I want to prove that $f = X^4 + \overline{2}$ is irreducible in $\mathbb{F}_{125}[X]$.

I know that $\mathbb{F}_{125}$ is the splitting field of $X^{125} - X$ over $\mathbb{F}_5$, and that this is a field extension of $\mathbb{F}_5$ of degree 3, and that $\mathbb{F}_{125} = \mathbb{F}_5[X]/(g)$ where $g = X^3 + X - \overline{1}$.

If we put $\alpha = X + (g)$ then $\{\overline{1}, \alpha, \alpha^2\}$ is a basis for $\mathbb{F}_{125}$ considered as a vector space over $\mathbb{F}_5$. Hence every element $x \in \mathbb{F}_{125}$ can be written $x = a_0 + a_1\alpha + a_2\alpha^2$, with $a_i \in \mathbb{F}_5$.

How do I proceed?

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    $\begingroup$ Suggestion: it's enough to show that $f$ has no roots over $\mathbb{F}_{125}$ and is irreducible over $\mathbb{F}_5$. $\endgroup$ – Pete L. Clark Apr 6 '13 at 19:40
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    $\begingroup$ Thanks for your answers. @Pete I can show that, probably. Can you explain (or give a reference that explains) why that is enough? $\endgroup$ – Ricardo Buring Apr 6 '13 at 20:28

The roots of the polynomial $x^4+2=x^4-3$ are sixteenth primitive roots of unity because $3$ is a fourth primitive root of unity.

The smallest extension field of $K=\mathbb{F}_5$ that contains sixteenth roots of unity is the field $E=\mathbb{F}_{625}$. This is because $16$ does not divide either $24=5^2-1$ nor $124=5^3-1$, but it does divide $624=5^4-1$. Consequently $$ f(x)=(x-\alpha)(x-\alpha^5)(x-\alpha^{25})(x-\alpha^{125}) $$ for some $\alpha\in E=K[\alpha]$, and $f$ is irreducible over $K$ as the minimal polynomial of $\alpha$.

So any factor of $f$ must have its coefficients in $E$, as the said coefficients are sums of products of conjugates of $\alpha$. Let $L=\mathbb{F}_{125}$. As $E\cap L=K$, any factor with coefficients in $L$ must have its coefficients in the prime field $K$. But we just saw that over $K$ this polynomial is irreducible, so no such factor can exist.

  • $\begingroup$ Why are the roots of $x^4 + 2$ sixteenth primitive roots of unity? I understand $E = \mathbb{F}_{625}$ is the smallest extension field of $K = \mathbb{F}_5$ that contains sixteenth roots of unity, and so $E$ must be the splitting field of $f$. Why does there exist an $\alpha$ such that $E = K[\alpha]$ and $f$ factors in that way? And why is $f$ then the minimum polynomial of $\alpha$? The rest of the argument is clear. Thank you. $\endgroup$ – Ricardo Buring Apr 6 '13 at 21:13
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    $\begingroup$ If $\alpha$ is a root of $f$, then (as you also saw) $\alpha^{16}=1$, so the order of $\alpha$ must be a factor of $16$, i.e. $1,2,4,8,$ or $16$. It can't be eight or lower because $\alpha^8=(\alpha^4)^2=3^2=4=-1\neq1$. $\endgroup$ – Jyrki Lahtonen Apr 6 '13 at 21:25
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    $\begingroup$ Continuing with the same $\alpha$ (a root of $f$). Because $\alpha$ is of order 16, and such elements can only exist in $E$ we must have $K[\alpha]=E$. As $[E:K]=4$, the minimal polynomial of $\alpha$ is quartic. The other roots are then conjugates of $\alpha$, so we get them by iterating the Frobenius automorphism. Are you familiar with that bit of Galois theory of extensions of finite fields? $\endgroup$ – Jyrki Lahtonen Apr 6 '13 at 21:35
  • $\begingroup$ Yes, I remember now. Your answer is now clear to me. I quite like this approach. Thanks a lot! $\endgroup$ – Ricardo Buring Apr 6 '13 at 21:46
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    $\begingroup$ YW. Often one can milk quite a lot out of the bits: 1) cyclicity of the multiplicative group, 2) uniqueness of fields of a fixed size, 3) completely understood Galois theory. $\endgroup$ – Jyrki Lahtonen Apr 6 '13 at 21:50

Let me justify my comment above. Actually it was a little off in the sense that it is not necessary to check or roots in the extension field. What I want to show is the following:

Proposition 1: Let $K$ be a field, $f \in K[t]$ an irreducible polynomial of degree $d$, and let $L/K$ be a field extension of degree $n$. Then if $d$ and $n$ are relatively prime, then $f \in L[t]$ remains irreducible.

Proof: Let $\alpha$ be a root of $f$ in an algebraic closure of $K$ which contains $L$. I claim that the natural map $\Phi: K[\alpha] \otimes_K L \rightarrow K[\alpha] L$ is an isomorphism. Indeed it is always surjective, and since $K[\alpha]L$ is a field extension of $K$ containing both $K$ and $L$, its degree must be divisible by both $d$ and $n$. Since $d$ and $n$ are relatively prime, we have $[K[\alpha]L:K] = dn = \dim_K K[\alpha] \otimes_K L$. So $\Phi$ is a surjection between two $F$-algebras of the same finite dimension, hence it is an isomorphism.

Thus, since $K[\alpha] \cong K[x]/(f)$, $K[\alpha] \otimes_K L \cong L[x]/(f)$. Because this is a field, $(f)$ is a maximal ideal of $L[x]$, hence a prime ideal, so $f$ is irreducible in $L[t]$.

Added later: here is a more elementary proof of the above result. Suppose that $f$ becomes reducible in $L$, thus over $L$ there is an irreducible factor $g$ of $f$ of degree $k \leq \frac{d}{2} < d$. Let $M = L[x]/(g)$. Then $f$ has a root $\alpha$ in $M$ and $[M:K] = kn$. On the other hand, since $f$ is irreducible over $K$, the map $K[t] \rightarrow M$ which sends $t$ to $\alpha$ has kernel $(f)$ and thus we get a field homomorphism $K[t]/(f) \rightarrow M$, which shows that $d \mid kn$. Since $\operatorname{gcd}(d,n) = 1$, we get $d \mid k$, and since $k < d$, this is a contradiction.


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