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

Question

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.

Answer 1

One proof of the theorem cited by nfdc23 goes through the case of hypersurfaces. Essentially, the proof goes as follows.

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.