Posts Tagged ‘radical of ideals’

Ideals of a polynomial ring R[x] which are of the form I[x] for some ideal I of R

Posted: December 26, 2023 in Basic Algebra, Rings
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Let R be a commutative ring with identity, and let I be an ideal of R. Let R[x] be the ring of polynomials over R, and let I[x] be the set of all polynomials in R[x] with all coefficients in I. So an element of I[x] has the form r_0+r_1x + \cdots + r_nx^n, \ n \ge 0, where r_i \in I for all i. It is quite easy to see that I[x] is an ideal of R[x]. Now, you may ask: how are algebraic properties of I and I[x] related? In this post, we look into some of those properties and relations.

Note. Regarding parts v), vi) of the following Problem, for the definitions of radical of an ideal and the nilradical of a commutative ring, see this post.

Problem. Let R be a commutative ring with identity, and let I be an ideal of R. Let R[x] be the ring of polynomials over R. Show that

i) I is a finitely generated ideal of R if and only if I[x] is a finitely generated ideal of R[x],

ii) R[x]/I[x] \cong (R/I)[x],

iii) I is a prime ideal of R if and only if I[x] is a prime ideal of R[x],

iv) I[x] is never a maximal ideal of R[x],

v) \sqrt{I[x]}=\sqrt{I}[x],

vi) if I is the nilradical of R, then I[x] is the nilradical of R[x].

Solution. i) If I is generated by r_1, \cdots , r_n, i.e. I=\sum_{i=1}^nRr_i, then it’s clear that I[x] is also generated by r_1, \cdots , r_n, i.e. I[x]=\sum_{i=1}^nR[x]r_i. Conversely, if I[x] is generated by p_1(x), \cdots , p_n(x), then I is generated by r_1, \cdots , r_n, where r_i is the constant term of p_i(x).

ii) Define the map f: A[x] \to (A/I)[x] by

f(r_nx^n+ \cdots +r_1x+r_0)=(r_n+I)x^n+ \cdots + (r_1+I)x+r_0+I, \ \ \ r_i \in R, \ n \ge 0.

See that f is an onto ring homomorphism and \ker f=I[x].

iii) I is a prime ideal of R if and only if R/I is a domain if and only if (R/I)[x] is a domain if and only if R[x]/I[x] is a domain, by ii), if and only if I[x] is a prime ideal of R[x].

iv) Suppose I[x] is a maximal ideal of R[x] for some ideal I of R. Then R[x]/I[x] is a field and so, by ii), (R/I)[x] is a field, which is nonsense because a polynomial ring can never be a field (because, for example, x has no inverse in there). Here’s an easier proof. If I=R, then I[x]=R[x], which is not maximal. If I \ne R, then I[x]+\langle x \rangle is a proper ideal of R[x] which contains I[x] properly, because x \notin I[x].

v) For any p(x)=\sum_{i=0}^nr_ix^i \in R[x], let \overline{p(x)}:=\sum_{i=1}^n(r_i+I)x^i \in (R/I)[x]. By ii), the function \tilde{f}: R[x]/I[x] \to (R/I)[x] defined by \tilde{f}(p(x)+I[x])=\overline{p(x)} is a ring isomorphism. So p(x) \in \sqrt{I} if and only if (p(x))^m \in I[x] for some positive integer m if and only if p(x)+I[x] is nilpotent in R[x]/I[x] if and only if \overline{p(x)} is nilpotent in (R/I)[x] if and only if r_i+I is nilpotent in R/I for all i, by the third part of the Problem in this post, if and only if r_i^{k_i} \in I for some positive integers k_i if and only if r_i \in \sqrt{I}.

vi) In v), choose I=(0). \ \Box

Exercise. By the second part of the above Problem, R[x]/I[x] is a polynomial ring. Suppose now that R is a commutative ring with identity, and J is an ideal of the polynomial ring R[x]. Suppose also that R[x]/J is a polynomial ring. Does that imply J=I[x], for some ideal I of R?
Hint. No. For example, choose R to be the polynomial ring \mathbb{R}[t] and J to be the ideal \langle x\rangle.

Intersection of all P-primary ideals of a Noetherian ring

Posted: March 10, 2022 in Basic Algebra, Rings
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Let R be a commutative ring with identity.

Let I be an ideal of R. Recall that \sqrt{I}, the radical of I, is defined as follows

\sqrt{I}=\{x \in R: \ \ x^n \in I, \ \ \ \text{for some positive integer} \ n\}.

it is immediately seen that \sqrt{I} is an ideal of R and I \subseteq \sqrt{I}. Note that \sqrt{(0)} is just the nilradical of R.

Definition 1. Let Q be an ideal of R. We say that Q is primary if

x,y \in R, \ xy \in Q \implies x \in Q \ \text{or} \ y \in \sqrt{Q}.

So the concept “primary ideal” is just a simple extension of “prime ideal”.

Lemma. Let Q be an ideal of R.

i) \sqrt{Q^k}=\sqrt{Q} for all positive integers k.

ii) If Q is prime, then \sqrt{Q}=Q.

iii) If Q is primary, then \sqrt{Q} is prime.

iv) If \sqrt{Q} is maximal, then Q is primary.

Proof. i) We have Q^k \subseteq Q and so \sqrt{Q^k} \subseteq \sqrt{Q}. Conversely, if x \in \sqrt{Q}, then x^n \in Q for some positive integer n and so x^{kn} \in Q^k implying that x \in \sqrt{Q^k}.

ii) If x \in \sqrt{Q}, then x^n \in Q for some positive integer n and hence, since Q is prime, x \in Q.

iii) Suppose that xy \in \sqrt{Q} for some x, y \in R. Then x^ny^n \in Q for some positive integer n and hence, since Q is primary, either x^n \in Q or y^n \in \sqrt{Q}, which implies y^{kn} \in Q for some positive integer k. So either x \in \sqrt{Q} or y \in \sqrt{Q}, which proves that \sqrt{Q} is prime.

iv) Suppose that xy \in Q for some x, y \in R, and assume that y \notin \sqrt{Q}. So we just need to show that x \in Q. Since y \notin \sqrt{Q} and \sqrt{Q} is maximal, we have \sqrt{Q}+Ry=R and hence z+ry=1 for some z \in \sqrt{Q}, \ r \in R. Thus z^n \in Q for some positive integer n and hence (1-ry)^n=z^n \in Q. Therefore, since (1-ry)^n=1+yt for some t \in R, we get that x+xyt=x(1-ry)^n \in Q and so x \in Q because xy \in Q. \ \Box

Definition 2. Let Q be a primary ideal of R. By the above Lemma, P:=\sqrt{Q} is prime. We call P the associated prime ideal of Q, and we say that Q is P-primary.

Note. The reference for the following Proposition, which I believe is a beautiful result, is Corollary 10.21 in Atiyah-MacDonald’s book “Introduction to Commutative Algebra”. Since the proof given in that book is not very easy to understand for many students, I decided to make it easy. I hope you like the proof.

Proposition. Let R be a commutative Noetherian ring with identity, and let P be a prime ideal of R. Let R_P be the localization of R at P, and let f: R \to R_P be the natural ring homomorphism defined by

\displaystyle f(x)=\frac{x}{1}, \ \ \ \forall x \in R.

Let I be the intersection of all P-primary ideals of R. Show that I=\ker f.

Proof (Y. Sharifi). We show that I \subseteq \ker f and \ker f \subseteq I.

i) \ker f \subseteq I.

Proof. Let x \in \ker f and suppose that Q is a P-primary ideal of R. Then \displaystyle 0=f(x)=\frac{x}{1} gives sx=0 for some s \notin P. Thus sx=0 \in Q and hence, since Q is P-primary, we have either x \in Q or s \in \sqrt{Q}=P. But s \notin P and so x \in Q. Hence \ker f \subseteq I. As you might have noticed, we did not need to assume that R is Noetherian to prove the inclusion. But we do need R to be Noetherian for the next inclusion.

ii) I \subseteq \ker f.

Proof. Let \mathfrak{m}=P_P, the maximal ideal of the Noetherian local ring R_P. Let Q' be any \mathfrak{m}-primary ideal of R_P, and let Q:=\{r \in R: \ f(r) \in Q'\}.

Claim: Q is a P-primary ideal of R.

Proof. Since Q' is an ideal of R_P and f is a ring homomorphism, it’s clear that Q is an ideal of R. Also, since \displaystyle \frac{r}{1}=f(r) \in Q' \subseteq \mathfrak{m}=P_P implies that r \in P, we have Q \subseteq P. Now, let r \in P. Then \displaystyle \frac{r}{1} \in \mathfrak{m} and, since \sqrt{Q'}=\mathfrak{m}, we get that \displaystyle \frac{r^n}{1} \in Q' for some positive integer n. Hence r^n \in Q, i.e. r \in \sqrt{Q}, which proves that P=\sqrt{Q}. Finally, to show that Q is primary, suppose that r_1r_2 \in Q for some r_1,r_2 \in R. Then we have f(r_1)f(r_2)=f(r_1r_2) \in Q' and therefore, since Q' is \mathfrak{m}-primary, either f(r_1) \in Q' or f(r_2) \in \mathfrak{m}. If f(r_1) \in Q', then r_1 \in Q, and if f(r_2) \in \mathfrak{m}, then r_2 \in P=\sqrt{Q}. The proof of the Claim is now complete.

We can now finish the proof of the Proposition. Let x \in I, and let k be a positive integer. By the above Lemma, \mathfrak{m}^k is an \mathfrak{m}-primary ideal of R_P and hence, by the Claim,

Q_k:=\{r \in R: \ f(r) \in \mathfrak{m}^k\}

is a P-primary ideal of R. Hence x \in Q_k, i.e. f(x) \in \mathfrak{m}^k and so f(x) \in \bigcap_{k \ge 1}\mathfrak{m}^k=(0), by the Krull intersection theorem (see Corollary 2, iv), in this post!). Thus x \in \ker f and so I \subseteq \ker f. \ \Box