I am trying to prove the following statement.
If a field $K$ is not algebraically closed, then any $K$-variety $V\subset\mathbb{A}^n$ can be written as the zero set of a single polynomial in $K[X_1,X_2,…,X_n].$
Suppose $V=V(f_1,f_2,…,f_n)$, and I would want to find a polynomial $g\in K[X_1,X_2,…,X_n]$ such that $V=V(g)$.
I know that if I can find a polynomial $\phi\in K[X_1,X_2,…,X_n]$ whose only zero is $(0,0,…,0)$, then it is done, since we could take $g=\phi(f_1,f_2,…,f_n)$.
For $n=1$, then I would just choose $\phi=X$, but I do not even know how to proceed to the case $n=2$. Maybe we could choose an irreducible polynomial of degree $>1$.
Any help will be appreciated.
Best Answer
It actually suffices to find such a $\phi$ in the case $n=2$. For instance, if you have such a $\phi$ for $n=2$, then $\phi(X_1,\phi(X_2,X_3))$ works for $n=3$, and $\phi(X_1,\phi(X_2,\phi(X_3,X_4)))$ works for $n=4$, and so on.
In the case $n=2$, what you can do is take a nonconstant polynomial $f(t)\in K[t]$ with no roots in $K$ and homogenize it. That is, if $f$ has degree $d$, define $\phi(X,Y)=Y^df(X/Y)$. You can check that $\phi$ is a homogeneous polynomial of degree $d$ in $X$ and $Y$ in which the coefficient of $X^d$ is nonzero (since that coefficient comes from the leading coefficient of $f$). If $\phi(a,b)=0$ for $a,b\in K$ and $b\neq 0$, then $b^df(a/b)=0$ so $a/b$ is a root of $f$, which is impossible. On the other hand, if $b=0$, then because the coefficient of $X^d$ in $\phi$ is nonzero, we must also have $a=0$. Thus $(0,0)$ is the only zero of $\phi$.