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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?

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    Note that you don't need to find an irreducible polynomial of degree 2, it is enough to show that there exists at least one. Can you simply count the reducible ones? – 2017-02-27
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    Didn't think about this. Number of polynomials of order 2 is $p^3$ and number of reducibles is $p^2$ i guess, then there exist $p^3-p^2$ irreducible ones right? – 2017-02-27
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    You should do any such count on monic polynomials. – 2017-02-27
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    @EitanKosman Not really. You should rather count the *monic* polynomials, because otherwise you may run into some nasty double-counting when considering the reducible ones. So, $p^2$ monic polynomials. Then, the number of reducible monic polynomials is $$\text{n}^\circ\text{ of monic squares}+\frac12\left(\text{n}^\circ\text{ of non-identical ordered pairs of monic linear polynomials}\right)$$ – 2017-02-27
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    Thanks! I just realized the number of reducible ones is ${{p+1}\choose 2}$ – 2017-02-27

4 Answers 4

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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.

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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.

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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 $a\in L\setminus\{0\}\implies a^{-1}=a^{p^2-1}$.

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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.