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

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?

• Take $f(x, y) = x$ or $f(x, y) = x - y$, for example. Oct 12, 2018 at 12:46

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!!

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.

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!