# Show that $F=Y^2+X^2(X-1)^2\in\mathbb{R}[X,Y]$ is irreducible, but $V(F)$ is reducible.

Problem 1.26 from Fulton's Algebraic Curves asks:

Show that $F=Y^2+X^2(X-1)^2\in\mathbb{R}[X,Y]$ is irreducible, but $V(F)$ is reducible.

Showing $F$ irreducible was no problem. Fulton, previous to the problem, gives the lemma $V$ irreducible if and only if $I(V)$ is prime. I'm convinced that $I(V(F))=\langle F\rangle$, since if $F$ is irreducible, so in a sense it's the `most basic thing' to vanish where it does. I can't see how we could get $V(F)$ from the unions of zero-sets of other polynomials. In my mind so far, unions correspond to products: $V(fg)=V(f)\cup V(g)$.

I'm also convinced that's my mistake. $\mathbb{R}[X,Y]$ is a UFD so irreducibles coincide with primes, thus $\langle F\rangle$ is prime, and so the lemma gives $V(F)$ irreducible, contradicting the problem! I'm stuck for a better way of determining what the ideal $I(V(F))$ should be.

• The implication: $F$ irreducible $\Rightarrow$ $I(V(F))=\langle F\rangle$ does not hold (it would be true over an algebraic closed field, but not over the reals) – emeu Oct 9 '15 at 22:31
• hint: determine first what V(F) is (it is actually a finite set) – emeu Oct 9 '15 at 22:34
• @emeu Ahh. The implicit assumption that we're over an algebraically closed field (out of habit!) was the problem. Thank you. – FireGarden Oct 10 '15 at 11:35

$V(F)=\{(a,b)\in\mathbb R^2:F(a,b)=0\}$ is a union of two points.