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Let $(R, \mathfrak{m})$ be a non-regular depth 1 noetherian local ring. Then if $x$ is any regular element of $R$ the module $R/(x)$ will have depth 0, and so it has $\mathfrak{m}$ as an associated prime. This means that there is an element $y\in R$ with $((x):(y)) = \mathfrak{m}$. Let $t = y/x$ in the localization $R_x$. This element is integral because multiplication by $t$ gives a map $\phi : \mathfrak{m}\to \mathfrak{m}$, and an application of Cayley-Hamilton gives a monic polynomial which $\phi$ satisfies, and by evaluating at $x$ we find that $t$ satisfies this polynomial as well.

My question is this: do there exist choices of $R, x, y$ such that $t$ does not satisfy any monic polynomial of degree 2? Every example I've tried has this property, but I see no reason for it to be true.

Edit: I forgot to specify that $x$ should be chosen outside of $\mathfrak{m}^2$. Then $t\mathfrak{m} \subseteq \mathfrak{m}$ because otherwise $t\mathfrak{m} = R$, since any proper ideal is contained in $\mathfrak{m}$, and thus $1 = tz = (yz)/x$ for some $z\in \mathfrak{m}$, which implies $x = yz \in \mathfrak{m}^2$, a contradiction (if $y\notin \mathfrak{m}$ then $\mathfrak{m} = ((x) : (y)) = ((x) : R) = (x)$, so $R$ is regular).

Edit: Disregard the last edit, $x \notin \mathfrak{m}^2$ is unnecessary for $t\mathfrak{m} \subseteq \mathfrak{m}$! Consider whether $y\mathfrak{m} \subseteq x\mathfrak{m}$. If this is the case then $t \mathfrak{m} \subseteq \mathfrak{m}$, as desired. Otherwise there is an element $a \in y\mathfrak{m}$ which is not in $x\mathfrak{m}$. Write $a = by$ for $b \in \mathfrak{m}$. We argue $\mathfrak{m} = (b)$, so $R$ is regular. Since $\mathfrak{m} = ((x) : (y))$ we have $a\in (x)$, so $a = cx$ for some $c\in R$. By choice of $a$ we have $c \notin \mathfrak{m}$, and thus $c$ is a unit. Thus $x = c^{-1} a = bc^{-1}y$. This implies $y$ isn't a zero divisor, since $x$ isn't. Given any $s\in \mathfrak{m}$ we have $sy\in (x)$, so $sy = dx$ for some $d\in R$, and then $sy = dx = dbc^{-1} y$. Since $y$ is regular, $s = dbc^{-1}\in (b)$. This shows $\mathfrak{m} \subseteq (b)$, and we know $b \in \mathfrak{m}$ already.

Arctic Char
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Brendan Murphy
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  • I think your proof is a little too complicated. It is obvious that $\frac yx\mathfrak m\subseteq R$. Suppose that $\frac yx\mathfrak m=R$. Then there is $a\in\mathfrak m$ such that $x=ay$. We show that $(a)=\mathfrak m$. Let $b\in\mathfrak m$. Then $by\in(x)$, so $by=acy$. From $x=ay$ we get that $y$ is regular, and thus $b=ac\in(a)$. – user26857 May 23 '21 at 18:43

1 Answers1

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Let $k$ be a field and set $R := k[[X^{3},X^{5},X^{7}]]$ and $x := X^{3}$ and $y := X^{5}$. Then $y X^{3} = x X^{5}$ and $y X^{5} = x X^{7}$ and $y X^{7} = x (X^{3})^{3}$, hence $((x) : (y)) = \mathfrak{m}$. Moreover $t := y/x = X^{2}$ satisfies the polynomial $T^{3}-X^{6} \in R[T]$. Suppose $p \in R[T]$ is a monic polynomial of degree $2$ such that $p(X^{2}) = 0$. Then we can write $p(T) = (T-a)(T-X^{2})$ for some $a \in k((X))$ (ring of formal Laurent series). Since $a+X^{2} \in R$, the coefficient of $X^{2}$ in $a$ must be $-1$; then the coefficient of $X^{4}$ in $aX^{2}$ is $-1$ (in particular nonzero), which contradicts $X^{4} \not\in R$.

Minseon Shin
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