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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.

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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.

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  • $\begingroup$ 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. $\endgroup$ – David Lampert Aug 28 '15 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! $\endgroup$ – Thomas Kahle Aug 31 '15 at 8:09
  • $\begingroup$ I edited my answer more detail (add a Fact). $\endgroup$ – Pham Hung Quy Aug 31 '15 at 10:44
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    $\begingroup$ Thank you! I've taken the liberty to edit your proof a little. As you say, it's not short, but quite elementary. Nice. $\endgroup$ – Thomas Kahle Aug 31 '15 at 19:35

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