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I am reading this article

and would like to ask if this section is correct:

If it can, then you would have

$f(x,y) = g(x,y) * h(x,y)$,

where g and h are polynomials of degree at least one (that is, not constants). It turns out that there will necessarily be at least one complex solution $(x,y)$ to the simultaneous equations

$g(x,y) = 0$ $h(x,y) = 0$

This is known from Bezout's Theorem.

Is this true? It seems like $g(x,y) = x^2 + y^2$ and $h(x,y) = x^2 + y^2 + 1$ is an obvious counterexample.

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your $h(x,y)$ has no real zeroes – Jean-Sébastien Nov 7 '12 at 15:53
@Jean-Sébastien The post and the article both specifically talk about complex zeroes. – Steven Stadnicki Nov 7 '12 at 15:54
@StevenStadnicki The way I understand it, he says that his functions are a counter example to the condition "there will be necessarily at least one complex ...". His function are not a counterexample of that. Perhaps I misinterpreted what was meant. – Jean-Sébastien Nov 7 '12 at 15:56
As far as I can tell, your counterexample is a valid one; even with a bit of digging it's not immediately clear what Dr. Math was talking about there, and it might be worth posting asking for clarification and giving your example. – Steven Stadnicki Nov 7 '12 at 15:58
Is it not the case that, since the Bezout's Theorem requires the curves to be projective, the theorem has been misused by Dr.Math? You should suppose the polynomials to be homogeneus on an algebraically closed field to have that statement. – Giovanni De Gaetano Nov 7 '12 at 17:09

Your example is correct and the proof from your link is not.

The polynomial $2xy + 3x^2 + y$ given there is irreducible just because it is irreducible as a polynomial in $y$ with coefficients in the ring $\mathbb C[x]$.

Sometimes one can use algebraic geometry tools to determine whether a homogeneous polynomial $f$ (in at least three variables) is irreducible: if its zero set is smooth ($f$ and its partial derivatives have no common zeros other than $(0,0,...,0)$), then it is irreducible.

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