# a certain set of polynomials forming a subspace

Just out of curiosity:

Why is it that all polynomials vanishing on an irreducible component in some affine space form a vector subspace?

Let $\mathbb{A}_k^n$ be affine $n$-space over some algebraically closed field $k$, and let $S\subseteq \mathbb{A}_k^n$ be any subset. Given $f,g\in k[x_1,\ldots,x_n]$, if $f(p)=0$ and $g(p)=0$ for all $p\in S$, then for any $a,b\in k$, $$(af+bg)(p)=af(p)+bg(p)=a\cdot0+b\cdot0=0.$$ Thus $I(S)$, the set of polynomials in $k[x_1,\ldots,x_n]$ which vanish on $S$, forms a vector subspace of $k[x_1,\ldots,x_n]$. In fact, it forms an ideal, because for any $h\in k[x_1,\ldots,x_n]$, $$(hf)(p)=h(p)\cdot f(p)=h(p)\cdot 0=0.$$ When $S$ is an irreducible algebraic set in $\mathbb{A}_k^n$ (a.k.a. an algebraic variety), the ideal $I(S)$ is a prime ideal, and conversely as well.