I am trying to solve the following problem.

Suppose that $F$ Is a field, $p(x) \in F[x]$ is a separable, irreducible polynomial of degree $3$ with roots $\alpha_1, \alpha_2, \alpha_3$. Show that if the characteristic of $F$ is not $2$ or $3$, then $F(\alpha_1, \alpha_2, \alpha_3) = F(\alpha_1- \alpha_2)$.

I know that since $p(x)$ is separable, then $\alpha_1$,$\alpha_2$, $\alpha_3$ are distinct. Thus, if we let $K =F(\alpha_1,\alpha_2,\alpha_3)$ be the splitting field of $p(x)$, we know that $K/F$ is a Galois Extension. TO solve the problem, I think I need to use the Fundamental theorem of Galois Theory, but I am not sure how I could use it to solve this problem. Can anyone give me a hint please? Thanks!

  • $\begingroup$ Try this: show that if $\sigma$ is an automorphism of $K$ fixing $\alpha_1-\alpha_2$ then it has to fix $\alpha_1,\alpha_2,\alpha_3$. Since $\sigma$ permutes the roots, we have few cases to consider. $\endgroup$ – Marco Sep 3 '18 at 1:01
  • $\begingroup$ Thanks Marco!! Can I ask, how did you come up with that approach? What did you notice to try that? Thanks! $\endgroup$ – user110320 Sep 3 '18 at 1:04
  • $\begingroup$ It's Galois correspondence. $F(\alpha_1-\alpha_2)$ sits inside $K$. If it is a proper subfield, then by the correspondence there should exist automorphisms of $K$ that fix $F(\alpha_1-\alpha_2)$ but not $K$. $\endgroup$ – Marco Sep 3 '18 at 1:08
  • $\begingroup$ Ok, so if I suppose such an automorphism that fixes $F(\alpha_1 - \alpha_2)$ exists, then such a supposition will end up in some sort of contradiction, right? And that would show that $F(\alpha_1 - \alpha_2)$ is actually the whole field. Is this right? Thanks for the answer by the way! $\endgroup$ – user110320 Sep 3 '18 at 1:12
  • 1
    $\begingroup$ Yes to start say $\sigma(\alpha_1)=\alpha_2, \sigma(\alpha_2)=\alpha_1$, and $\sigma(\alpha_3)=\alpha_3$. See how this leads to a contradiction. There are four more cases likes this. $\endgroup$ – Marco Sep 3 '18 at 1:16

The point is in my opinion (as also commented by Marco under the OP) that no non-identity permutation $\sigma\in S_3$ of the roots can stabilize the difference $\delta=\alpha_1-\alpha_2$. We can check this for all the permutations of $S_3$ as follows. Assume that $\sigma(\delta)=\delta$:

  • If $\sigma=(12)$ then $\delta=\sigma(\delta)=-\delta$ implying that $2\delta=0$. So either $2=0$, in violation of the assumption about the characteristic, or $\delta=0$ and $\alpha_1=\alpha_2$ in violation of the separability assumption.
  • If $\sigma=(23)$ then $\alpha_1-\alpha_2=\delta=\sigma(\delta)=\alpha_1-\alpha_3$ implying that $\alpha_2=\alpha_3$ again in violation of separability. A similar argument rules out the possibility that $\sigma=(13)$ would stabilize $\delta$.
  • If either $(123)$ or $(132)$ stabilizes $\delta$ then so does the other because they are inverses of each other. So in this case $\delta=\alpha_1-\alpha_2=\alpha_2-\alpha_3=\alpha_3-\alpha_1$ implying (as pointed out by both Balaji sb and nguyen quang do) that $$3\delta=(\alpha_1-\alpha_2)+(\alpha_2-\alpha_3)+(\alpha_3-\alpha_1)=0.$$ As we are not in characteristic three this implies $\delta=0$, again in violation of the separability assumption.

So $\alpha_1-\alpha_2$ has a trivial stabilizer, and hence, by Galois correspondence, generates all of $K$.

For the sake of completeness I want to show why the assumptions about the characteristic are essential.

  • In characteristic three the difference $\alpha_1-\alpha_2$ may actually be an element of the base field $F$. The easiest to describe example is the case of an Artin-Schreier polynomial $$p(x)=x^3-x-1\in\Bbb{F}_3[x].$$ From the characteristic three "freshman's dream", $(a+b)^3=a^3+b^3$, it follows immediately that if $p(\alpha)=0$ then also $p(\alpha\pm 1)=0$. So in this case $\delta=\alpha_1-\alpha_2=\pm1$ is an element of the prime field. Yet, the splitting field of $p(x)$ is $\Bbb{F}_{27}$. The polynomial $p(x)$ is irreducible over $F=\Bbb{F}_3$ for lack of zeros in the prime field.
  • In characteristic two it's a bit trickier. From the first part of this answer we see that the only possible non-trivial stabilizer can be the 2-cycle $(12)$ in the Galois group $S_3$. This rules out the use of finite fields, for their extensions are always cyclic. So let, instead, $F=\Bbb{F}_2(t)$ be the field of rational functions. Let $$p(x)=x^3-t.$$ It clearly has no zeros in $F$, so, being cubic, it is irreducible over $F$. Let $\omega$ be a cubic root of unity. It is an element of $\Bbb{F}_4$, and also satisfies the equation $1+\omega+\omega^2=0$. If $\alpha_1$ is a zero of $p(x)$, the other two are $\alpha_2=\omega\alpha_1$ and $\alpha_3=\omega^2\alpha_1$. Here $$\alpha_1-\alpha_2=\alpha_1+\alpha_2=(1+\omega)\alpha_1=\omega^2\alpha_1=\alpha_3.$$ It follows that $F(\alpha_1-\alpha_2)=F(\alpha_3)$ is only a cubic extension of $F$, but the splitting field is $K=F(\alpha_1,\omega)=\Bbb{F}_2(\alpha_1,\omega)$, a degree six extension of $F$.
  • $\begingroup$ Yes, I edited my error in the assertion b) $\endgroup$ – nguyen quang do Sep 4 '18 at 21:35

If $F=F_q$ ($F_q$ is a finite field) then the roots of $p(x)$ are $\alpha_i,\alpha_i^{q},\alpha_i^{q^2}$ because $p(x)=(x-\alpha_1)(x-\alpha_2)(x-\alpha_3)$ and $p(x)^{q^j}=p(x)$. Hence $F(\alpha_1,\alpha_2,\alpha_3)=F(\alpha_i) = F[x]/<p(x)>$. Hence $F(\alpha_1,\alpha_2,\alpha_3)$ is an extension of degee $3$ over $F$. Also $1,\alpha_1-\alpha_2,(\alpha_1-\alpha_2)^2$ are linearly independent. If not the minimal polynomial of $\alpha_1-\alpha_2 $ over $F$ has degree one or two.

Let $\alpha_1^q=\alpha_2$ and $\alpha_1^{q^2}=\alpha_3$.

1) If $\alpha_1-\alpha_2 = (\alpha_1-\alpha_2)^{q}$: then $\alpha_1-\alpha_2 = \alpha_1^{q}-\alpha_2^{q}=\alpha_2 -\alpha_3$ and also $(\alpha_1-\alpha_2)^{q} = (\alpha_1-\alpha_2)^{q^2}$ which implies $\alpha_1^q-\alpha_2^{q} = \alpha_1^{q^2}-\alpha_2^{q^2}$ Hence $\alpha_2-\alpha_3 = \alpha_3-\alpha_1$.

Hence $\alpha_2-\alpha_3 = \alpha_3-\alpha_1 = \alpha_3-(2 \alpha_2-\alpha_3)$. Hence $\alpha_2 = \alpha_3$ which is a contradiction.

2) If $\alpha_1-\alpha_2 = (\alpha_1-\alpha_2)^{q^2}$: then $\alpha_1-\alpha_2 = \alpha_1^{q^2}-\alpha_2^{q^2}=\alpha_3 -\alpha_1$ and also $(\alpha_1-\alpha_2)^{q} = (\alpha_1-\alpha_2)^{q^3}$ which implies $\alpha_1^q-\alpha_2^{q} = \alpha_1^{q^3}-\alpha_2^{q^3}$ Hence $\alpha_2-\alpha_3 = \alpha_1-\alpha_2$.

Hence $ \alpha_1-\alpha_2 =\alpha_2-\alpha_3 = \alpha_2-(2 \alpha_1-\alpha_2)$. Hence $\alpha_1 = \alpha_2$ which is a contradiction.

Hence $\alpha_1-\alpha_2,(\alpha_1-\alpha_2)^q,(\alpha_1-\alpha_2)^{q^2}$ are distinct. Hence minimal polynomial of $\alpha_1-\alpha_2$ is of degree 3.

Hence $F(\alpha_1,\alpha_2,\alpha_3)= F(\alpha_i) = F(\alpha_1-\alpha_2)$.

if $F$ is not a finite field replace the operation $\alpha^q$ by $\sigma(\alpha)$ where $\sigma$ is an element of automorphism group which fixes $F$.

  • $\begingroup$ You don't seem to use the assumption about the characteristic anywhere? Given that the claim is false in both characteristics 2 and 3, this suggests that something is wrong. $\endgroup$ – Jyrki Lahtonen Sep 3 '18 at 4:30
  • $\begingroup$ For example, when $F=\Bbb{F}_3$ and $p(x)=x^3-x-1$ we have $\alpha_2=\alpha_1+1$ and $\alpha_3=\alpha_2+1$. In that $\alpha_2-\alpha_1=1$ does not generate a cubic. So there is no contradiction in having $\alpha_2-\alpha_1=\alpha_3-\alpha_2$ if we allow characteristic three. In this case the minimal polynomial of $\alpha_2-\alpha_1$ has degree one. $\endgroup$ – Jyrki Lahtonen Sep 3 '18 at 4:32
  • $\begingroup$ I am using characterisitc is not 2 and 3...when i prove $\alpha_1=\alpha_2$, i actually get $3\alpha_1=3\alpha_2$. Hence i use $3$ is invertible. $\endgroup$ – Balaji sb Sep 3 '18 at 6:45
  • $\begingroup$ Yes, that's right. But, IMHO the answer would be better if you spelled that out. $\endgroup$ – Jyrki Lahtonen Sep 3 '18 at 6:56
  • $\begingroup$ What does IMHO mean ? $\endgroup$ – nguyen quang do Sep 4 '18 at 13:02

I keep your notations, adding $L=F(\alpha_1-\alpha_2)$. In any characteristic, it is classically known that $G=Gal(K/F)$ is either $S_3$ or $A_3$, and in characteristic $\neq 2, G=A_3$ iff the discriminant of $g$ is a square in $F$. A preliminary remark is that $L\neq F$ : if $L=F$, i.e. $\alpha_1-\alpha_2 \in F$, one would get, on taking the trace in $K/F$, that $Tr(\alpha_1 - \alpha_2)=3(\alpha_1 -\alpha_2)$ or $6(\alpha_1 -\alpha_2)$; but $Tr(\alpha_1)=Tr(\alpha_2)$ because $\alpha_1$ and $\alpha_2$ are conjugate, hence $\alpha_1 =\alpha_2$ because 6 is invertible by hypothesis, a contradiction. Now :

a) If $G=A_3 (=C_3), [L:K]=3$ and $L=K$

b) If $G=S_3 (=D_6)$, its non trivial elements are the 3-cycle $\gamma=$(1 2 3), $\gamma^{-1}$, and the three transpositions $\tau_{1,2}, \tau_{1,3}, \tau_{2,3}$. One can check easily that none of these fixes $\alpha_1-\alpha_2$, so $L=K$

One can generalize the result to an arbitrary prime $p$ as follows:

Result. Let $F$ be a field of characterisic $\neq p$, $K$ the splitting field over $F$ of a separable irreducible polynomial $g \in F[X]$ of degree $p$, and $\alpha_1,\alpha_2$ be two distinct roots of $g$. Then $K$ is the normal closure of the field $L=F(\alpha_1-\alpha_2)$.

Proof. We must show that the normal closure $N$ of $L$ coincides with $K$. By construction, $[K:F]$ is a multiple of $p$, hence, by Sylow, $G=Gal(K/F)$ contains an automorphism $\gamma$ of order $p$, which is necessarily a $p$-cycle on the roots $\alpha_i$ of $g$, say $\gamma=$(1 2...p). Since $N/F$ is normal, $\gamma$ stabilizes $N$, so that $\alpha_1-\alpha_2, \alpha_2-\alpha_3$ etc. all belong to $N$, hence $\alpha_1 - \alpha_i \in N$ for all $i$. Adding up, we get that $p\alpha_1 - \sum \alpha_i \in N$. But the sum of all the $\alpha_i$ is the trace in $K/F$ of $\alpha_1$, which belongs to $F$. It follows that $p\alpha_1\in N$, and also $\alpha_1\in N$ because $p$ is invertible by hypothesis, and finally, by normality, all the $\alpha_i$ belong to $N$, i.e. $N=K$. QED

NB. The answer given by @Balaji sb in the case of finite fields deals only with cyclic Galois groups, so it seems to me that it cannot be extended without explaining further the content of the last sentence.

  • $\begingroup$ A useful generalization! $\endgroup$ – Jyrki Lahtonen Sep 4 '18 at 21:43

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