Q:

Find a maximal ideal in $F_7[x]/(x^2+5)$

$F_p=Z/(pZ)$
I'm stuck at this problem.  I tried to find the irreducible factorisation of $x^2+5$, but that didn't help too much.  Any ideas?  and also could anyone explain the general philosophy with regard to solving this kind of problem?  

A:

sorry, but the further explanation in the comments would actually give the solution, so I just write it here.] 
We know for that $\mathbb{F}_7[x]/(x^2+5)$ some things:
1.$(x^2+5)=(x^2-9)=(x-3)(x+3)$
2.$\mathbb{F}_7[x]/(x^2+5)\cong \mathbb{F}_7 \oplus [x]\mathbb{F}_7$ as vectorspaces. 
3.The above isomorphism makes $\mathbb{F}_7[x]/(x^2+5)$ into a $\mathbb{F}_7$ algebra.  
Now we get that any ideal is a sub-$\mathbb{F}_7$-vectorspace of $\mathbb{F}_7[x]/(x^2+5)$ not containing $1$ in particular it cannot contain any element of $\mathbb{F}_7 \hookrightarrow \mathbb{F}_7[x]/(x^2+5)$. Now for dimensionreasons this means that our maximal ideal has maximal dimension one, which means that any nontrivial ideal is already maximal, so it suffices to find a nontrivial nonunit element and take the ideal generated by it.  So a nontrivial zero-devisor has to do the trick of which we clearly already have 2 obvious possible choices: $(x-3)$ and $(x+3)$.
And we are done, actually no proper algebra involved, only linear algebra.