1

Suppose $i: X \hookrightarrow \mathbb{P}^n_k$ is a projective scheme, over some field $k$ (I'm not sure if algebraically closed is necessary here). So $X$ comes together with an ideal sheaf and a short exact sequence $$ 0 \rightarrow \mathcal{I}_X \rightarrow \mathcal{O}_{\mathbb{P}^n} \rightarrow i_* \mathcal{O}_X \rightarrow 0.$$ Topologically $X$ consists of finitely many irreducible components. I would like to know if it is always possible to write $X$ as a scheme-theoretic union of irreducible subschemes $X_i \hookrightarrow X$, i.e. $\mathcal{I}_X = \bigcap_i \mathcal{I}_{X_i}$.

I think this works if $X$ is reduced, then one can just take the irreducible components together with the reduced structure as $X_i$, and $\mathcal{I}_X = \bigcap_i \mathcal{I}_{X_i}$, because the intersection of radical ideals is again radical, and two radical ideals with the same vanishing locus are equal, by Hilbert's Nullstellensatz.

Does this have anything to do with primary decomposition? Wikipedia states that there is some scheme theoretic interpretation of primary decomposition, but suppose $\mathcal{I}_X$ is the intersection of non-primary ideals, then why should it be possible to write it as the intersection of primary ideals?

red_trumpet
  • 8,515
  • It's a theorem that in a noetherian ring every ideal has a primary decomposition (c.f., Atiyah-MacDonald, Thm 7.13), which deals with the case of affine schemes. I think there is a similar result for graded rings in Eisenbud, which would mean the same is true for projective schemes. – Viktor Vaughn Jan 29 '19 at 17:54
  • 1
    Yeah I'm still trying to figure out primary ideals. Is it true that an ideal $\mathfrak{a} \subset A$ is primary if and only if $\text{Spec}(A/\mathfrak{a})$ is irreducible? – red_trumpet Jan 29 '19 at 17:57
  • Yes, I think that follows from this question: https://math.stackexchange.com/q/541943 – Viktor Vaughn Jan 29 '19 at 18:46
  • I don't think this is true anymore :D See the discussion here, and the counterexample here. – red_trumpet Jan 29 '19 at 18:49
  • But...you're not using the definition of primary, just a strictly weaker property. – Viktor Vaughn Jan 29 '19 at 19:02
  • Correctly. So there are ideals for which $A/\mathfrak{a}$ has irreducible spectrum, but $\mathfrak{a}$ is not primary. – red_trumpet Jan 29 '19 at 19:04
  • Ah, I see. Only one implication is true. Thanks! – Viktor Vaughn Jan 29 '19 at 19:14
  • For reduced schemes, the minimal decomposition is unique and does give you the irreducible components on the nose, as you said. For non-reduced, you'll have to look at the minimal primes in such decomposition to see the irreducible components; the other factors will correspond to embedded components, and are not unique. As @André3000 mentions, that a primary decomposition always exists is a theorem for any finitely generated module over a Noetherian ring. And clearly primary decomposition determines a scheme theoretic decomposition. I'm not sure if your question cuts deeper than this. – Artur Araujo Feb 07 '19 at 16:25
  • As an aside, the point of primary decomposition is, I guess, that the embedded components will not only be irreducible, but will only have nilpotents, as pointed out in the questions you link to. – Artur Araujo Feb 07 '19 at 16:27

0 Answers0