Is the claim $\sqrt{J_1+J_2}\neq J_1+J_2$ with $J_1=\sqrt{J_1},J_2=\sqrt{J_2}$ correct? and possible proof?

Asked Oct 15 '22 at 19:37

Active Oct 15 '22 at 19:37

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I am trying to show that in a commutative ring $R$ and for the radical ideals $J_1,J_2\subset R$ we do not have an equality between $\sqrt{J_1+J_2}$ and $J_1+J_2$. Is this necessarily an inequality in a polynomial ring over an algebraically closed field?

All my attempts to proof this claim of an inequality in the general case of a commutative ring $R$ have failed so far.

asked Oct 15 '22 at 19:37

user823

1

What happens if $R=k[x,y]$, $J_1=\langle y\rangle$ and $J_2=\langle x^2-y\rangle$? But it should also be possible that there is equality. What are you trying to prove actually? – Jyrki Lahtonen Oct 15 '22 at 19:41
In this case, $x\in\sqrt{J_1+J_2}$, but $x\not\in J_1+J_2$, but $x^2$ is in the latter. This is already a counterexample. – user823 Oct 15 '22 at 19:49
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$J_1=J_2$ is an example for $\sqrt{J_1+J_2}=J_1+J_2$. – Kenta S Oct 16 '22 at 01:54
@JyrkiLahtonen : I was trying to find examples why we need to take radicals on the right hand side of $I(Y_1\cap Y_2)=\sqrt{I(Y_1)+I(Y_2)}$ where $Y_1$ and $Y_2$ are affine algebraic sets. – user823 Oct 16 '22 at 17:47
Does this answer your question? What is an example of a radical of sum of ideals not being equal to the sum of radicals?. As it happens, the answer there uses the same example I cooked up. Users who really understand AG can comment. I think a common mechanism leading to this is that the two varieties make a higher order contact in their intersection (like two curves sharing a common tangent). – Jyrki Lahtonen Oct 17 '22 at 03:12

Is the claim $\sqrt{J_1+J_2}\neq J_1+J_2$ with $J_1=\sqrt{J_1},J_2=\sqrt{J_2}$ correct? and possible proof?

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