# Proving the irreducibility of a polynomial in a field extension.

Let $$L$$ be splitting field of the polynomial $$f(x)=(x^3+2x+1)(x^3+x^2+2)(x^2+1) \in \mathbb{F}_3[x]$$. How many proper subfield does $$L$$ have?

This is a question from an old qual at my university, and while the problem doesn't seem tricky by any means, the way I can see forward in solving this is computationally exhaustive and would probably take far too long to do on a timed exam.

The way I can see forward is to find the Galois group of the splitting field, $$L$$, and count its subgroups. We see that $$f_1(x)=x^3+2x+1$$ is irreducible in $$\mathbb{F}_3[x]$$, and if $$\alpha$$ is a root of $$f_1$$, then so is $$\alpha+1$$. So, the splitting field of $$f_1(x)$$ is $$\mathbb{F}_3(\alpha)$$. Using plain "intuition," proving that $$f_2(x)=x^3+x^2+2$$ is irreducible in $$\mathbb{F}_3(\alpha)[x]$$ would be cumbersome, and so one can "guess" that it probably will split in $$\mathbb{F}_3(\alpha)$$, and you'd be right. $$\alpha^2+\alpha$$ and $$\alpha^2+2\alpha$$ are both roots of $$f_2(x)$$. The last polynomial, $$f_3(x)=x^2+1$$ is the tricky one.

I can't seem to find a root in $$\mathbb{F}_3(\alpha)$$, but the only way I can think to prove that $$f_3$$ doesn't reduce in the extension is to plug in an arbitrary element and show it's never $$0$$. Since $$\alpha$$ is a root of $$f_1$$, we have that $$\alpha^3=\alpha+2$$. Thus, any element of $$\mathbb{F}_3(\alpha)$$ can be written as $$a_0+a_1\alpha+a_2\alpha^2$$. We could then plug in a general term into $$f_3(x)$$: $$(a_0+a_1\alpha+a_2\alpha^2)^2+1=0$$ and then try to solve the enormous system of equations that pops out the other side to prove that $$f_3$$ is irreducible (or possibly reducible, if we find a solution that works in $$\mathbb{F}_3$$).

On a homework assignment that I'm not bound by time on, I suppose this is a way of doing it. On a timed test, this feels completely unfeasible and unreasonable.

So, my question, is there a way to approach this problem that doesn't involve blindly guessing the roots for $$f_3(x)$$ or trying to prove that it's irreducible by solving an enormous system of equations? Or does this qual question simply boil down to needing insane arithmetic skills in order to solve in a timely manner?

• The first factor is $x^3-x+1$ which has no roots in $\Bbb{F}_3$ by Little Fermat, so it is irreducible and has a root generating the extension $\Bbb{F}_{27}$. That is a normal extension, so the said factor splits completely in there. The last factor $x^2+1$ has no roots in the prime field either (those would be roots of unity of order four), so its splitting field if $\Bbb{F}_9$. Thus the splitting field $L$ contains both $\Bbb{F}_{27}$ and $\Bbb{F}_9$ as subfields. The smallest possibility is thus the field $\Bbb{F}_{3^6}$. Apr 18 at 5:26
• (cont'd) Irrespective of whether the middle factor has a root in the prime field it will also split in $\Bbb{F}_{3^6}$ because its irreducible factors have degrees $1,2$ or $3$ (we don't care). Apr 18 at 5:27

For each $$q$$, there is at most one field of size $$q$$ in any field, since they are all the roots of the polynomial $$x^q-x$$. Therefore you don't have to do any calculation to conclude $$x^3+x^2+2$$ is completely reducible in $$\mathbb F_{27}=\mathbb F_3[\alpha]$$.
The main point is not only all finite fields of the same order are isomorphic to each other, but in any field $$F$$, if we have subfields $$F_1, F_2\subset F$$ with $$|F_1|=|F_2|<\infty$$, then $$F_1=F_2$$.
Also if $$[K:F]<\infty$$ and $$|F|<\infty$$, then it's easy to show $$|K|=|F|^{[K:F]}$$ .The splitting field of $$x^2+1$$ has order $$9$$, and if it's a subfield of $$\mathbb F_{27}$$, then $$27$$ must be a power of $$9$$. The same technique can be used to show more generally $$\mathbb F_{p^n}\cap \mathbb F_{p^m}=\mathbb F_{p^{\gcd(m,n)}}$$.
• It doesn't track for me, noting that there's only one field of size $9$ to conclude that $x^3+x^2+2$ reduces in $\mathbb{F}_3(\alpha)$. Why is that the case? And does that help us with the $x^2+1$ polynomial? Apr 18 at 4:33
• @TyPerkins If you have to, pick the algebraic closure $F$ of $\mathbb F_3$, then both the splitting field of $x^3+2x+1$ and $x^3+x^2+2$ will be a subfield of order $27$ of $F$, and as said, there can only one such subfield in $F$. The way to approach $x^2+1$ is also added. Apr 18 at 4:36
• @TyPerkins Oh, the fact we just discussed easily resolve this without Galois theory: for different roots $\alpha, \beta$ of irreducible $p(x)$ over $F$, $|F[\alpha]|=|F[\beta]|=|F|^{\deg p}$, so $F[\alpha]=F[\beta]$. Apr 18 at 5:15