1. 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?

2. Originally Posted by Sampras
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

3. Originally Posted by Sampras
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.