# Quotient ring, field, irreductible

• Sep 27th 2008, 04:14 PM
roporte
Quotient ring, field, irreductible
Let $\displaystyle K$ be a field and $\displaystyle f \in K[x]$. Prove that $\displaystyle K[x]/ \langle f \rangle$ is a field if and only if $\displaystyle f$ is irreductible.

How can I build a field with nine elements?

Prove that $\displaystyle R[x]/ \langle x^2 + 1 \rangle \cong \mathbf{C}$

Thanks!
• Sep 27th 2008, 04:31 PM
ThePerfectHacker
Quote:

Originally Posted by roporte
Let $\displaystyle K$ be a field and $\displaystyle f \in K[x]$. Prove that $\displaystyle K[x]/ \langle f \rangle$ is a field if and only if $\displaystyle f$ is irreductible.

Say that $\displaystyle f$ is irreducible then we need to show for any $\displaystyle [g] \in K[x]/(f)$ with $\displaystyle [g]\not = [0]= (f)$ (i.e. non-zero) we can find $\displaystyle [h]$ such that $\displaystyle [g][h] = [gh]= [1]$. Now since $\displaystyle f,g$ are relatively prime since $\displaystyle f$ is irreducible it means there exists $\displaystyle q,h$ so that $\displaystyle fq+hg=1$ and so $\displaystyle hg\equiv 1 ~ (f)$ i.e. $\displaystyle [hg] = [1]$.

Now you try proving the converse by assuming that $\displaystyle f$ is reducible.

Quote:

How can I build a field with nine elements?
$\displaystyle \mathbb{F}_3 / (x^2+1)$

Quote:

Prove that $\displaystyle R[x]/ \langle x^2 + 1 \rangle \cong \mathbf{C}$
Define $\displaystyle \theta : R[x] / (x^2+1) \to \mathbb{C}$ by $\displaystyle \theta(1) = 1$ and $\displaystyle \theta (x) = i$. Now show this extends to a function which is in fact a field isomorphism.