# Quotient ring, field, irreductible

• Sep 27th 2008, 05:14 PM
roporte
Quotient ring, field, irreductible
Let $K$ be a field and $f \in K[x]$. Prove that $K[x]/ \langle f \rangle$ is a field if and only if $f$ is irreductible.

How can I build a field with nine elements?

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

Thanks!
• Sep 27th 2008, 05:31 PM
ThePerfectHacker
Quote:

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

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

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

Quote:

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

Quote:

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