12
$\begingroup$

Let $R$ be a commutative Noetherian ring and $I\subset R$ an ideal that is irreducible in the sense that if $I = J_1 \cap J_2$, then $I=J_1$ or $I=J_2$. Is (the ideal generated by) $I$ irreducible in the polynomial ring $R[x]$?

The answer to the question is "yes" since the number of irreducible components has a homological characterization which can be found in Computational methods in commutative algebra and algebraic geometry by W. V. Vasconcelos.

Proposition 3.1.7. Let $(R,\mathfrak{m})$ be a local commutative Noetherian ring. Let $\mathfrak{p}$ be an associated prime of an $R$-module $M$ and denote $\Delta_\mathfrak{p}(M)$ the submodule of $M$ whose elements are annihilated by $\mathfrak{p}$. The number of irreducible $\mathfrak{p}$-primary components in an irredundant irreducible decomposition of $0\subset M$ is $\dim_{k(\mathfrak{p})}(\Delta_\mathfrak{p}(M))_\mathfrak{p}$.

The minimal number of irreducible intersectands of $I$ equals the $R_{P}/P_P$-vector space dimension of the socle of $R[x]_{P}/I_{P}$. Now when adjoining indeterminates the field $R[x]_P/PR[x]_P$ grows, but the socle dimension is the same.

Now the real question is: Is there a simpler proof (without localization(?)) or do we have to use homological invariants of ideal decompositions?

Background: This comes from Exercise 3.6 in Eisenbud's book on commutative algebra which asks for a characterization of irreducible monomial ideals. I'm wondering if a reader who has only read Chapters 1,2,3 would be able to do that exercise. This is a repost of my math.se question which can't be migrated because it is too old.

$\endgroup$

1 Answer 1

9
$\begingroup$

I prove your question for (not necessarily Noetherian) commutative ring. Irreducible ideals in non-Noetherian ring are complicate (see this question). For Noetherian ring, see our paper for a study of the index of reducibility. The proof is elementary but quite long.

Edit: I realise that the proof also work for another question. If$(0)$ is an irreducible ideal of $R$ then it is a graded irreducible ideal of the graded ring $R[X]$.

We can assume that $I = 0$. Suppose $ 0 \in R[X]$ is reducible as the intersection of two proper ideals. We can assume these two ideals are principal, so $0 = (f) \cap (g)$ with $$f = X^r (a_0 + a_1X + \cdots + a_nX^n), a_0 \neq 0,$$ $$g = X^s (b_0 + b_1X + \cdots + b_mX^m), b_0 \neq 0.$$ Choose $f$ and $g$ so that $m+n$ is minimal.

Fact: Let $f = X^rf'$ and $g = X^sg'$. Then $(f) \cap (g) = 0$ if and only if $(f') \cap (g') = 0$.

Proof. The "if part" is clear since $(f) \subseteq (f')$ and $(g) \subseteq (g')$. For the "only if" part, suppose $(f') \cap (g') \neq 0$. We have $(X^{r+s} f') \cap (X^{r+s}g') = X^{r+s}((f') \cap (g')) \neq 0$. Thus $(f)\cap (g) \neq 0$ since $(X^{r+s}f') \subseteq (f)$ and $(X^{r+s}g') \subseteq (g)$.

Using the above fact we can assume $$f = a_0 + a_1X + \cdots + a_nX^n, a_0 \neq 0$$ $$g = b_0 + b_1X + \cdots + b_mX^m, b_0 \neq 0.$$

If $m+n = 0$ then $f, g \in R$ and this contradicts the assumption that $(0)$ is irreducible there.

So $m+n> 0$. Without loss of generality we assume that $m \ge n$ and thus $m>0$. Choose $0 \neq c \in (a_0) \cap (b_0)$ in $R$. Then $c = da_0 = eb_0$ for some $d, e \in R$. Replacing $f$ and $g$ by $df$ and $eg$, respectively, we can assume henceforth that $a_0 = b_0$.

By the minimality of $m+n$ we have the following.

Claim 1: Let $r$ be an element of $R$ such that $ra_0 = 0$. Then $rf = 0$ and $rg = 0$.

Proof. Assume $ra_0 = 0$ and $rf \neq 0$. Since $(f) \cap (g) = 0$ also $(rf) \cap (g) = 0$. However, since $ra_0 = 0$, $rf = X^tf'$ with $\mathrm{deg}(f')<n$. By the Fact, $(f') \cap (g) = 0$ in contradiction to minimality of $m+n$. The same argument applies to $g$.

Applying Claim 1 inductively one can prove:

Claim 2: If a polynomial $h = c_0 + c_1X + \cdots + c_k X^k$ satisfies $hf = 0$ (resp. $hg = 0$), then each coefficient $c_i$ satisfies $c_if = 0$ (resp. $c_ig = 0$).

Combining Claims 1 and 2 we have

Claim 3: $hf = 0$ if and only if $hg = 0$.

We continue the proof. Let $g' = g-f$. Since $a_0 = b_0$, we get $g' = X^{m-m'}g' '$ for some $m' < m$ and polynomial $g' '$ of degree $m'$. By minimality of $m+n$ and the Fact, we have $(f) \cap (g' ') \neq 0$. Thus there are polynomials $u,v,w$ such that $0 \neq w = uf = vg' '$, and thus $X^{m-m'}uf = vg' = v(g-f)$. Now, if $vg \neq 0$ then $vg = (u+v)f \in (f) \cap (g)$, a contradiction. Therefore $vg = 0$. By Claim 3 we have $vf = 0$ so $w = 0$. This is also a contradiction. The proof is complete.

$\endgroup$
4
  • $\begingroup$ Nice! Just a minor clarification at the end: $g'=X^{m-m'}g'', (f) \cap (g'') \neq 0$ by minimality, $uf=vg'', X^{m-m'}uf=v(g-f)$, then finish as above. $\endgroup$ Aug 28, 2015 at 19:15
  • $\begingroup$ Thank you for your answer. There are a number of things that I don't understand yet. Can you please explain: Why can you saturate at X ("since X is indeterminate")? In Claim 1, are you saying that you want to choose f,g so that Claim 1 is satisfied? What do you mean by "then we replace"/why can you replace $f$? And then David Lamperts comment. It would be great if you could fill in some more detail. Thanks! $\endgroup$ Aug 31, 2015 at 8:09
  • $\begingroup$ I edited my answer more detail (add a Fact). $\endgroup$ Aug 31, 2015 at 10:44
  • 1
    $\begingroup$ Thank you! I've taken the liberty to edit your proof a little. As you say, it's not short, but quite elementary. Nice. $\endgroup$ Aug 31, 2015 at 19:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.