$F_p=Z/(pZ)$
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?
$F_p=Z/(pZ)$
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?
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.