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I am reading this article

http://mathforum.org/library/drmath/view/75056.html

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.

  • 0
    your $h(x,y)$ has no real zeroes2012-11-07
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    @Jean-Sébastien The post and the article both specifically talk about _complex_ zeroes.2012-11-07
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    @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.2012-11-07
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    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.2012-11-07
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    @Jean-Sébastien In what way is his example not a counter-example? Both $g(x,y)$ and $h(x,y)$ have complex zero-curves ($g$'s of course is trivial), but there are no complex $x$ and $y$ that are simultaneous solutions of both.2012-11-07
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    @Jean-Sébastien I'm interested in factoring in $\mathbb{C}[x,y]$. The common zero can be complex and not real.2012-11-07
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    Yeah I misread the word simultaneous, I'm sorry2012-11-07
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    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.2012-11-07

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