Show that $ F_{7}[i]=\{a+bi : a,b \in F_{7} \}$ is a field but that $F_5[i]$ is not

Question

(i) Show that $ F_{7}[i]=\{a+bi : a,b \in F_{7} \}$ is a field with 49 elements.

(ii) Show that $ F_{5}[i] =\{a+bi : a,b \in F_{5} \} $ is not a field.

My answer: (i) I have shown that $ F_{7}[i] $ has 49 elements in the following way: Since $a$ has $7$ choices and for each of these choices $b$ has $7$ choices, we have $7\cdot 7=49$ choices in total. So $ F_{7}[i] $ has $49$ elements. But now how can I be sure that $ F_{7}[i] $ is a field?

Please help me the last part and hints of part (ii) if possible

Answer 1

$$F_7[i] \cong \frac{F_7[x]}{(x^2+1)}$$

Therefore $F_7[i]$ will be a field if and only if $x^2+1$ is irreducible over $F_7$. Since it is a quadratic, you only need to check that it has no roots.

For $F_5[i]$, note that $x^2+1$ is $\textbf{not}$ irreducible over $F_5$.

Answer 2

Hint It's true that $F_7[i]$ has $49$ elements, but not all rings with $49$ elements are fields! (For example, consider the direct product $F_7 \times F_7$.) One way to proceed is as follows. By $F_7[i]$ one presumably means the quotient $F_7[x] / \langle x^2 + 1 \rangle$, where $i$ just denotes the image of $x$ under the quotient $F_7[x] \to F_7[x] / \langle x^2 + 1\rangle$. So, if you can show that $x^2 + 1$ is irreducible in $F_7[x]$, the ideal $\langle x^2 + 1 \rangle$ is maximal.

What goes wrong if we try to carry out the same procedure with $F_5[i]$?