Stone Weierstrass, Maximal Ideals

The Closure of a Separating Subring

Recall the somewhat unorthodox definition of a subring of c(S). It's a ring, yet it need not contain 1. It is also an R vector space.

Subrings need not be ideals, but every ideal in c(S) is a subring.

If T is a subset of S, the functions that vanish on T (mapping T to 0) form an ideal. If T is a single point p, the quotient map is isomorphic to R, a field, hence the ideal is maximal.

Assume S is compact, and W is a subring that separates points. Let W′ be the closure of W in c(S). Now W′ could be all of c(S). If not, it is the ideal that vanishes on a particular point p.

Suppose every point p has some function fp in W that does not vanish on p. Each function determines an open set Op that keeps fp near fp(p), and away from 0. Finitely many open sets cover S, so let f be the sum of the squares of these finitely many functions. Note that f is nowhere 0.

Since S is compact f(S) is closed and bounded. Suppose the image lies in [u,v]. We have seen that the polynomials approximate any continuous function uniformly. Define a continuous function on [0,v] that consists of two line segments. One slopes from the origin up to u,1, and the next runs horizontally from u,1 to v,1. Choose a polynomial approximation within ε/2, hence the constant term has absolute value below ε/2. Drop the constant and find a polynomial within ε of this function, hence it is within ε of 1, over the domain [u,v]. Since W is a linear space and a ring, we can compose f with this polynomial to find a function that is everywhere within ε of 1. A sequence of functions approaches 1, and 1 is in W′. Since W′ is a ring, W′ = c(S).

Failing this, assume all the functions of W vanish on p, for a particular point p, hence all the functions of W′ vanish on p. We need to show all the functions that vanish on p are in W′. Let V be the ring generated by W and the constant functions. The members of W can already be scaled by R, so V is actually the direct sum of W and the constant functions. Now let f be any function that vanishes on p. Remember that V separates points, and it has the constant functions, hence V′ = c(S). The functions of V approximate f uniformly. The approximations that are within 1 of f must have constant term 0, to approach f(p) = 0. Therefore all the approximating functions actually come from W. A function vanishes on p iff it belongs to W′.

Note that W′ does not vanish on some other point q, because W separates points, so something in W maps p to 0 and q to a nonzero value. The functions of W′ are precisely those functions that vanish on p.

In summary, if W is a subring that separates points, W′ = c(S), or W′ is the set of continuous functions that vanish on p.

Maximal Ideals

Assume S is compact and hausdorff, hence normal. Let M be a maximal ideal in c(S). Suppose M does not separate x and y, and M does not vanish on x and y. There is some function g in M with g(x) = g(y) nonzero. Let f be a function that is 0 on x and 1 on y. Now fg separates x and y, and belongs to M. This is a contradiction, hence M separates pairs of points that it does not always map to 0.

If M contains the function described earlier, that is everywhere nonzero, multiply by its inverse to get 1. Therefore M has no such function, and M vanishes on at least one point; call it p. Bring in every function that vanishes on p and find a possibly larger ideal. But there is no larger ideal, hence M is the set of functions that vanish on p.

In summary, maximal ideals correspond to the points of S.

Zariski Topology

Take the zariski topology of c(S), and restrict to the subspace of maximal ideals. A base closed set ve is the set of maximal ideals containing the function e, or the set of functions that vanish on x where e(x) = 0. The preimage of 0 is closed, so ve consists of maximal ideals corresponding to a closed set in S. Conversely, if C is a closed set in S and x is not in C, find a function that is 0 on C and 1 on x. Let E be the collection of all these functions, and vE consists of functions that vanish on C. These are housed in the maximal ideals that vanish on the points of C. Closed sets in S and in the spectrum of maximal ideals correspond, and the spaces are homeomorphic.

Prime Ideals

If S is discrete then every function from S into R is continuous. Thus c(S) is the direct product of copies of R, one for each point in S. Let P be a prime ideal in S. If P includes a function that carries x to something nonzero, multiply by the function that carries x to 1 and everything else to 0. Thus P inclludes the copy of R associated with x. In other words, an ideal consists of the direct product of some of the copies of R, corresponding to some of the points of S. But remember, P is a prime ideal. If it vanishes on x and y, multiply a function that is 1 on x by a function that is 1 on y and get 0. This contradicts the definition of P. Therefore P is the set of functions that vanish on x, which is a maximal ideal. Every prime ideal is maximal.

Of course S is not usually discrete. Let S be [0,1], and build a multiplicatively closed set consisting of all the nonzero polynomials in x with real coefficients. Drive 0 up to a prime ideal P missing these polynomials. Suppose P is maximal, the functions that vanish on a certain value u. Yet x-u vanishes on u, so this is a contradiction. There are many prime ideals that are not maximal.