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## Return to Answer

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Consider the ring $\prod k[x,y]/(x^n,x^{n-1}y,...,xy^{n-1},y^n)$. First some notation. If $f$ is an element, then $f_n$ is the part of it in $k[x,y]/(x^n,x^{n-1}y,...,xy^{n-1},y^n)$. The degree of $f_n$ is the highest power of the ideal $(x,y)$ that it lies in. Since this is not reduced, the ring we will work with is this modulo the ideal of nilpotents. An element $f$ is nilpotent if $f_n^k$ has degree at least $n$ for some fixed $k$, or, equivalently, if $f_n$ has degree at least $n/k$. Since $n/k$ is positive, if $f_n$ is degree $0$ then $f$ cannot be nilpotent.

This will allow us to check that the ring is classical. If $f_n$ has positive degree, then $f$ is a zero-divisor. Take $g$ such that $g_n=1$, $g_m=0$ for $m\neq n$. $g_n$ is degree $0$ so $g$ is not nilpotent. $f_mg_m$ has degree at least $m/n$, so $f_mg_m$ is nilpotent. Thus a regular element is one which is degree $0$ everywhere, and such an element is invertible.

The localization at $y$, however, is not classical. $f y^l=0$ if the degree of $f_n y^l$ is at least $n/k$, equivalently, if the degree of $f_n$ is at least $n/k-l$. This is the kernel of the localization. If $xf$ is in this ideal then $f$ is in the ideal, since $f$ would have degree $n/k-l-1$. So $x$ is regular. We need to check that $x$ is not invertible, which would happen when $xa+b=y^d$ for some $d$ and some $b$ in the ideal. (then the inverse would be $a/y^d$.) Choose some $n$ high enough that the degree of $b_n$ is more than $d$. Then the equation $xa_n+b_n=y^d$ is impossible.

This completes the problem.

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An element $x/p^k$ of the localization of $\prod_n \mathbb Z/p^n$ is zero if $p^{n-l}$ divides $x_n$ for all $n$ and some $l$. Thus it is a zero divisor if the power of $p$ dividing $x_n$ is unbounded. So a regular element is one where the power of $p$ dividing $x_n$ is bounded, say by $m$. Then there exists some $y$ such that $x_ny_n=p^m$ everywhere, so $xy=p^m$, so $x^{-1}=yp^{k-m}$.

So

Consider the localization is classical.(This argument goes through for any product of Artinian quotients of DVRs.)

We can fix this by using a ring of dimension two, say $\prod k[x,y]/(x^n,x^{n-1}y,...,xy^{n-1},y^n)$. First some notation. If $f$ is an element, then $f_n$ is the part of it in $k[x,y]/(x^n,x^{n-1}y,...,xy^{n-1},y^n)$. The degree of $f_n$ is the highest power of the ideal $(x,y)$ that it lies in. Since this is reduced, the ring we will work with is this modulo the ideal of nilpotents. An element $f$ is nilpotent if $f_n^k$ has degree at least $n$ for some fixed $k$, or, equivalently, if $f_n$ has degree at least $n/k$. Since $n/k$ is positive, if $f_n$ is degree $0$ then $f$ cannot be nilpotent.

This will allow us to check that the ring is classical. If $f_n$ has positive degree, then $f$ is a zero-divisor. Take $g$ such that $g_n=1$, $g_m=0$ for $m\neq n$. $g_n$ is degree $0$ so $g$ is not nilpotent. $f_mg_m$ has degree at least $m/n$, so $f_mg_m$ is nilpotent. Thus a regular element is one which is degree $0$ everywhere, and such an element is invertible.

The localization at $y$, however, is not classical. $f y^l=0$ if the degree of $f_n y^l$ is at least $n/k$, equivalently, if the degree of $f_n$ is at least $n/k-l$. This is the kernel of the localization. If $xf$ is in this ideal then $f$ is in the ideal, since $f$ would have degree $n/k-l-1$. So $x$ is regular. We need to check that $x$ is not invertible, which would happen when $xa+b=y^d$ for some $d$ and some $b$ in the ideal. (then the inverse would be $a/y^d$.) Choose some $n$ high enough that the degree of $b_n$ is more than $d$. Then the equation $xa_n+b_n=y^d$ is impossible.

This completes the problem.

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So the localization is classical.(This argument goes trough through for any product of Artinian quotients of DVRs.)

Other questions:

We can fix this by using a ring of dimension two, say $\prod k[x,y]/(x^n,x^{n-1}y,...,y^n)$k[x,y]/(x^n,x^{n-1}y,...,xy^{n-1},y^n)$. Localize First some notation. If$f$is an element, then$f_n$is the part of it in$k[x,y]/(x^n,x^{n-1}y,...,xy^{n-1},y^n)$. The degree of$f_n$is the highest power of the ideal$(x,y)$that it lies in. Since this is reduced, the ring we will work with is this modulo the ideal of nilpotents. An element$f$is nilpotent if$f_n^k$has degree at least$y$, n$ for some fixed $k$, or, equivalently, if $f_n$ has degree at least $n/k$. Since $n/k$ is positive, if $f_n$ is degree $0$ then $x$ f$cannot be nilpotent. This will allow us to check that the ring is no longer classical. If$f_n$has positive degree, then$f$is a zero divisor because zero-divisor. Take$xf=0$if g$ such that $g_n=1$, $g_m=0$ for $m\neq n$. $g_n$ is degree $0$ so $g$ is not nilpotent. $f_mg_m$ has degree at least $m/n$, so $f_mg_m$ is nilpotent. Thus a regular element is one which is degree $0$ everywhere, and only such an element is invertible.

The localization at $y$, however, is not classical. $f y^l=0$ if the degree of $yf=0$, but nothing times f_n y^l$is at least$x$n/k$, equivalently, if the degree of $f_n$ is a power at least $n/k-l$. This is the kernel of the localization. If $y$ so xf$is in this ideal then$f$is in the ideal, since$f$would have degree$n/k-l-1$. So$x$does is regular. We need to check that$x$is not have an invertible, which would happen when$xa+b=y^d$for some$d$and some$b$in the ideal. (then the inverse would be$a/y^d$.) Choose some$n$high enough that the degree of$b_n$is more than$d$. Then the equation$xa_n+b_n=y^d\$ is impossible.

This completes the problem.

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