Polynomial in $k[x_1, \ldots,x_n]$ has finitely many roots?

Question

Let $ f \in k[x_1, \ldots,x_n]$, $k$ a field then it seems to me that $f$ has only finitely many roots in $k^n$. I was trying induction but did not really work:

$n=1$ follows from Euclidean algorithm. Suppose holds $ \le n-1$, write polynomial with coefficients in $k[x_1]$.For each fixed choice of $x_1 = a \in k$, there are finitely many solutions...

Any hints?

Answer 1

This statement is false in general. Consider the following polynomials in $\mathbb{C}[X,Y]$

The polynomials $ XY $ and $ X + Y $ have infinitely many zeros in $ \mathbb{C}^{2} $.

The zero-set of $XY$ is union of $(\{0\}×\mathbb{C})$ and $(\mathbb{C}×\{0\})$, while the zero-set of $X+Y$ is $\{(a,−a) | a\in \mathbb{C}\}$.

Both are uncountable sets!!

Answer 2

Look at

$p(x, y) = x^2 + y^2 - 1 \in \Bbb R[x, y]; \tag 1$

this polynomial has an uncountable infinity of zeroes

$(x, y) = (\cos \theta, \sin \theta) \in \Bbb R^2, \; \theta \in [0, 2\pi). \tag 2$

The problem is that, though for each value of $x \in [-1, 1]$ there are precisely two values of $y$ such that $p(x, y) = 0$, there are ucountable such $x$, so we are really looking at an uncountable collection of real polynomials $p(x, y) \in \Bbb R[x][y]$.

Similar situations occur in higher dimensions.

Answer 3

There is a characterization of zero-dimensional ideals in the polynomial ring $R={\Bbb K}[x_1,\ldots,x_n]$ using Gröbner bases. An ideal $I$ is zero-dimensional if its zero locus $V_{\Bbb K}(I)$ is finite. Equivalently, let $G=\{g_1,\ldots,g_m\}$ be a reduced Gröbner basis for $I$ w.r.t. the lex order $>$ with $x_1>\ldots>x_n$ such that ${\rm LT}_>(g_i)>{\rm LT}_>(g_{i+1})$ for all $i$. Then for each $1\leq i\leq n$, there exists $j=j_i$ such that ${\rm LT}_>(g_{j_i})=x_i^{e_i}$ for some $e_i\geq 1$; i.e., the initial term is a pure power of $x_i$. This can be directly decided by inspection!