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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?

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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
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.

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

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.

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