Exercise 1.27

Let $k$ be an algebraically closed field and let

be a set of polynomial equations in $n$ variables with coefficients in $k$. The set $X$ of all points $x=(x_1,\dots ,x_n)\in k^n$ which satisfy these equations is an affine algebraic variety.

Consider the set of all polynomials $g\in k[t_1,\dots ,t_n]$ with the property that $g(x)=0$ for all $x\in X$. This set is an ideal $I(X)$ in the polynomial ring, and is called the ideal of the variety $X$. The quotient ring

is the ring of polynomial functions on $X$, because two polynomials $g,h$ define the same polynomial function on $X$ if and only if $g-h$ vanishes at every point of $X$, that is, if and only if $g-h\in I(X)$.

Let ${\xi }_i$ be the image of $t_i$ in $P(X)$. The ${\xi }_i$ $(1\le i\le n)$ are the coordinate function on $X$: if $x\in X$, then ${\xi }_i(x)$ is the $i$th coordinate of $x$. $P(X)$ is generated as a $k$-algebra by the coordinate functions, and is called the coordinate ring (or affine algebra) of $X$.

As in Exercise $26$, for each $x\in X$ let ${𝔪}_x$ be the ideal of all $f\in P(X)$ such that $f(x)=0$; it is a maximal ideal of $P(X)$. Hence, if $\tilde{X}=\mathrm{Max}<!--\; FUNCTION\; APPLICATION\; -->(P(X))$, we have defined a mapping $\mu :X\to \tilde{X}$, namely $x\mapsto {𝔪}_x$.

It is easy to show that $\mu$ is injective: if $x\ne y$, we must have $x_i\ne y_i$ for some $i$ $(1\le i\le n)$, and hence ${\xi }_i-x_i$ is in ${𝔪}_x$ but not in ${𝔪}_y$, so that ${𝔪}_x\ne {𝔪}_y$. What is less obvious (but still true) is that $\mu$ is surjective. This is one form of Hilbert’s Nullstellensatz (see Chapter 7).