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.

  • 1
    $\begingroup$ $p$ 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

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.

  • $\begingroup$ @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

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.