A field of order $32$

I was working on this problem from an old qual exam and here is the question. In particular this is not for homework.

True or False: There are no fields of order 32. Justify your answer.

Attempt: From general theory I know that any finite field has prime power order and conversely given any prime power there exists a finite field of that order. So of course such fields exist. But now I need to explicitly construct such a field. If I could somehow construct $\mathbb{Z}_2[x]/(p(x))$ where $p(x)$ is a polynomial of degree 5 which is irreducible over $Z_2$ I am done. But wait, how do I come up with a degree 5 polynomial that is irreducible over $Z_2$. My normal methods don't work here because $p(x)$ does not have order 2 or 3. In which case it is easy to check for irreducibility.

My question is in these kinds of situations, is there a general way to proceed.

Note: I have not learnt Galois' theory or anything like that. Does this problem require more machinery to solve?

Please help.

Solution 1:

No more machinery. Make yourself a table of irreducible polynomials of degrees up to 5 by thinking about how to recognize polynomials over $\mathbb Z_2$ with $0$ or $1$ as a root, then proceeding by doing a sieve of Eratosthenes (crossing out polynomials that are divisible by lower-degree irreducibles).

Solution 2:

Polynomials of degree $2$ or $3$ are reducible if and only if they have a root. You want to find a degree $5$ irreducible. Can you see that a degree $5$ polynomial will be irreducible if it has no roots and is not the product of an irreducible cubic with an irreducible quadratic? Now just list the irreducible cubics and quadratics (there aren't many of them) and form all possible products. This will give you a list of reducible degree $5$ polynomials. Now pick any degree $5$ polynomial with no roots not on that list.

Solution 3:

Let $p$ be prime. Consider the polynomial $f(x) = x^{p^n} - x$ over $\mathbb F_p$. Its derivative is $f'(x) = -1 \ne 0$; hence $f(x)$ is separable. Therefore it has $p^n$ roots. Let $\alpha$, $\beta$ be two roots of $f(x)$. By using the Frobenius map $x \mapsto x^p$ we see that $a^{-1}$, $\alpha + \beta$ and $\alpha \beta$ are also roots of $f(x)$. Thus, the set of roots of $f(x)$ is closed under addition, multiplication and inversion. Hence, the splitting field consists entirely of the $p^n$ roots.

This shows that for any prime $p$ and any positive integer $n$, a finite field of order $p^n$ exists.

(In fact, this field is unique up to an isomorphism. This can be proved using the facts that the splitting field is unique up to an isomorphism, and the multiplicative group of a finite field is cyclic.)

Conversely, let $\mathbb F$ be a finite field. Let $\mathbb F_p$ be its prime subfield. We have $[\mathbb F : \mathbb F_p] = n$ for some positive integer $n$. Thus the order of $\mathbb F$ is $p^n$.

This shows that all finite fields have order $p^n$ for some prime $p$ and positive integer $n$.