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Thread: idempotents

  1. #1
    Senior Member Sampras's Avatar
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    idempotents

    What are three idempotents of the quotient ring $\displaystyle \mathbb{Q}[x]/(x^4+x^2) $?

    So an element is of the form $\displaystyle a+x^2+x^4 $ where $\displaystyle a \in \mathbb{Q}[x] $. So $\displaystyle (a+x^2+x^4)^2 = a+x^2+x^4 $.


    So one idempotent would be: $\displaystyle 0 $. The others would be just guess and check?
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    Quote Originally Posted by Sampras View Post
    What are three idempotents of the quotient ring $\displaystyle \mathbb{Q}[x]/(x^4+x^2) $?

    So an element is of the form $\displaystyle a+x^2+x^4 $ where $\displaystyle a \in \mathbb{Q}[x] $. So $\displaystyle (a+x^2+x^4)^2 = a+x^2+x^4 $.


    So one idempotent would be: $\displaystyle 0 $. The others would be just guess and check?

    Besides 0 I'd check 1 (!!) and $\displaystyle x^2+1$

    Tonio
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    Quote Originally Posted by Sampras View Post
    What are three idempotents of the quotient ring $\displaystyle \mathbb{Q}[x]/(x^4+x^2) $?

    So an element is of the form $\displaystyle a+x^2+x^4 $ where $\displaystyle a \in \mathbb{Q}[x] $. So $\displaystyle (a+x^2+x^4)^2 = a+x^2+x^4 $.


    So one idempotent would be: $\displaystyle 0 $. The others would be just guess and check?
    use Chinese Remainder Theorem for rings:

    since $\displaystyle x^2$ and $\displaystyle x^2+1$ are coprime, we have $\displaystyle \frac{\mathbb{Q}[x]}{<x^4+x^2>} \cong \frac{\mathbb{Q}[x]}{<x^2>} \times \frac{\mathbb{Q}[x]}{<x^2+1>}.$ obviously for any two rings $\displaystyle R,S,$ an element $\displaystyle (a,b) \in R \times S$ is an idempotent iff $\displaystyle a$ and $\displaystyle b$ are idempotents.
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