# Is irreducibility of an affine $k$-scheme, an open condition?

About 2 weeks ago, I posted a question about irreducibility of a scheme over a completed local ring, on whether this is a continuous property or a limit property. I didn't succeed in answering it, but I got a bit more elementary question. I guess it should have been known already, as this is a basic question, but I had difficulties in locating a good reference. So, let me ask.

[The question was corrected a bit reflecting comments.]

To motivate, suppose $k$ is a field of characteristic $0$ (or something more general). Let $y_1, y_2, y_3$ be variables, and for nonzero constants $a_1, \cdots, a_4 \in k$, consider the equation $V_{\alpha_0}: a_1 y_1 + a_2 y_1 y_2 + a_3 y_1 y_3^2 + a_4 = 0.$ The shape of the equation does not matter, but it is a finite linear combination of monomials in $y_i$. Roughly put, the question is: Suppose the affine $k$-scheme $V_{\alpha_0}$ is integral. If we take small changes" of $a_i$ to obtain a new affine scheme $V_{\alpha}$, then is $V_{\alpha}$ at least irreducible?

Here, it is important that we do not turn a monomial" with $0$ coefficient into something nonzero, i.e. we modify only the coefficients that are nonzero.

I tried to reformulate the question as follows: replace the nonzero constants $a_1, \cdots, a_4$ by variables $x_1, \cdots, x_4$, and consider the general equation $V: x_1 y_1 + x_2 y_1 y_2 + x_3 y_1 y_3 ^2 + x_4 = 0$ in $\mathbb{A}^4 \times \mathbb{A}^3$ (with $(x_1, \cdots, x_4, y_1, y_2, y_3)$ as the coordinates). Consider the projection $pr_1: V \to \mathbb{A}^4$ to the $x$-coordinates, and we are given that for $\alpha_0= (a_1, \cdots, a_4) \in \mathbb{A}^4$, the fiber $V_{\alpha_0} = pr^{-1} (\alpha_0)$ is integral.

Then I ask whether one can find an open neighborhood $U \subset \mathbb{A}^4$ of $\alpha_0$ such that for each $\alpha \in U$, the fiber $V_{\alpha} = pr^{-1} (\alpha)$ is irreducible.

Any suggestions or ideas or discussions would be appreciated.

The situation I'm eventually interested in is the case when I'm given a system of algebraic equations, for which a similar question can be formulated.

• Are you asking about irreducibility of a polynomial in a polynomial ring $k[y_1,y_2,y_3]$, or are you asking about irreducibility of a Zariski closed subset of affine space? These are not the same thing. – Jason Starr Jul 21 '17 at 11:48
• @JasonStarr Sorry for the confusion. I am asking about irreducibility of the affine scheme given by it. I just edited the text. Thank you for asking for clarification. – Jinhyun Park Jul 21 '17 at 11:50
• In that case, the answer to your question is "no, irreducibility is not preserved by small deformations." For instance, in $\mathbb{A}^1 = \text{Spec}\ k[y]$, the zero scheme of $(1+y)^2 = 1 + 2y+1y^2$ is irreducible. However, for a generic choice of $(a_0,a_1,a_2)\in k\times k\times k$, the zero scheme of $a_0+a_1y+a_2y^2$ is reducible. – Jason Starr Jul 21 '17 at 11:59
• Geometric irreducibility is a constructible condition but irreducibility is not. – Damian Rössler Jul 21 '17 at 12:45
• In the presence of properness (which for a single polynomial one could pass to via homogenization) and subject to the more "natural" condition of geometric integrality, an affirmative answer is provided by EGA ${\rm{IV}}_3$ 12.2.1(x). – nfdc23 Jul 21 '17 at 12:48

Let $V_d$ be the "space" of all polynomials in $x_1, \ldots, x_n$ of total degree at most $d$. This is an affine space. Multiplication of polynomials defines a map $$\mu_i : V_i \times V_{d - i} \longrightarrow V_d$$ Let $W_i$ be the image. By Chevalley's theorem $W_i$ is a constructible subset.
Now suppose you have a point $v$ of $V_d$ over a field $k$, i.e., $v \in V_d(k)$ in standard notation. Then $v$ corresponds to a polynomial $P$ of total degree $d$ in $x_1, \ldots, x_n$ with coefficients in $k$. If $v \in W_i$, then there exists an extension $k'/k$ of fields (!), and points $v' \in V_i(k')$, $v'' \in V_{d - i}(k')$ such that $\mu_i(v', v'') = v$. In other words, we have $P = P' P''$ where $P'$ and $P''$ are the polynomials corresponding to $v'$ and $v''$. And in fact, by the Hilbert nullstellensatz, it is easy to see that if $v \in W_i$, then you can pick $k'/k$ to be a finite extension of fields.
Thus if $P$ has no such factorization over the algebraic closure of $k$, then $v$ is not in $W_i$ for $i = 1, \ldots, d - 1$. Well, then $E = V_d \setminus \bigcup_{i = 1, \ldots, d - 1} W_i$ is a neighbourhood of $v$ where the same is true. In other words, if $e \in E(k)$, then $e$ corresponds to a polynomial which does not factor.
• As written, the set $E$ is constructible in $V_d$ rather than open, so to assert that it is a neighborhood of $v$ in $V_d$ seems to require further argument. – nfdc23 Jul 22 '17 at 16:25
• In fact, take $n=1$ and $d=2$. Thus we consider the space $V_2$ of polynomials $a+bx+cx^2$. Now $x$ is irreducible, but any neighborhood of it in $V_2$ contains some polynomial of the form $x+cx^2$ with $c\neq0$, which is reducible. – Laurent Moret-Bailly Jan 18 '18 at 19:08