# irreducibility of a scheme; a “continuous” property or a “limit” property?

I bumped into the following question while working on a research problem on closed subschemes over a completed DVR. I hope someone could possibly answer it, or give a hint toward its resolution. To make the problem simple, let me state the relevant question in the following form.

Let $A$ be the local ring $\mathcal{O}_{\mathbb{A}^1_k, 0}= k[t]_{(t)}$ at the origin of the affine line over a field $k$ of characteristic $0$ and let $\widehat{A}=k[[t]]$ be its completion.

Let $I \subset \widehat{A} [y_1, \cdots, y_n]$ be a prime ideal of the ring of polynomial ring in $n$ variables with the coefficients in $\widehat{A}$. Let $Z$ be the closed subscheme given by this prime ideal $I$, which is in particular irreducible.

As in the title, I wonder if this "irreducibility" is a "continuous" property or a "limit" property, by which I mean, I wonder whether any of the following questions hold, in which case I would like to say "irreducibility is a continuous property in a sense".

(1) Under the above assumptions, is there a sufficiently large integer $N>0$ such that the image of the ideal, $\bar{I} \subset (\widehat{A}/(t^N))[y_1, \cdots, y_n]$ defines an irreducible scheme?

(2) Under the above assumptions, express $I= (f_1, \cdots, f_r)$ for some $f_i \in \widehat{A}[y_1, \cdots, y_n]$, which is possible because the ring is noetherian. For each integer $n>0$, let $f_{in}$ be the polynomial in $k[t][y_1, \cdots, y_n]$ obtained by ignoring the terms of $f_i$ with $\deg_t \geq n$. Then is there a sufficiently large integer $N>0$ such that the ideal $J_1 = (f_{1N}, \cdots, f_{rN}) \subset A[y_1, \cdots, y_n]$ defines an irreducible scheme?

(3) Under the above assumptions and notations, is there a sufficiently large integer $N>0$ such that the ideal $J_2= (f_{1N}, \cdots, f_{rN}) \subset \widehat{A}[y_1, \cdots, y_n]$ defines an irreducible scheme?

• The answer to (1) is no. Take the prime ideal $(t-y_1y_2)$. In this case the closed fiber is reducible. – Manish Kumar Jul 5 '17 at 5:24
• @ManishKumar Hi Manish, I think what you suggested does not answer (1). For instance, when you take $N=2$, you can check (with some computations) that $t-y_1y_2$ is irreducible. Similar arguments show that this is irreducible for all questions of (1), (2), (3) for $N=2$. – Jinhyun Park Jul 5 '17 at 17:55
• @JinhyunPark: If $R = \hat A[y_1,\ldots,y_n]/I$, then don't all $\operatorname{Spec}(R/t^N)$ have the same underlying topological space? You're just killing some nilpotents... – R. van Dobben de Bruyn Jul 5 '17 at 21:08
• Dear Jinhyun, the example obviously doesn't work for (2) and (3). But it does work for (1) as R. van also points out. Also it is not difficult to see that (3) implies (2). And I think (3) should hold. Though I don't have a proof for (3) some variant of Artin approximation may work. – Manish Kumar Jul 6 '17 at 18:43
• @ManishKumar Hi Manish. Thanks for your thoughts. When I saw your first comment, I thought about what R. van Dobben de Bruyn says, but on the other hand I was massively confused because when I tried to do the computations (over N=2) directly, I didn't get a decomposition into factors... Do you see what kind of factorization one may have for $(t-y_1y_2)$ for $N=2$? I am still not sure how to prove (3), though, while I believe it should be true. But I do not see that Artin approximation could provide its proof. It was somewhat more delicate so far. I would appreciate anyone's hint for (3). – Jinhyun Park Jul 9 '17 at 17:50