Assume K is a field that is not algebraically closed, then how to prove that for any m>0, there is a polynomial with m variables over K that possess a unique zero point?

  • $\begingroup$ Doesn't $\prod_{i=1}^m (x_i-1)$ work? $\endgroup$ – Prahlad Vaidyanathan Oct 19 '13 at 8:23
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    $\begingroup$ @PrahladVaidyanathan No, for $m>1$ this has many roots, e.g. $x_1=1$, $x_2$ arbitrary. $\endgroup$ – Hagen von Eitzen Oct 19 '13 at 8:23
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    $\begingroup$ Of course not, your polynomial has too many roots: they form a variety! $\endgroup$ – Lao-tzu Oct 19 '13 at 8:25
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    $\begingroup$ For $K=\mathbb R$ we could take $x_1^2+\ldots +x_m^2$, but how do we get this polynomial from merely knowing that $X^2+1$ has no root? And what if for general $K$ we just have an arbitrary $f$ without roots? $\endgroup$ – Hagen von Eitzen Oct 19 '13 at 8:27
  • $\begingroup$ Yeah, this is just what I want to know, even I know this is true, since it is an exercise in a commutative algebra book. I think the main point is that K is a field that is not algebraically closed, and we can find a polynomial (with just one variable) of degree>0, which is nowhere vanish, but I don't know how to use it. $\endgroup$ – Lao-tzu Oct 19 '13 at 8:30

As $K$ is not algebraically closed, there exists $p\in K [X]$ with no root in $K$ and $\deg p\ge1$. Let $f(X,Y)=Y^{\deg p}p\left(\frac XY\right)$, that is, if $\deg p=d$ and $$p(X)=a_0+a_1X+\ldots +a_dX^d $$ we let $$ f(X,Y)=a_0Y^d+a_1XY^{d-1}+\ldots +a_dX^d. $$

Then clearly $f(0,0)=0$. If $v\ne 0$ then $f(u,v)=v^dp(\frac uv)\ne 0$ and if $v=0, u\ne 0$ then $f(u,v)=a_du^d\ne 0$. Thus $f(u,v)=0\iff u=v=0$. For $m\ge 2$ let $f_2=f$ and recursively $f_{m+1}(X_1,\ldots, X_{m+1}) = f(f_m(X_1,\ldots,X_m),X_{m+1})$. Then it follows quickly by induction that for $u_1,\ldots ,u_m\in K$ we have $f_m(u_1,\ldots ,u_m)=0\iff u_1=\ldots u_m=0$.

  • $\begingroup$ This is in principle the same answer as by @user33433, but without Galois theory. $\endgroup$ – Hagen von Eitzen Oct 19 '13 at 10:47
  • $\begingroup$ Yes, this is the simplest proof I guess! Thank you! $\endgroup$ – Lao-tzu Oct 19 '13 at 10:49
  • $\begingroup$ Remark: $f$ is known as the homogenization of $p$. $\endgroup$ – Martin Brandenburg Oct 19 '13 at 11:34
  • $\begingroup$ Fascinating! The situation is very similar with that of the affine space and the projective space. $\endgroup$ – Lao-tzu Oct 19 '13 at 12:12

Take any $\alpha \in \overline{K}\setminus K$, let $m_\alpha$ be its minimal polynomial, and let $n$ be its degree.

For $m=2$, let $f(X,Y) = Y^n m_\alpha (X/Y)$. Then $f(X,Y)\in K[X,Y]$ and $f$ has only $(0,0)$ as a root in $K^2$.

For $m=3$, write $g(X,Y,Z) = f(f(X,Y),Z)$. If $g(a,b,c)=0$ for $a,b,c\in K$, then $c=0$ and $f(a,b)=0$, so $a=b=0$.

And so on.

  • $\begingroup$ What do you mean by "conjugates" in the algebraic closure of K? $\endgroup$ – Lao-tzu Oct 19 '13 at 9:22
  • $\begingroup$ @Lao-tzu I have clarified the answer. I mean the Galois conjugates. What I have written now is not exactly the same, as there may be duplicates, but that's not a big deal here. $\endgroup$ – Slade Oct 19 '13 at 9:30
  • $\begingroup$ OK, I know your meaning, thank you very much! $\endgroup$ – Lao-tzu Oct 19 '13 at 9:32
  • $\begingroup$ Note that with $K=\mathbb R$ and $\alpha=i$ this construction indeed gives us the "obvious" example $x_1^2+\ldots +x_m^2$. - Looking closely at the product definig $f$ one sees: Another way to obtain $f(X,Y)$ is to simply take $f(X,Y)=Y^d p(\frac XY)$ where $p$ is irreducible (and without root) of degree $d$ $\endgroup$ – Hagen von Eitzen Oct 19 '13 at 10:32
  • $\begingroup$ Don't you need that $L/K$ is Galois here? What happens when $K$ is not perfect? $\endgroup$ – Martin Brandenburg Oct 19 '13 at 11:32

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