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Transcript
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