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Thread: Radical

  1. #1
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    Radical

    Let $\displaystyle F$ be a field and let $\displaystyle I$ and $\displaystyle J$ be ideals in a ring $\displaystyle F[x_1,...,x_n]$.
    Generally, $\displaystyle \sqrt{\sqrt{I}+\sqrt{J}}=\sqrt{I}+\sqrt{J}$.

    Give an example in $\displaystyle F[x_1,...,x_n]$ where $\displaystyle \sqrt{I+J} \ne \sqrt{I}+\sqrt{J}$.
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  2. #2
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    Quote Originally Posted by KaKa View Post

    Let $\displaystyle F$ be a field and let $\displaystyle I$ and $\displaystyle J$ be ideals in a ring $\displaystyle F[x_1,...,x_n]$.
    Generally, $\displaystyle \sqrt{\sqrt{I}+\sqrt{J}}=\sqrt{I}+\sqrt{J}$. false!
    who told you that? for example in $\displaystyle \mathbb{R}[x,y]$ let $\displaystyle I=<x^2+y^2>, \ J=<y>.$ then both $\displaystyle I,J$ are prime ideals and hence $\displaystyle \sqrt{I}=I, \sqrt{J}=J$ and so $\displaystyle \sqrt{I}+\sqrt{J}=I+J=<x^2,y>.$

    but $\displaystyle \sqrt{\sqrt{I}+\sqrt{J}}=\sqrt{I+J}=\sqrt{<x^2,y>} =<x,y> \neq <x^2,y>.$


    Give an example in $\displaystyle F[x_1,...,x_n]$ where $\displaystyle \sqrt{I+J} \ne \sqrt{I}+\sqrt{J}$.
    same as above.
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