I have a few troubles with this problem, I would appreciate for some help, thanks

Let $f(x)=3x^4+3x^3+x+1$, find a splitting field $L$ of $f(x)$ over $\Bbb{F_{5}}$

That's what I got:

It is clearly that $f(x)=3x^4+3x^3+x+1$ =$(x+1)(3x^3+1)$, then $f(-1)=0$ $\in \Bbb{Q}$ and his others 3 roots are "$ \sqrt[3]{\frac{-1}{3}}$,$\frac{-1}{\sqrt[3]{3}}$,$-\frac{-1^{(2/3)}}{\sqrt[3]{3}}$"

Note that $i,\sqrt[3]{3},\sqrt[6]{3} \notin \Bbb{Q}$ then $L=\Bbb{Q}(i,\sqrt[3]{3},\sqrt[6]{3})$ is an splitting field of $f(x)$ Over $\Bbb{Q}$, with $deg(Irr(i+\sqrt[3]{3}+\sqrt[6]{3}))=12$

Then $|Gal(L/ \Bbb{Q})|=[L:\Bbb{Q}]$ this because $L/\Bbb{Q}$ is a Normal and separable extension

$\Rightarrow$ $Gal(L/\Bbb{Q}) \simeq G_{12}$ where $G_{12}$ is a group of order 12, the groups of order 12 are the 5 next; $ \lbrace \Bbb{Z_{12}}, A_{4},\Bbb{D_{12}},\Bbb{Z_{6}}\times \Bbb{Z_{2}}$, $\Bbb{Dic_{12}} \rbrace$,

Here is my firts question ¿which one of these is the indicated?,obviously if I'm right till this point

Now it's clear that $f(x)$ its irreducible over $\Bbb{F_{5}}$=$\lbrace {0,1,2,3,4 \rbrace}$ with $f(0)=6,f(1)=8,f(2)=3,f(3)=328,f(4)=193$

Then let $L_{2}$ an splitting field of $f(x)$ over $\Bbb{F_5}$, as $\Bbb{F_5}$ is finite,There must be a correspondence between $Gal(L_{2}/\Bbb{F_5})$ and $Gal(L/\Bbb{Q})$ here is where I am stuck.

I would appreciate any help, and please tell me if I am well or if my procedure is wrong, in any case it would help me to improve, greetings and thanks.

  • $\begingroup$ Every Galois group over a finite field is cyclic. $\endgroup$ – Lord Shark the Unknown Aug 1 '17 at 5:53
  • $\begingroup$ $f(x) = (x+1)(3x^3+1)$ is valid in $\Bbb F_5$ as well, so you should have $f(4) = 0$. Also, why use so large numbers for $f(3)$ and $f(4)$ in $\Bbb F_5$? $\endgroup$ – Arthur Aug 1 '17 at 5:54
  • $\begingroup$ @Arthur yep I have seen my mistake $f(-1)=0$ and $f(3)=23$ i'll work on it $\endgroup$ – 316316J316 Aug 1 '17 at 6:09
  • $\begingroup$ @Arthur what should i do..¿ edit the post or delete it and fix it all and post it again? I am new on this page and I dont know what to do in these situations, thanks. $\endgroup$ – 316316J316 Aug 1 '17 at 6:51
  • 1
    $\begingroup$ To get the splitting field over $\Bbb{F}_5$ all you need to do is to observe that A) the quadratic extension field $\Bbb{F}_{25}$ contains a primitive third root of unity, and B) every element of the field $\Bbb{F}_5$ has a cube root IN THE SAME FIELD. This is because the multiplicative group is (cyclic) of order four, so cubing, as an endomorphism of the multiplicative group, has trivial kernel and hence is a bijection. To wit, $x=2$ is a zero of $3x^3+1$ in $\Bbb{F}_5$ because $3\cdot2^3+1=25=0$. So $3x^3+1$ factors, but the quadratic factor is irreducible over the prime field. $\endgroup$ – Jyrki Lahtonen Aug 1 '17 at 7:26

I spotted a few problems in your attempt (props for posting the details as that makes it easier to gauge where you are in your studies).

The factor $x+1$ can pretty much be ignored as it won't affect neither the splitting fields nor the Galois groups. You are really only looking at the splitting fields of the other factor $p(x)=3x^3+1$. This cubic has as its complex zeros $x_j=\omega^j\root3\of{-1/3}$, where $j=0,1,2$ and $\omega=(-1+i\sqrt3)/2$ is a primitive third root of unity. So a splitting field $L$ of $p(x)$ over $\Bbb{Q}$ is $L=\Bbb{Q}(\root3\of3,\omega)$. In other words, this is the same splitting field that the polynomial $x^3-3$. The Galois group is isomorphic to $S_3$, and $[L:\Bbb{Q}]=6$. The arguments proving those facts are undoubtedly familiar to you from the standard example of the splitting field of $x^3-2$ over $\Bbb{Q}$ so I won't rehash them. You seem to have made the mistake of including $i$ and $\root6\of3$ into the splitting field. That is not unnatural given that $\sqrt{-3}$ and $\root3\of3$ are in there, but you cannot actually get the real sixth root of $3$ by combining those (but you do get a sixth root of $-3$).

Over the field $\Bbb{F}_5$ a splitting field $L_2$ is much smaller. You apparently missed (due to numerical errors) that $p(2)=25=0\in\Bbb{F}_5$, so we get a factorization $$ p(x)=(x-2)(3x^2+x+2)\in\Bbb{F}_5[x]. $$ That quadratic is irreducible over $\Bbb{F}_5$. To see that you can either verify that it has no zeros in $\Bbb{F}_5$, or you can check that the discriminant $$\Delta=b^2-4ac=1^2-4\cdot3\cdot 2=-23=2$$ has no square root in $\Bbb{F}_5$. I outlined a different way of reaching the same conclusion in the comments: because $\Bbb{F}_5^*$ is cyclic of order $4$ the element $-1/3$ has a single cube root in there.

Anyway, over $\Bbb{F}_5$ the splitting field is thus $L_2=\Bbb{F}_5(\sqrt\Delta)=\Bbb{F}_{25}$. Therefore $Gal(L_2/\Bbb{F}_5$ is cyclic of order two.

You were also interested in the relation between these two Galois groups, $G_1=Gal(L/\Bbb{Q})$ and $G_2=Gal(L_2/\Bbb{F}_5)$. A relation can be described using concepts of algebraic number theory. Let $\mathcal{O}_L$ be the ring of algebraic integers in $L$. It turns out that in that ring the ideal $(5)$ can be written as a product of three prime ideals $$ (5)=\mathfrak{p}_1\mathfrak{p}_2\mathfrak{p}_3. $$ All those prime ideals $\mathfrak{p}_j, j=1,2,3,$ have $5$ as an element, and all the quotient rings $\mathcal{O}_l/\mathfrak{p}_j$ are actually isomorphic to $L_2$. Furthermore, if $\sigma\in G_1$ then $\sigma$ must also permute the three prime ideals. It then turns out that the point stabilizers of that action are all isomorphic to $G_2$. If $\sigma(\mathfrak{p}_1)=\mathfrak{p}_1$, then $\sigma$ induces an automorphism $\overline{\sigma}$ of $\mathcal{O}/\mathfrak{p}_1\simeq L_2$. A non-trivial fact is that the mapping $\sigma\mapsto \overline{\sigma}$ is surjective onto $G_2$. The kernel of this mapping measures the ramification of a prime. The fact that five is not a factor of the discriminant of $p(x), \Delta=-972,$ is the key here. It implies that $p=5$ is unramified. This allows us to identify $G_2$ as a subgroup of $G_1$.

  • $\begingroup$ I know too little Galois theory to be sure if there's a convention I'm overlooking, but how can $L=\Bbb{Q}(\root3\of3,\omega)$ if $L$ is by definition a splitting field over $\mathbb F_5$? An extension of $\mathbb Q$ would have the wrong characteristic! $\endgroup$ – Henning Makholm Aug 1 '17 at 10:25
  • $\begingroup$ @Henning I thought that the OP used $L$ for the splitting field of this polynomial over $\Bbb{Q}$ and $L_2$ for the splitting field over $\Bbb{F}_5$. Upon re-examining the OP it appears that their notation is not consistent. In other words, sorry about being somewhat misleading. Edited. $\endgroup$ – Jyrki Lahtonen Aug 1 '17 at 11:50

$3x^4+3x^3+x+1 = 3 (x + 1) (x + 3) (x^2 + 2 x + 4)$ in $\Bbb{F}_{5}[x]$.

Therefore, the splitting field of $f$ is the same as the splitting field of $g(x)=x^2 + 2 x + 4$.

Since $g$ is irreducible in $\Bbb{F}_{5}[x]$ and has degree $2$, its splitting field is $\Bbb{F}_{5}[x]/(g) \cong \Bbb{F}_{25}$.


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