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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.

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  • $\begingroup$ This should answer your question: math.stackexchange.com/questions/152880/… $\endgroup$ – Dane Aug 4 '13 at 21:18
  • $\begingroup$ See this math.SE thread: math.stackexchange.com/q/340149/264 (possible duplicate?) $\endgroup$ – Zev Chonoles Aug 4 '13 at 21:19
  • $\begingroup$ It's been a while since I've thought about stuff like this, but how about this: In any finite field $F$, the polynomial $f(x) = 1 + \prod_{c\in F} (x-c)$ is irreducible. Extend the field to contain roots of $f(x)$, and if that's not big enough, do it again with that bigger field. $\endgroup$ – Michael Hardy Aug 4 '13 at 21:39
  • $\begingroup$ If you know something from general theory and you can prove it, why not justify your answer by proving that result from general theory and then saying the conclusion that you need follows from that? Of course, it might also be of interest to construct the field, but from the way the question is phrased, it seems as if what I suggest above should be enough to answer it. $\endgroup$ – Michael Hardy Aug 4 '13 at 21:41
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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).

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    $\begingroup$ (+1) To OP: When you do this you will quickly find out that $x^2+x+1$ is the only irreducible quadratic polynomial over $\mathbb{F}_2$. Combine that with the observation that if $p(x)$ is of degree five, then it is either irreducible or has a factor of degree $\le2$. So a quintic $p(x)$ is irreducible, iff it passes the three tests: i) $p(0)\neq0$, ii) $p(1)\neq0$ (no linear factors), iii) $p(x)$ is not divisible by $x^2+x+1$ (no quadratic factors). Hint: Try $x^5+x^2+1$. $\endgroup$ – Jyrki Lahtonen Aug 5 '13 at 6:35
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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.

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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 $n$.

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Over $\mathbb F_p$, the factors polynomial $x^{p^n}-x$ are exactly the primes of degree $d\mid n$.

In particular, the primes of degree $5$ must be factors of $\frac{x^{32}-x}{x^2-x}$, and any such factor must be a prime of degree $5$. So you just need a factor of $1+x+\dots + x^{30}$.

Such factors definitely exist - there are exactly six prime polynomials of degree $5$ over $\mathbb F_2$.

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