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I single variable polynomial splits completely in some field extension. Say $f(x,y,z)$ has a root $(a_0, b_0, c_0) \in \Bbb{Q}^3$.

In a single variable we can say that if $a_0$ is a root then the linear polynomial $x-a_0$ divides $f$. In particular the polynomial $x-a_0 = 0 \iff x = a_0$. In several variables though there are polynomials that satisfy that, namely:

$p(x,y,z) = (x- a_0)^{2k} + (y-b_0)^{2k} + (z-c_0)^{2k}$ for some integer $k \geq 1$.

I.e. $p(x,y,z) = 0 \iff (x,y,z) = (a_0, b_0, c_0)$. So if the latter is a root of $f$, does the polynomial $p$ have any useful relation to $f$?

I'm aware of how to compute Grobner bases. I don't know how this relates though.

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    $\begingroup$ $p$ has an entire surface of "roots" in ${\bf C}^3$. $p(x,y,z) = 0 \iff (x,y,z) = (a_0, b_0, c_0)$ is true for rational $x,y,z$, but that doesn't say that much in terms of decomposition. In general, the relationship between a polynomial and its zeroes over a field which is not algebraically closed is rather loose. On the other hand, for algebraically closed fields, nullstellensatz guaratnees that polynomials are, to a very large degree, determined by their zero sets. $\endgroup$ – tomasz Dec 7 '13 at 22:00
  • $\begingroup$ $p$ has a surface of roots? In $\Bbb{R}$, though it's only the single root. $\endgroup$ – Shine On You Crazy Diamond Dec 7 '13 at 22:10
  • $\begingroup$ $x^2+1$ has no real roots, and neither does $x^4+1$. But they have no common factors! For a "real" example: $x^2-y^3$ has many real "roots", but it's actually irreducible even over complex numbers. $\endgroup$ – tomasz Dec 7 '13 at 22:17
  • $\begingroup$ I'm not looking for whether $p$ divides $f$ necessarily, but some generalization. $\endgroup$ – Shine On You Crazy Diamond Dec 7 '13 at 22:20
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I think the proper generalization is as follows: if $p=p(X_1,X_2,\ldots,X_n)$ is a polynomial over an algebraically closed field $F$ and $q$ is a nonzero polynomial over $F$ in $n$ variables which is irreducible, and such that $q(X)=0\implies p(X)=0$, then $q$ divides $p$. This follows from Hilbert's Nullstellensatz.

To see that this is a generalization, notice that for polynomials of single variable over algebraically closed fields, the irreducible polynomials are exactly the polynomials of degree $1$ and a polynomial in one variable of degree $1$ has only a single point in its zero set.

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  • $\begingroup$ @EnjoysMath: yeah, I meant that. The part about $q$ dividing $p$ I wrote correctly. It follows directly from Nullstellensatz applied to the principal ideal $(q)$. $\endgroup$ – tomasz Dec 7 '13 at 22:18
  • $\begingroup$ Please show me that my $p$ divides an example polynomial, for clarity. Thanks. $\endgroup$ – Shine On You Crazy Diamond Dec 7 '13 at 22:40
  • $\begingroup$ @EnjoysMath: you can't use this to show that $p$ divides something, only that something divides $p$. As for your example, I think that it's most likely irreducible in characteristic $0$, at least for $k=1$... I can't think of a practical application for the generalisation, actually. But then again, maybe some number theorists could help with this one, alas, I'm far from being a number theorist. :) $\endgroup$ – tomasz Dec 7 '13 at 23:07

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