Radical of ideals in local one dimensional rings

Question

Let $R$ be a local one dimensional ring. I want to show that for all $ a,b\in R$, $\sqrt{Ra+Rb}$ is equal to $\sqrt{Ry}$ for some $y\in Ra+Rb$ or is equal to $R$.

Answer 1

Suppose $R$ is noetherian. Let $I$ be a radical ideal of $R$, we want to show that there exists $y\in R$ such that $I=\sqrt{yR}$.

If $I=R$, take $y=1$. So we can suppose $I\ne R$. Let $p_1, \dots, p_n$ be the (pairwise distinct) minimal prime ideals of $R$ (here we use the hypothesis $R$ is noetherian). As $\dim R=1$, the only prime ideals of $R$ are the $p_i$'s and the maximal ideal $m$ of $R$.

Suppose $I=m$. Let $y\in I \setminus (\cup_i p_i)$. Then the only prime ideal of $R$ containing $y$ is $m$, so $\sqrt{yR}=m$ by Krull's intersection theorem.

Suppose $I\subsetneq m$. For each $p_i$, there exists $y_i\in p_i\setminus (\cup_{j\ne i} p_j)$ (if $n=1$, take $y_1=0$). Similarly to the above case, we have $p_i=\sqrt{y_iR}$. We finally get $$ I =\bigcap_{p_i\supseteq I} p_i=\sqrt{\prod_{p_i\supseteq I} p_i}=\sqrt{(\prod_{p_i\supseteq I} y_i)R}.$$ So we can take $y$ equal to the product of the $y_i$ for the $i$ such that $p_i\supseteq I$.