Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

As the title speaks for itself, is the polynomial ring in infinitely many variables over a field normal or not?

Can someone provide a reference/proof? Thanks

Also, what about a directed union of normal subrings? Is that normal?

Tangential to this, what's a criterion for an element in $K[x_{1},x_{2},\ldots]$ to be nilpotent? We know that in $K[x]$ we have that $f$ is nilpotent if and only if the coefficients are nilpotent and similarly in finitely many variables. Does that hold in our case too?

share|improve this question
    
Since your ring is the union of polynomial rings in finitely many variables, you should have no trouble fescribing the nilpotent elements! –  Mariano Suárez-Alvarez Nov 18 '12 at 7:25
    
Likewise for normality, really. It would be best if you told us what you did and explained where you goy stuck. –  Mariano Suárez-Alvarez Nov 18 '12 at 7:27
    
For the question about normality, as a hint note that any element of the fraction field $K(X_1,...)$ will be an element of $K(X_1,...,X_N)$ for some $N$. Similarly, any polynomial with coefficients in $K[X_1,...]$ will be a polynomial with coefficients in $K[X_1,...,X_M]$ for some $M$. –  Rankeya Nov 18 '12 at 8:55
    
Dear @Mariano, when you write "where you goy stuck", are you blaming the OP for not being Jewish? –  Georges Elencwajg Nov 18 '12 at 9:18
add comment

2 Answers

up vote 0 down vote accepted

If $R$ is an UFD (unique factorization domain), then $R$ is integrally closed (see here). The polynomial rings in finitely many indeterminates are UFDs. Moreover, $K[X_1,\dots,X_n,\dots]$ is also an UFD (why?), so it is integrally closed.

A direct union of integrally closed rings is integrally closed.

Remark. If $K$ is a field, as I suppose to be, then $K[X_1,\dots,X_n,\dots]$ is an integral domain and its only nilpotent element is $0$. In general, if $K$ is a commutative ring, then you have the same criterion as for the polynomial rings in finitely many indeterminates.

share|improve this answer
    
Thank you for the answer, everything is clear; I forgot that UFDs are normal –  Dquik Nov 18 '12 at 23:56
add comment

Since the polynomial ring in finitely many variables over a field is a UFD, it is normal. Hence it suffices to prove that a directed union of normal subrings is normal.

Let $A$ be a directed union of a family of normal subrings $(A_i)_{i \in I}$. Since every $A_i$ is an integral domain, $A$ is also an integral domain. Let $K$ be the field of fractions of $A$. Suppose $\alpha \in K$ is integral over $A$. There exists $c_1, \cdots, c_n \in A$ such that $\alpha^n + c_1\alpha^{n-1} + \cdots + c_n = 0$. $\alpha$ can be wriiten as $\alpha = \frac{a}{b}$, where $a, b \in A$. Since $(A_i)_{i\in I}$ is directed, there exists $i \in I$ such that $a, b, c_1,\dots,c_n \in A_i$. Hence $\alpha$ is contained in the field of fractions of $A_i$ and it is integral over $A_i$. Since $A_i$ is normal, $\alpha \in A_i$. Hence $A$ is normal.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.