Quotient ring, field, irreductible

**roporte** · September 27th 2008, 04:14 PM

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!

**ThePerfectHacker** · September 27th 2008, 04:31 PM

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.

How can I build a field with nine elements?

$\mathbb{F}_3 / (x^2+1)$

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.

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