0
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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) $\endgroup$ – emeu Oct 9 '15 at 22:31
  • 2
    $\begingroup$ hint: determine first what V(F) is (it is actually a finite set) $\endgroup$ – emeu Oct 9 '15 at 22:34
  • $\begingroup$ @emeu Ahh. The implicit assumption that we're over an algebraically closed field (out of habit!) was the problem. Thank you. $\endgroup$ – FireGarden Oct 10 '15 at 11:35
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.