# Why is a variety over a non-alg. closed field a hypersurface?

Exercise $3$ on page $8$ of Kunz's Introduction to Commutative Algebra and Algebraic Geometry is as follows:

If the field $K$ is not algebraically closed, then any $K$-variety $V \subset A^n(K)$ can be written as the zero set of a single polynomial function in $K[X_1,\dots,X_n].$

This follows easily from his hint (which I don't know how to prove):

For any $m>0$ there is a polynomial $\varphi \in K[X_1,\dots,X_m]$ whose only zero is $(0,\dots,0)\in A^m(K).$

I have tried proving this using induction, but haven't gotten anywhere. Does anyone have a proof of this hint (or a counterexample)? I would really appreciate any help!

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Let $F(x)$ be a one-variable polynomial of degree at least one with no roots in $K$. Suppose it is of degree $d$. Consider the homogenization of this polynomial, $G(x,y)$ obtained from $F(x)$, such that $G(x,1) = F(x)$ and $G$ is homogeneous of degree $d$. The claim is that $G$ has no roots except at the origin. For any root of $G$ must, first of all, have at least one coordinate zero: otherwise we would get a nontrivial zero of $F$ by homogeneity.

So we need to show that if $G(x,0) = 0$, then $x =0$, and similarly for $G(0, y)$. Now $G(x,0)$ is, however, a nontrivial polynomial of the form $cx^d$ for some $c \in K$, so this is clear. $G(0,y)$ is also such a polynomial, though (corresponding to the lowest term of $F$---this is where we need that $F$ has more than one term). So $G(0,y)$ doesn't have roots except where $y=0$.

Thus the result is clear in two variables: the origin is a hypersurface. We can now "bootstrap" to higher dimensions inductively as follows. Consider affine $n$-space $\mathbb{A}^n$ and consider a polynomial $H(x_1, \dots, x_{n-1})$ with no nontrivial roots other than the origin. We can use the function $G(H(x_1, \dots, x_{n-1}), x_n)$ to obtain a polynomial that vanishes only at the origin in $\mathbb{A}^n$.

Edited to take into account Georges Elencwajg's correction.

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Beautiful solution! – curious Jul 11 '11 at 18:29
Thank you very much, Akhil! This question had been bothering me for a while. Your solution is crystal-clear! – Wes Jul 11 '11 at 18:55
Dear Akhil, if you take any non-constant polynomial $f(x)$ without zero and homogeneize it, you will already get a polynomial $\Phi(x,y)$ whose sole zero is (x,y)=(0,0).Your polynomial $G(x,y)$ , on the other hand, unfortunately doesn't work since it has all points on the $y$-axis as zeros: $G(0, \eta)=0$ for all $\eta \in K$. – Georges Elencwajg Jul 11 '11 at 19:50
@Georges: Thanks. You're right. I've just made the correction. – Akhil Mathew Jul 12 '11 at 1:17
This is great. Thinking about how to go from $X^2 + 1$ (here $K = \mathbf{R}$, if you like) to $X^2 + Y^2$ really clarified things. Is it correct to think of this as saying that if $f$ defines an empty affine variety, then its projective closure is also empty? – Dylan Moreland Jul 12 '11 at 2:30

First let's try the following lemma:

Lemma: If $k$ is not algebraically closed field and $n\in{\mathbb{N}}$, then there exists $f\in{k[x_ {1},x_{2},\dots{x_{n-1},x_{n}}]}$ such that $V(f)=\{(0,0,\dots,0,0)\}\subseteq{\mathbb{A}_{k}^{n}}$.

Proof: Let $k^{a}$ the algebraic closure of $k$. (Reasoning Inductively) First we note that if $n=1$ then there exists $f(x_{1})=x_{1}$ in $k[x_{1}]$ such that $V(f)=\{0\}\subseteq{\mathbb{A}_{k}^{1}}$. By induction hypothesis suppose there is $g(x)\in{k[x_{1},x_{2},\dots{x_{n-1},x_{n}}]}$ such that $V(g)=\{(0,0,\ dots,0,0)\}\subseteq{\mathbb{A}_{k}^{n}}$. Then consider the following cases:

Case(I). If $k$ is of characteristic zero. In that case are $\alpha\in{k^{a}\smallsetminus{k}}$ and $p_{\alpha,k}(x)$ its minimal polynomial. Then as $p_{\alpha,k}(x)$ is separable (for $k$ is of characteristic zero), note that $k(\alpha,\alpha_{1},\ldots,\alpha_{r-1},\alpha_{r})\mid{k}$ where $\{\alpha_{0},\alpha_{1},\ldots,\alpha_{r-1},\alpha_{r}\}$ are the roots of $p_{\alpha,k}(x)$ in $k^{a}$ is a Galois extension whose Galois group will be denoted by $G$, then we notice that:

i) $f(x_{1},x_{2},\ldots,x_{n},x_{n+1}):=\displaystyle\prod_{\sigma\in{G}}\sigma(g(x_{1},x_{2},\ldots,x_{n-1},x_{n})+\alpha{x_{n+1}^{r}})$ is a polynomial in $k[x_{1},x_{2},\ldots,x_{n},x_{n +1}]$. In effect it as $\sigma(f)=f$ for all $\sigma\in{G}$ we have that $f$ coefficients are in the fixed field $k^{G}=k$.

ii) $V(f)=\{(0,0,\dots,0,0)\}\subseteq{\mathbb{A}_{k}^{n +1}}$. Indeed: First notice that $f(0,0\ldots,0,0)=\displaystyle\prod_{\sigma\in{G}}\sigma(g(0,0,\ldots,0,0)+\alpha{0^{r}})=0$. On the other hand if $(a_{1},a_{2},\ldots,a_{n},a_{n+1})\in{V(f)}$, there exists $\sigma\in{G}$ such that $\sigma(g(a_{1},a_{2},\ldots,a_{n-1},a_{n})+\alpha{a_{n+1}^{r}})=0$, $g(a_{1},a_{2},\ldots,a_{n-1},a_{n})+\sigma(\alpha){a_{n+1}^{r}}=0$, then $a_{n+1}=0$ (pues si $a_{n+1}\not=0$, $\sigma(\alpha)=-\displaystyle\frac{g(a_{1},a_{2},\ldots,a_{n-1},a_{n})}{a_{n+1}^{r}}$, taking $\sigma^{-1}$ in both members have $\alpha=-\displaystyle\frac{g(a_{1},a_{2},\ldots,a_{n-1},a_{n})}{a_{n+1}^{r}}$ ($\Rightarrow\Leftarrow$)), then $g(a_{1},a_{2},\ldots,a_{n-1},a_{n})=0$; then $a_{1},a_{2},\ldots,a_{n-1},a_{n}=0$.

Case (II). If $k$ has characteristic $p$ with $p$ prime. In that case is $k_{0}$ the separable closure of $k^{a}$ of $k$. If $k\not=k_ {0}$ then repeat what was done in the case (I) taking $\alpha\in{k_{0}\smallsetminus{k}}$. If $k=k_{0}$, then taking $\alpha\in{k^{a}\smallsetminus{k}}$ we have that there is $m\in{\mathbb{N}}$ such that $\alpha^{p^{m}}\in{k}$ (for $\alpha$ is purely inseparable over $k_{0}= k$), then we note that: $F(x_{1},x_{2},\ldots,x_{n},x_{n +1}): = (g(x_{1},x_{2},\ldots,x_{n-1},x_{n})+\alpha{x_{n+1}})^{p^{m}}$ is a polynomial in $k[x_{1},x_{2},\ldots,x_{n},x_{n+1}]$.

$V(f)=\{(0,0,\dots,0,0)\}\subseteq{\mathbb{A}_{k}^{n +1}}$. Indeed: First notice that $f(0,0,\dots,0,0)=(g(0,0,\ldots,0,0)+\alpha{0})^{p^{m}}=0$. On the other hand if $(a_{1},a_{2},\ldots,a_{n},a_{n+1})\in{V(f)}$, then $g(a_{1},a_{2},\ldots,a_{n-1},a_{n})+\alpha{a_{n+1}}=0$, then as $a_{n+1}=0$ (because $\alpha\not\in{k}$) we have $g(a_{1},a_{2},\ldots,a_{n-1},a_{n})=0$, then $a_{1},a_{2},\ldots,a_{n-1},a_{n}=0$.///

Thus, by the above we have that for all $n\in\mathbb{N}$ exists $f\in{k[x_{1},x_{2},\dots{x_{n-1},x_{n}}]}$ such that $V(f)=\{(0,0,\dots,0,0)\}\subseteq{\mathbb{A}_{k}^ {n}}$.

Let $X$ an affine algebraic set affine space $\mathbb{A}_{k}^{n}$, then there exists an ideal $I$ of $k[x_{1},x_{2},\dots{x_{n-1},x_{n}}]$ such that $X=V(I)$, and $I$ is an ideal finitely generated (for $k [x_{1},x_{2},\dots{x_{n-1},x_{n}}]$ is a NoEther-Ring) have to exist $f_{1},f_{2},\ldots,f_{m-1},f_{m}\in{k[x_{1},x_{2},\dots{x_{n-1},x_{n}}]}$ such that $I=\langle{f_{1},f_{2},\ldots,f_{m-1},f_{m}}\rangle$, then $V(I)= V(\{f_{1},f_{2},\ldots,f_ {m- 1}, f_{m}\})$, then it exists $f\in{k[x_{1},x_{2},\dots{x_{n-1},x_{n}}]}$ such that $V(f)=\{(0,0,\dots,0,0)\}$ must be $X=V(I)=V(f(f_{1},f_{2},\ldots,f_{n-1},f_{n}))$.

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