Finding mistake in a proof - irreducible variety after identification of variables

Question

I am looking for help in identifying a mistake in the following proof:

Let $P(x,y,z)$ be a (complex) irreducible polynomial in three variables. In other words, the algebraic variety $$Z_P:=\{(x,y,z)\in \mathbb C^3:P(x,y,z)=0\}$$ is irreducible. Now, consider the set $$Z_P':=\{(x,z):P(x,x,z)=0 \}\subset \mathbb C ^2.$$ This is basically the variety you get upon identifying the first two variables in $P$, projected down to $\mathbb C^2$ in the natural way. I want to say that $Z_P'$ is irreducible as well. Here is the idea why:

Consider the mapping $f:\mathbb C^3 \rightarrow \mathbb C^2$ given by: $$f(x,y,z)=(x,z)$$ Then it is easy to see that $f(Z_P)=Z_P'$. Since $f$ is a regular mapping between algebraic varieties, it maps irreducible subvarieties into irreducible ones, meaning that $Z_P'$ is irreducible.

The thing that bothers me is the following - had I considered the set $$Z_P'=\{x:P(x,x,x)=0\}\subset \mathbb C,$$ the same argument would show that this new $Z_P'$ is irreducible. On the other hand, $Z_P'$ is the zero set of a polynomial in one variable, and this is always reducible (assuming its degree is higher than one). But where is the issue in my original argument?

Thanks in advance.

Answer 1

This won't work - setting $y=x$ is equivalent to intersecting with the hyperplane $V(y-x)$, and there's no reason for intersecting with a hyperplane to preserve irreducibility. For instance, if $P(x,y,z)=x^2+y^2+z^2$, then $P$ is irreducible, but $P(x,x,z)=2x^2+z^2=(\sqrt{2}x+iz)(\sqrt{2}x-iz)$ is reducible.

Answer 2

@GSofer Does not answer the question, but has relevance to it and the secondary question raised by you in the comments.

Take a Zariski-closed irreducible affine variety and a Zariski closed irreducible affine curve(even a line) embedded in an affine space, say $V, C\subset \mathbf{A}^n$. The intersection being a closed subvariety of a curve has the following possibilities:

(i) empty

(ii) $S$ (i,e the case $C\subset V$)

(iii) a singleton

(iv) finite set with at least 2 points.

Constructing examples for all cases would be instructive. Easy intuitive case over the field of real numbers which would often fail over C: A circle and a line lying in the Euclidean plane need not intersect, or could be tangent or, a secant line giving 2 intersection points making it reducible.

An interesting case over C: Take SL(2,C) regarded as an affine subvariety: $V\subset \mathbf{A}^4$ defined by the equation $xy-zw-1=0$.

Take the following infinite collection curves $L_{a,b,c}$ (actually lines) corresponding to parametric constants $a,b,c,\in \mathbf{C}$:

$L_{a,b,c}= \{(x,ax,bx,cx)\mid x\in \mathbf{C}\}$

Now $L_{a,b,c}\cap V$ is the subset of $L_{a,b,c}$ with $x$ satisfying $(c-ab)x^2-1=0$ as easily seen by substituting in the defining equation of $V$. So the intersection has just 2 points, namely those corresponding to the two square roots of $1/(c-ab)$ (avoid choice of constants making $c-ab=0$!).

In case $V$ were replaced by varieties defined by equations of degree $n$ (here $n=2$) we can find examples with intersections having upto $n$ points.