Showing that if $F,G \in k[x_1, ..., x_n]$ are two polynomials such that $V(F,G)$ is irreducible, that dim$(V(F,G))$ does not have to be $n - 2$

Question

My solution is to consider the polynomials $F = x$ and $G = x - 1$. Then $V(F,G) \in k^2$ and $V(F,G) = \emptyset$, and dim$(\emptyset) = -1$. Since $2 - 2 = 0 \neq -1$, we are done.

My friend says this is not valid, since $I(V(F,G)) = k[x_1, x_2]$, and $k[x_1, x_2]$ as an ideal of itself is not prime, ergo $V(F,G)$ is not irreducible.

Answer 1

As Hamed notes in a comment, an irreducible algebraic set must be nonempty.

For a reference, see page $10$ of this handout:

$\qquad$http://www.math.uwaterloo.ca/~moraru/764Week3_2015.pdf

But that restriction doesn't present much of an obstacle . . .

As a simple example, let $F,G\in k[x_1,...,x_n]$ be given by $F=G=x_1$.

To camouflage it a little, you can take $F=x_1^2x_2,\;G=x_1^3(x_2+1)$.