# What is the Galois group of the extension $\mathbb F_3(x^4)\subset\mathbb F_{3^2}(x)?$

Square brackets $[\;]$ will denote taking the ring of polynomials, and round brackets $(\;)$ will denote taking the field of rational functions.

My homework assignment from about a month ago had the following problem in it.

Find the Galois group of the field extension $\mathbb F_3(x^4)\subset\mathbb F_{3^2}(x).$

I didn't do it then and now I'm trying to do it to prepare myself for the exam.

The problem looks very exotic to me and I'm having trouble starting to do it. I haven't determined any Galois groups by myself yet. I first tried to see what I know about this extension.

First of all, we have $$\mathbb F_3(x^4)\subset \mathbb F_{3}(x)\subset \mathbb F_{3^2}(x).$$

The first extension has degree at most $4$ because $x$ is a zero of $f\in \mathbb F_{3}(x^4)[y],$ where $$f(y)=y^4-x^4.$$ But $f$ is irreducible by Eisentein's criterion: $x^4$ is a prime element in $\mathbb F_3[x^4]$. Therefore $f$ is the minimal polynomial of $x^4$ and the first extension has degree $4$.

The extension $\mathbb F_3\subset \mathbb F_{3^2}$ has degree $2$ by an element count. Therefore by what's been said here, the second extension has degree $2$. Thus, the extension in the problem is finite and has degree $8$.

Unfortunately, $\mathbb F_3(x^4)$ isn't finite, nor is it of characteristic $0$. If it were one of these things, I could say that the order of the Galois group of the extension is $\leq 8.$ The best I'm able to see is that there is an eight-element $\mathbb F_3(x^4)$-basis of $\mathbb F_{3^2}(x),$ for example $(1,x_1,x_2,\ldots,x_7)$. So $$\mathbb F_{3^2}(x)=\mathbb F_3(x^4)(x_1,x_2,\ldots,x_7).$$

I could try to give an upper bound to the order of the Galois group of the extension by saying that each of $x_i$ must be mapped by any automorphism to some root of its minimal polynomial. But I don't think I can get a satisfactory bound this way.

In general, most of the study of field automorphisms done in classes was in characteristic zero and for finite fields, so most of the theorems don't apply here. I think this must mean that the problem must have an elementary solution, but I don't see it.

• The order of the automorphism group is always less than or equal to the degree of the extension, in any characteristic. So I don't understand why you imply that you do not know that the automorphism group has order at most $8$. The first extension is Galois, because $y^4-x^4$ is separable over a field of characteristic $3$, so any automorphism of the big that fixes $F_3(x^4)$ field must map $F_3(x)$ to itself, and must map $x$ to one of the roots of $y^4-x^4$, which are $\pm x$ and $\pm 2x$. – Arturo Magidin May 5 '12 at 23:04
• Both of your extensions are separable. Usually the problems that occur in fields of characteristic $p$ occur because you have the possibility of inseparability, which does not occur here. All the theorems you want to apply should apply in this situation as well. – Arturo Magidin May 5 '12 at 23:08
• Any automorphism must map $x$ to a root of $y^4-x^4$, whether or not this is the minimal polynomial, because the coefficients of that polynomial are fixed by the action of the automorphism. Likewise, any automorphism must map a root of $y^2-2$ to a root of $y^2-2$ (this root generates the extension $F_3\subseteq F_9$. So there are at most $8$ possibilities, regardless of whether these polynomials are still the minimal polynomials or not. – Arturo Magidin May 5 '12 at 23:15
• @ArturoMagidin I have asked a question here about what I understand you said in your first comment. Could you please take a look? – user23211 May 6 '12 at 11:33
• @ymar Would you mind if I asked where this problem came from? That is, might I be able to find it in a book or in some online source? – David K. May 11 '12 at 20:39

Let $\alpha$ be a root of $y^2-2$ in $F_9$; then $F_9 = F_3(\alpha)$, and so $F_9(x) = F_3(x,\alpha)$. Any automorphism is completely determined by what it does to $x$ and to $\alpha$.

Any automorphism of $F_9(x)$ over $F_3(x^4)$ must map $x$ to a root of $y^4-x^4$, because the coefficients of the latter are fixed. So an automorphism of $F_9(x)$ over $F_3(x^4)$ must map $x$ to one of $x$, $-x$, $\alpha x$, or $-\alpha x$, (the four roots of $y^4-x^4$ in $F_9(x)$). Any automorphism of $F_9(x)$ over $F_3(x)$ must map $\alpha$ to either $\alpha$ or $-\alpha$ (the two roots of $y^2-2$, because $y^2-2$ is fixed by the automorphisms; thus, there are at most $8$ automorphisms.

This does not rely on these polynomials being the minimal polynomials. However, it is worth noting that your extension is the splitting field of $(y^4-x^4)(y^2-2)$, and this polynomial is separable; since your extension is a splitting field of a separable polynomial, it follows that it is a Galois extension. Therefore, the order of the Galois group equals the degree of the extension, so we will have the eight automorphisms mentioned above actually occur.

In general, whether the extension $K/F$ is Galois or not, whether the extension is simple or not, we have that $|\mathrm{Aut}_F(K)|\leq [K:F]$. In fact, we have that the order of the automorphism group is less than or equal to the separable degree of $K$ over $F$. Your extensions here are all separable, so the issue of finite characteristic simply does not show up. In fact, the Primitive Element Theorem holds for any finite separable extensions, and your extension is finite and separable, so the Primitive Element Theorem holds. You will find that the only time that there are potential problems is when the degree of the extension is a multiple of the characteristic, which is not the case here.

• I've consulted my notebook and eureka! The proof of the version of the primitive element theorem we've been told does only use the property of the extension that minimal polynomials are separable. This is very good to know, thank you very much! I have to re-think the problem and try to actually determine which 8-element group this is. I'll ask again when I encounter problems. – user23211 May 6 '12 at 0:01
• I can't see why $x,-x,2x,-2x$ are four roots of $y^4-x^4$ in $\mathbb F_3(x).$ I get $$\begin{eqnarray}(y-x)(y+x)(y-2x)(y+2x)&=&(y^2-x^2)(y^2-4x^2)\\&=&(y^2-x^2)^2\end{eqnarray}$$ Or, in other words, $2x=-x$ in characteristic $3.$ We have $$y^4-x^4=(y-x)(y+x)(y^2+x^2),$$ and I think the last factor is irreducible, because $-1$ isn't a square in $\mathbb F_3$. – user23211 May 6 '12 at 17:30
• @ymar: You are right; it should be $x$, $-x$, $\beta x$ and $-\beta x$, where $\beta$ is a square root of $-1$. The extension is cyclic, generated by the map sending $x$ to $\beta x$. I'll fix the post, but the counting is still correct (at most 8 automorphisms). – Arturo Magidin May 6 '12 at 19:01
• OK, I understand why $\mathbb F_9$ has a square root of $-1$ in it. I think now I can say that the group I'm looking for is $\mathbb Z/4\mathbb Z\times\mathbb Z/2\mathbb Z$ just by looking at it, right? The isomorphism is given by $$(\overline 1,\overline 0)\mapsto (x\mapsto \beta x,\alpha\mapsto \alpha)$$ and $$(\overline 0,\overline 1)\mapsto (x\mapsto x,\alpha\mapsto -\alpha)$$ – user23211 May 6 '12 at 19:23
• @ymar: In fact, $\alpha=\beta$! A square root of $-1$ is the root of $x^2-2=x^2+1$. So you have to be careful: the composition of those two maps does not commute. – Arturo Magidin May 6 '12 at 19:35