2

Let $F$=$\mathbb{Z}/p\mathbb{Z}$ where $p \in \mathbb{Z}$ is prime

I need to show there is a field with $p^2$ elements.

I am not sure where to start with this.

Seth
  • 9,393
  • 1
  • 27
  • 64
cf12418
  • 869

3 Answers3

3

If $p=2$, consider $\mathbb{F}_p[x]/{(x^2+x+1)}$. If $p$ is odd, there is $b\in\mathbb{F}_p$ such that $b^2-1$ is a quadratic non-residue - if all the quadratic non-residues were followed by a quadratic non-residue, $0$ would be a quadratic non-residue, contradiction. Hence $x^2-2bx+1$ is an irreducible quadratic polynomial over $\mathbb{F}_p$, and $\mathbb{F}_p[x]/(x^2-2bx+1)$ is a field with $p^2$ elements.

Jack D'Aurizio
  • 353,855
3

Another approach, a little more advanced than the basic one in Jack's answer: take the polynomial $\;x^{p^2}-x\in\Bbb F_p[x]\;$ and let $\;K\;$ be its splitting field over $\;\Bbb F_p\;$.

Now just prove that $\;\left|K\right|=\left|\Bbb F_{p^2}\right|=p^2\;$ and we're done by the uniqueness of finite fields.

DonAntonio
  • 211,718
  • 17
  • 136
  • 287
  • Notice that the point is to prove the existence of a field with $p^2$ elements so we cannot prove in the middle that «$|K|=|\mathbb F_{p^2}$»! :-) – Mariano Suárez-Álvarez May 25 '14 at 18:47
  • @MarianoSuárez-Alvarez, that's precisely what the above is proving: by means of splitting fields to get a field with exactly $;p^2;$ elements. That is not "in the middle": that's the proof! If you want you may take away that "middle" $;\Bbb F_{p^2};$ ... – DonAntonio May 25 '14 at 19:10
  • You cannot refer to $\mathbb F_{p^2}$ is you do not know that there exist fields with $p^2$ elements! And you don't, because the whole point of this question is to prove that they do exist! (You seem to have understood a different question by the way as evinced by your «and we are done by the uniqueness of finite fields») – Mariano Suárez-Álvarez May 25 '14 at 19:17
  • Common, @MarianoSuárez-Alvarez: it's just a way to call that group $;K;$ whose construction is hinted in the above answer, and it's just a way to hint that it actually is the field we're looking for. It never matters, though: whoever doesn't like it erase that $;\Bbb F_{p^2};$ in that last line. – DonAntonio May 25 '14 at 19:19
  • Ok,if you do not care about making sense (or answering the question asked by the OP) I don't mind! – Mariano Suárez-Álvarez May 25 '14 at 19:20
  • @MarianoSuárez-Alvarez, I see you're utterly unable to understand what I meant. Oh, well. – DonAntonio May 25 '14 at 19:21
  • I understand what you meant: I am suggesting you changee what you wrote so that it matches that. – Mariano Suárez-Álvarez May 25 '14 at 19:23
0

Hint :

suppose $p\neq 2$ and consider $\eta : \mathbb{F}_p^*\rightarrow \mathbb{F}_p^*$ defined as $\eta(x)=x^2$

  • Prove that $\eta$ is not injective (This does not behave properly in $p=2$ case)
  • Thus, $\eta$ is not surjective (Why ??)
  • So, you can find some $a\in \mathbb{F}_p^*$ that is not a square. (Why ??)
  • Consider $\mathbb{F}_p[\sqrt{a}]=\{m+n\sqrt{a} : m,n \in \mathbb{F}_p\}$ (Is this a field?)
  • What is the cardinality of $\mathbb{F}_p[\sqrt{a}]$?

It is your turn to construct field of order $4$ (you may even get some answers in this same site).

Another way of looking at the general case is...

You can surely say there does exists an irreducible polynomial in $\mathbb{F}_p[x]$ of degree $2$

Just check no of polynomials of degree $2$ in $\mathbb{F}_p[x]$ and look for all possible combinations like $(x-a)(x-b)$ (be careful in counting)

so, you would know that there does exist some irreducible polynomial $f(x)\in \mathbb{F}_p[x]$ and consider $\mathbb{F}_p[x]/(f(x))$