Prove that for any prime number (p) there exist a field with $p^2$ elements While practicing this subject, I got stuck on this question, and I don't know if my solution is correct. I'd like to have your input:
I define $I=\langle x^2+p\rangle$
Clearly: $x^2+p \neq 0\pmod{p}$ because p is prime, hence $I$ is a maximal ideal in $\mathbb{Z}_{p}[x]$, and is a kernel of some isomorphism from $\phi :\mathbb{Z}_{p}[x] \setminus I \rightarrow F$.
Now we know that in that field, there are $p^2$ elements. Is that complete?
 A: There are two glaring flaws in this proof:
Firstly, $x^2+c=x^2\pmod{c}$, so this actually is not prime in any of the $\mathbb{Z}_c[x]$
Secondly, $(x^2)=(x^2+p)$ in $\mathbb{Z}_p[x]$ and this ideal is not maximal, because it is contained in $(x)$.
You need to find a different ideal to use. I find it easier to think of it as $\mathbb{Z}/I\cong \mathbb{F}_{p^k}$ for appropriately chosen $I$, personally.
A: What you want to do is find an irreducible polynomial over the field $\Bbb F_p$ with $p$ elements. In case $p=2$, both $x^2+1$ and $x^2+0$ are squares, so reducible. The only other polynomial of degree two is $x^2+x+1$, which works.
For $p>2$, since the multiplicative group $\Bbb F_p^\times$ has $p-1$ elements, an even number, the subgroup of squares is proper, and so not every element of $\Bbb F_p$ is a square. Let $a$ be one such, and then $x^2-a$ is a good irreducible quadratic polynomial.
A: The polynomial you chose does not work, because "$x^2+p$" is not prime in $\Bbb F_p[x]$. In fact, "$x^2+p$", in $\Bbb F_p[x]$, is the polynomial $x^2$, which is not prime. Or, in your terms, $0\in\Bbb F_p$ is a root.
If you can take for granted that every field has an algebraic closure, you could do this, though: consider an algebraic closure of $\Bbb F_p$ and call it $\overline{\Bbb F}_p$. Consider $$L=\{x\in\overline {\Bbb F}_p\,:\,x^{p^2}-x=0\}$$
Since the derivative of $f(x)=x^{p^2}-x$ is $-1$, which is coprime with $f$ in $\overline{\Bbb F}_p[x]$, the polynomial $x^{p^2}-x$ has exactly $p^2$ distinct roots in $\overline{\Bbb F}_p$. You know that $0\in L$ and you can check (use Newton's formula) that $a,b\in L\implies ab\in L\wedge a-b\in L$. Now, observe that $a\in L\setminus\{0\}\implies a^{-1}=a^{p^2-1}$.
A: Well, half a year late, but this question was raised again. This time I found my own proof so I thought about sharing it here:
Consider the group $Z_p^*$.
Define $f : z_p^* \to z_p^*$ by $f(x)=x^2$, we're going to show that it's not onto.
Let $x \in ker(f)$, then $x^2=1 \Rightarrow x=+-1$ and $ker(f)\ne \{1\}$.
So $f$ isn't one-to-one, hence not onto. That means, there exist $z \in z_p^*/ Im(f)$, and for every $x \in z_p^* : x^2 \ne z$
Define: $p(x)=x^2-z$
By the above claim, $\forall x \in z_p^* : x^2 \ne z \Rightarrow x^2-z \ne 0$
and that means that $p(x)$ is irreducible, hence $(p(x))$ is a maximal ideal, and $z_p[x]/(x^2-z)$ is a field with $p^2$ elements.
