the other hand, clearly hV i ⊆ hT i. Therefore hT i = hV i = h{tij }ij i. Non-example 4.5. Let C(R, R) be the ring of all continuous functions from R to R, where the addition and multiplication are defined pointwise, that is: f + g is defined by (f + g)(r) = f (r) + g(r) for r ∈ R and f × g is defined by (f × g)(r) = f (r) × g(r) for r ∈ R. (Exercise: check that this is a ring and is commutative. What is the identity element? Is it a field? Is it a domain?) Let W be any subset of R and define IW = {f ∈ C(R, R) : f (r) = 0 for all r ∈ W } (the set of functions which are zero at every point of W ). It is an exercise (do it!) to show that this is an ideal of C(R, R). Also it is immediate that if W ⊆ V then IV ⊆ IW . Furthermore, if W ⊂ V and if W is a closed subset of R (for example, a closed interval) then there is a function in IW and not in IV (why?) and hence IV < IW . So if W0 ⊃ W1 ⊃ · · · ⊃ Wn ⊃ . . . is a strictly decreasing sequence of closed sets then the corresponding sequence 23 IW0 < IW1 < · · · < IWn < . . . of ideals is strictly increasing and so C(R, R) is not a noetherian ring. Non-example 4.6. Suppose that K is a field and let K[X1 , X2 , . . . , Xt , . . . ] be the polynomial ring over K in infinitely many indeterminates. Exercise: show that K[X1 , X2 , . . . , Xt , . . . ] is not noetherian. 4.2 Noetherian rings: Hilbert’s Basis Theorem From now on we assume that R is commutative. Theorem 4.7. Let K be a field. Then the polynomial ring K[X1 , . . . Xt ] is noetherian. To prove this we first prove a result which looks more general but also more particular. Theorem 4.8. (Hilbert’s Basis Theorem) Suppose that R is a commutative noetherian ring. Then the polynomial ring R[X] over R in one indeterminate is noetherian. Proof. Before we begin the proof we recall some notation from §8.1. Suppose that f ∈ R[X]; then lt(f ) denotes the leading term of f and lc(f ) denotes the leading coefficient of f . Let I be an ideal of R[X]. Since the zero ideal clearly is finitely generated we may assume that I 6= 0. Choose a non-zero polynomial f1 ∈ I of minimal degree. Then hf1 i ⊆ I. If hf1 i = 6 I then choose f2 ∈ I \ hf1 i, again of minimal degree in I \ hf1 i (so deg(f2 ) ≥ deg(f1 )). Then hf1 , f2 i ⊆ I. If hf1 , f2 i 6= I then choose f3 ∈ I \ hf1 , f2 i of minimal degree. Continue in this way. If this process stops at some point then it is because we have hf1 , f2 , . . . , fn i = I for some polynomials {f1 , f2 , . . . , fn } and, in that case, I is finitely generated. So assume, for a contradiction, that the process does not stop. Note that deg(f1 ) ≤ deg(f2 ) ≤ .... Set ai = lc(fi ) and consider the ideal ha1 , a2 , . . . , an , . . . i of R [sic]. Since R is noetherian this ideal is finitely generated and so, by 4.4, there Pmis m such that ha1 , a2 , . . . , an , . . . i = ha1 , a2 , . . . , am i. In particular, am+1 = i=1 ri ai for some ri ∈ R. By construction d = deg(fm+1 ) ≥ deg(fi ) for each i = 1, . . . , m. We use Pm f1 , . . . , fm to subtract off the leading term of fm+1 . Set g = i=1 ri fi X d−deg(fi ) . Notice that the leading term ri fi X d−deg(fi ) is ri lt(fi )X d−deg(fi ) = ri ai X deg(fi ) X d−deg(fi ) = Pof m d ri ai X . Therefore lt(g) = i=1 ri ai X d = am+1 X d = lt(fm+1 ). Hence fm+1 − g has degree <Pd. m But g = i=1 ri X d−deg(fi ) fi ∈ hf1 , f2 , . . . , fm i and so if we had fm+1 − g ∈ hf1 , f2 , . . . , fm i then we would have fm+1 ∈ hf1 , f2 , . . . , fm i - contradiction. But then we have that fm+1 − g is an element of I \ hf1 , f2 , . . . , fm i and of smaller degree than fm+1 - a contradiction to the choice of fm+1 (as having least degree among polynomials in I \ hf1 , f2 , . . . , fm i). Thus, in either case, we arrive at a contradiction. We conclude that the ideal I is finitely generated and hence that R is noetherian, as required. 24 Proof. of Theorem 4.7 As observed in the previous section, any field K is noetherian. So by HBT (4.8), K[X1 ] is noetherian. So by HBT again, K[X1 ][X2 ] is noetherian. But K[X1 ][X2 ] ≃ K[X1 , X2 ] (in fact we have identified these rings), so K[X1 , X2 ] is noetherian. We continue and conclude, by induction, that K[X1 , . . . Xt ] is noetherian, as required. This proof just given applies to any noetherian ring R in place of K and so we obtain the following. Corollary 4.9. If R is a commutative noetherian ring then so is R[X1 , . . . , Xt ]. Example 4.10. The rings Z[X1 , . . . , Xt ] and Zn [X1 , . . . , Xt ] are noetherian. Corollary 4.11. (of HBT) If K is a field and S is any set of polynomials in K[X1 , . . . , Xt ] then there is a finite subset S ′ of S such that hSi = hS ′ i. Proof. By 4.7, K[X1 , . . . , Xt ] is noetherian, so the ideal hSi generated by S is finitely generated. By 4.4, we can take the finite generating subset to be a subset of S, as required. Question 4.12. How can we find a generating set for an ideal? That is, given a set S of polynomials, consider the ideal I = hSi that they generate. We know, by the above, that some finite subset S ′ of S suffices to generate I. How can we find such a set? More realistically, even if S is finite it may be that a much smaller (sub)set S ′ generates the same ideal: how can we find a smaller or even “minimal” generating set for this ideal? Also, suppose we have a generating set for each of the ideals I and J. There are various constructions that we can perform to produce new ideals (such as forming I + J and I ∩ J): are there methods for finding generating sets for these new ideals from the generating sets of I and J? The answer is “yes”, as we will see in the next chapter. Theorem 4.13. Suppose that R is a commutative noetherian domain. Then every non-zero non-invertible element of R has a factorisation as a product of irreducible elements. Proof. The idea of the proof has been seen already in the proof of 3.14. We use the maximum condition rather than an inductive construction to shorten the proof. It hides the key idea a little but it gets straight to the point. Suppose, for a contradiction, that some non-zero non-invertible element of R does not have a factorisation as a product of irreducibles. Let I be the set of all principal ideals of R of the form hai where a is non-zero and non-invertible and doesn’t have a factorisation as a product of irreducibles; by assumption I is not empty. Since R is noetherian there is a maximal member of I, say hai where a is non-zero, non-invertible and has no factorisation as a product of irreducibles. In particular, a is not itself irreducible, so a = bc where neither b nor c is an associate of a. It follows (by 2.6 and 2.1) that both hbi and hci properly contain a, hence neither lies in I, hence each of b, c does factorise as a product of irreducibles. But then, placing these factorisations side-by-side, we obtain a factorisation of a as a product of irreducibles, a contradiction as required. We finish this section with a lemma and another question the answer to which can be found with the methods we will see in Chapter 5. 25 Lemma 4.14. Let R be a commutative ring and let I be an ideal in R[X, Y ]. Then I ∩ R[X] is an ideal in R[X]. Proof. Note that I ∩ R[X] is the set of polynomials in I in which Y does not appear. Certainly 0 ∈ I ∩ R[X] 6= ∅, since 0 ∈ I. If f, g ∈ I ∩ R[X] then f + g ∈ I(since I is an ideal) and f + g is again a polyomial in X only, hence is in I ∩ R[X]. If f ∈ I and h ∈ R[X] then hf ∈ I (since I is an ideal) and hf is again a polynomial in X only, hence is in I ∩ R[X]. Example 4.15. Let I = hXY + 1, Y 2 + Xi - an ideal of Q[X, Y ]. What is I ∩ Q[X]? The only obvious element of I which is a polynomial in just X is 0: are there any others? Note that the following are in I: X 2 Y 2 + X 3 , X 2 Y 2 + XY ; hence X 3 − XY ; hence X 3 + 1. So that is one non-zero polynomial in I ∩ Q[X] and so we have hX 3 + 1i ⊆ I ∩ Q[X]. Do we have equality here or are there some other elements of I ∩ Q[X] apart from multiples of X 3 + 1? 4.3 Prime and maximal ideals Let R be a commutative ring. Recall that an ideal P of R is said to be prime if whenever a, b ∈ R with ab ∈ P then either a ∈ P or b ∈ P . Proposition 4.16. Suppose that R is a commutative ring and let I be an ideal of R. Then I is a prime ideal iff the factor ring R/I is a domain. Proof. Suppose I is prime. Let a + I, b + I be elements of R/I such that (a + I)(b + I) = I(= 0 + I). That is, ab + I = I, that is ab ∈ I. Since I is prime, either a or b is in I, say a ∈ I. Then a + I = I - the zero element of R/I. Hence R/I is a domain. For the converse, suppose that R/I is a domain and let a, b ∈ R be such that ab ∈ I. Thus (a + I)(b + I) = ab + I = I = 0 + I. Since R/I is a domain, either a + I = I or b + I = I. That is, either a ∈ I or b ∈ I and we conclude that I is indeed prime. An ideal I of the commutative ring R is maximal if it is a proper ideal but there is no ideal strictly between it and R: I < R and, if I ≤ J ≤ R then either J = I or J = R (that is, if it is maximal in the set of proper ideals). Proposition 4.17. Any maximal ideal is prime. Proof. Let I be a maximal ideal of R. Let a, b ∈ R and suppose that neither a nor b is in I: we must show that ab ∈ / I. Let Ra + I = {ra + c : r ∈ R, c ∈ I} - this is an ideal (it is closed under addition, and multiplication by elements of R), in fact (clearly) it is just the ideal ha, Ii generated by a together with all of I. It is an ideal strictly containing I since it contains I and since a ∈ / I. So, by maximality of I, Ra + I = R. In particular 1 ∈ Ra + I: say 1 = ra + c with r ∈ R, c ∈ I. Similarly with Rb + I = {rb + c : c ∈ I} we have Rb + I = R and we obtain, say, 1 = sb + d with s ∈ R, d ∈ I. Then 1 = 1.1 = (ra + c)(sb + d) = rsab + rad + csb + cd. The three terms rad, csb, cd are in I since c, d ∈ I. So if ab ∈ I then we obtain 1 = rsab + rad + csb + cd ∈ I - contradiction (I is a proper ideal). We deduce that ab ∈ / I and hence that I is a prime ideal, as required. 26 Examples 4.18. (1) What are the maximal ideals in the ring Z of integers? By 4.17 we should look among just the prime ideals. Clearly the zero ideal is not maximal (every non-zero ideal lies between it and Z) and it’s not difficult to check that all the other prime ideals hpi are maximal (or use 4.19 below). (2) Let R = Q[X, Y, Z]. Of the (prime) ideals that we considered at 3.8 only hX + 1, Y + 0.5, Z − 2i is maximal Proposition 4.19. Suppose that R is a commutative ring and let I be an ideal of R. Then I is a maximal ideal iff the factor ring R/I is a field. Proof. Suppose I is maximal. By 4.17, I is a prime ideal and hence, by 4.16, we know already that R/I is a domain. We must show that every non-zero element of R/I is invertible. So take a + I 6= I (that is, a ∈ / I). Since a ∈ / I we have (as in the proof of 4.17) that Ra + I = R and so 1 = ra + c for some r ∈ R, c ∈ I. Then (r + I)(a + I) = ra + I = (1 − c) + I = (1 + I) − (c + I) = (1 + I) + (0 + I) (since c ∈ I) = 1 + I. Thus r + I = (a + I)−1 , so a + I is invertible and we conclude that R/I is a field. (Note that the proof shows how we might go about actually computing the inverse of a + I.) For the converse, suppose that R/I is a field and let J be an ideal of R with I ≤ J ≤ R. Consider the canonical projection π : R −→ R/I. The images under π of these three ideals of R are ideals π(I) ≤ π(J) ≤ π(R) of R/I, with π(I) being the zero ideal and π(R) being R/I. Since R/I is a field it has no ideals other than these two, so either π(J) = π(I) or π(J) = π(R). But then, since π gives a bijection between the ideals of R which contain I and the ideals of R/I (see page 4 of Section 1 and Exercise 3 for that section), we conclude that either J = I or J = R and so I is a maximal ideal, as required. Example 4.20. Consider the ring R = R[X] and the ideal I = hX 2 + 1i. Is I maximal? Consider the factor ring R/I = R[X]/hX 2 + 1i. We claim that this is actually isomorphic to the field C of complex numbers. I don’t give the details since this was done in Algebraic Structures 2 but recall that the isomorphism is given by taking 1 + I to 1 ∈ C and X + I to i ∈ C where i denotes a square root of -1. Given that, it follows by 4.19 that hX 2 + 1i is maximal. Example 4.21. Let R = Z3 [X] and I = hX 2 −1i. Then in the factor ring R/I we have the non-zero elements (X + 1) + I and (X − 1) + I (why are they non-zero?) with product (X + 1)(X − 1) + I = (X 2 − 1) + I = I. So R/I is not a domain and so I is not prime. Of course you can see more directly that I is not prime since X + 1, X − 1 ∈ / I but their product X 2 − 1 ∈ I. 4.4 The radical of an ideal Suppose √ that R is a commutative ring and let I be an ideal of R. The radical of I is I = {r ∈ R : rn ∈ I for some n ≥ 1} - the set of all those elements of R which, although √ not necessarily lying in I, have some power in I. Note that I is a subset of I (take n = 1). Proposition 4.22. Suppose √ that R is a commutative ring and let I be an ideal of R. Then the radical of I, I, is √ an ideal of R. If P is a prime ideal of R then P = P . 27 √ Proof. We have to show that I is closed under addition and multiplication by elements of R. √ Pm+n So let a, b ∈ I: say am , bn ∈ I. Consider (a + b)m+n = i=0 ci ai b(m+n)−i where ci is a binomial coefficient. Notice that each term of this sum is divisible by either am or bn and hence each term lies in the ideal√I. Therefore the whole sum, which equals (a + b)m+n , lies in I and so a + b ∈ I, as required. m m m Now suppose √ that r ∈ R√and that a is as above. Then (ra) = r a ∈ I and so ra ∈ I. Therefore I is, indeed, an ideal of R.√ m Finally, suppose that P is a prime ideal and √ let a ∈ P - say a ∈ P . Since P is prime it follows that a ∈ √ P and hence P ⊆ P . Since the other inclusion is always true we obtain P = P . Example 4.23. Consider the ring, Z, of integers. The radical of h8i consists of all elements n such that some power of n is divisible by 8. Since 8 is just a power √ of 2, this (clearly?) is the set of all elements n ∈ Z such that 2 | n. That is, h8i = h2i. √ Now consider the ideal h360i - what is h360i? First factorise 360, as 23 32 5. √ So n ∈ h360i iff some power of n is divisible by 360. Because 2,3 and 5 are coprime this will happen iff some power of n is divisible by 23 and 32 and 5. This, in turn, will happen iff n is divisible by 2 and by 3 and by 5, hence (since 2, 3 and 5 are coprime) iff n is divisible by 2.3.5=30. Thus we obtain √ h360i = h30i. Example 4.24. Consider Z] and the ideal I = hX 2 Y 4 + XY 2 Z 4 , √ the ring Q[X, Y, 3 3 3 XY Z i. What √ is I? Since (XY Z) is divisible by XY 3 Z 3 and hence in I we have XY Z ∈ I. Also note that Y (X 2 Y 4 + XY 2 Z 4 ) = X 2 Y 5 + XY 3 Z 4 = 3 3 2 5 5 5 X 2 Y 5 + Z(XY (XY )5 = X√ Y ∈ I and √ Z ) and hence X Y ∈ I. Therefore √ hence XY ∈ I. So far, we have got hXY i ≤ I. Is hXY i = I? We can argue as follows: each generator of I is divisible by XY . Hence every element of I will be divisible by XY (make sure you see why this is so). Therefore if an ∈ I then an will be divisible by X and by Y . This can only happen if a is divisible by X and Y (fairly clearly and, more precisely because hXi and hY i are prime ideals so if an ∈ hXi then √ a ∈ hXi and similarly for Y ), hence a ∈ hXY i. This argument shows that I ≤ hXY √ i and so, since we already have the opposite containment, we conclude that hX 2 Y 4 + XY 2 Z 4 , XY 3 Z 3 i = hXY i. We used rather ad hoc arguments at the beginning of this example and so the general question arises as to whether there is a more systematic way of proceeding to find the radical of an ideal in a polynomial ring, given a set of generators for that ideal (once again, this is something that can be done with Gröbner bases, though it’s not one of the applications we will cover). 28