From: wcw@math.psu.edu (William C Waterhouse)
Subject: Re: polynomial rings
Date: 18 Nov 1999 22:28:58 GMT
Newsgroups: sci.math
Keywords: isomorphic polynomial rings have the same dimension

In article <80rs5r$23d$1@nnrp1.deja.com>, 
Robin Chapman <rjc@maths.ex.ac.uk> writes:
>...
> > 	R[x,y] is not isomorphic to R[x,y,z].

> 
> No. THis is done in elementary texts on algebraic geometry and/or
> commutative algebra. The ring R[x_1,x_2, ..., x_n] (I presume R
> is the real field) has a chain of n+1 prime ideals
> 0 <= <x_1> <= <x_1, x_2> <= ... <= <x_1, x_2, ..., x_n>
> but no longer chain (this is not obvious). Hence n is determined
> by the ring structure of R[x_1, ..., x_n].

Here's a proof that doesn't use any results from dimension theory.

Theorem Let  F  and  F' be fields. Then the polynomial rings
 F[X,Y] and F'[U,V,W] cannot be isomorphic as rings.

Proof.  Suppose we have an isomorphism  h from F[X,Y] onto F'[U,V,W].
It must take invertible elements to invertible elements, so it
induces an isomorphism of F onto F'.  We may thus replace F' by F 
and assume that the isomorphism is the identity on constants.

Look now at the F-homomorphism  k:F[X,Y] -> F sending X and Y to 0.
Then composition with the inverse of h gives an F-homomorphism 
from  F[U,V,W] to F. It must send U,V,W to constants a,b,c. 
There is an automorphism of F[U,V,W] taking U-a to U, etc, so
we can change our choice of variables in the second ring and
assume that the composite sends U,V,W to 0. 

Now the kernel of k is the ideal M generated by X and Y.  Our h
must take this to the ideal M' generated by U,V,W. But then h
must also take M^2 to (M')^2. This however is impossible, since
M/M^2 is an F-space of dimension 2 and M'/(M')^2 has dimension 3.


Corollary. Let R be a (commutative) ring.  Then there is no
** R-algebra** isomorphism from R[X,Y] onto R[U,V,W].

Proof. Let N be a maximal ideal in R, with R/N = F (a field).
Then any R-algebra isomorphism from R[X,Y] onto R[U,V,W] will
induce an isomorphism from F[X,Y] onto F[U,V,W].

Remark.  There can sometimes be ring isomorphisms, as for 
example when R itself is already a polynomial ring in infinitely
many variables.


William C. Waterhouse
Penn State