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Let $k$ be a field and let $k[x,y]$ be the polynomial ring in two variables. Why this ring has trivial Jacobson radical?

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3 Answers 3

up vote 5 down vote accepted

Let me note the following more general approach to this question:

Exercise 1: Let $A$ be a commutative ring and let $A[x]$ be the polynomial ring in one variable over $A$. Prove the following assertions:

(a) The polynomial $f(x)=a_0+a_1x+\cdots+a_nx^n\in A[x]$ is a unit in $A[x]$ if and only if $a_0$ is a unit in $A$ and $a_1,\dots,a_n$ are nilpotent elements of $A$.

(b) The polynomial $f(x)=a_0+a_1x+\cdots+a_nx^n\in A[x]$ is a nilpotent element of $A[x]$ if and only if $a_0,a_1,\dots,a_n$ are nilpotent elements of $A$.

(Hint: In (a), note that an element of a commutative ring is a unit if and only if it is not contained in any maximal ideal of the commutative ring. In (b), note that the nilradical of a commutative ring is the intersection of all prime ideals of the commutative ring.)

Theorem 1 If $A$ is a commutative ring and if $A[x]$ is the polynomial ring in one variable over $A$, then the Jacobson radical of $A[x]$ is equal to the nilradical of $A[x]$.

Proof. We know that the nilradical of $A[x]$ is contained in the Jacobson radical of $A[x]$. Let us prove the reverse inclusion. If $f$ is an element of the Jacobson radical of $A[x]$, then $1+fx$ is a unit of $A[x]$. Exercise 1 (a) implies that the coefficients of $f$ are all nilpotent elements of $A$. Therefore, Exercise 1 (b) implies that $f$ is a nilpotent element of $A[x]$.

Let me recall that a commutative ring $A$ is reduced if the nilradical of $A$ is trivial.

Exercise 2 Prove that if $A$ is a reduced ring, then the Jacobson radical of the polynomial ring $A[x_1,\dots,x_n]$ over $A$ in $n$ variables is trivial. (Hint: Use Theorem 1 and Exercise 1 (b).)

I hope this helps!

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Let $J$ denote the Jacobson radical, $f\in J$, then $1+xf$ is a unit! So $f=0$.

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A splendidly crisp proof, wxu! –  Georges Elencwajg Sep 12 '11 at 20:37

One can classify all prime ideals of this ring, if $k$ is alg. closed: they are:

1) The zero ideal

2) Those of the form $(f)$ with $f\in k[X,Y]$ irreducible

3) The maximal ideals $(X-a,Y-b)$ with $a,b$ in $k$.

Now try to solve the problem again.

Edit: in my first answer I forgot to mention that $k$ is alg. closed.

Nevertheless, in any case (i.e. $k$ arbitrary) the ideals in 3) are maximal as the residue field is $k$. They suffice to conclude because if a polynomial in $k[X,Y]$ lies in all maximal ideals, it lies in all of the form in 3).

As Georges points out for this solution of course you have to assume $k$ infinite.

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You're missing some maximal ideals when $k$ is not algebraically closed. As for user10: the ideals you've seen are enough to answer your question. –  Hurkyl Sep 12 '11 at 8:26
    
Dear Descartes, dear Hurkyl, if $k$ is a finite field with $q$ elements, the rational maximal ideals $(X-a,Y-b)$ you mention do not suffice to prove that the Jacobson radical is zero, since for example $X^q-X$ is in the intersection of all those rational maximal ideals. –  Georges Elencwajg Sep 12 '11 at 11:34

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