Let $K$ be a field. Is $I=\langle xy^2, y^2-y\rangle$ a radical ideal in $K[x,y]$ ?
It is easy to see that $xy\in I$. This is because $$-xy=xy^2+x(y^2-y)\in I.$$
Is $\langle xy^2, y^2-y\rangle$ a radical ideal?
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ring-theory
commutative-algebra
ideals
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1I assume you mean for $I$ to be an ideal of $K[x,y]$? Well, the immediate concern is that that $(xy)^2$ is in $I$... it's not immediately clear whether $xy \in I$ or not! There may still be more subtle things that need to be tested too. You should simplify the description of $I$... or simplify the ring $K[x,y] / I$. Or find a Gröbner basis if you know about such things. – 2017-01-15
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1@Hurkyl Since $x(y^2-y)$ and $xy^2$ are both in $I$, so is $xy$. And as far as simplifying the quotient, it is probably easiest to view the ring as $R[x]/(xy)$ where $R=K[y]/(y-y^2)\cong k\times k$ with a vector space basis of $y$ and $1-y$. – 2017-01-15
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0$\sqrt I$ is the intersection of minimal primes over $I$, that is, $\sqrt I=(y)\cap(x,y-1)=(xy,y(y-1))$ (the last equality holds since $(y)$ and $(x,y-1)$ are comaximal) hence $\sqrt I=I$. – 2017-01-15
1 Answers
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Let's compute the quotient ring $K[x,y]/(xy^2, y^2-y)$. As $I=(xy,y(y-1))$, we have: \begin{align}K[x,y]/I&\simeq K[y]/\bigl(y(y-1)\bigr)[x]/(xy)\simeq K[y]/(y)[x]/(xy)\times K[y]/(y-1)[x]/(xy)\\ &\simeq K[x]\times K [x]/(x)\simeq K[x]\times K. \end{align} Thus the quotient ring is reduced, which means $I$ is a radical ideal.