# Thread: Galois group of f(x)=x^4-5x^2+6

1. ## Galois group of f(x)=x^4-5x^2+6

I am trying to figure out the logic behind this worked out example in a textbook.

The definition given of the Galois group, Gal(F/K) is the set of all automorphisms of F that fix K. It then proves if F is the splitting field of f(x) over K, then these automorphisms must permute the roots of f(x).

The example it gives is this: $f(x) = x^4-5x^2+6 = (x^2 - 3)(x^2-2)$. Now $F = \mathbb{Q}(\sqrt(2),\sqrt(3))$ is the splitting field of $f(x)$ over $\mathbb{Q}$.

Let $\theta \in Gal(F/\mathbb{Q})$. Then (according to the book) we must have $\theta(\sqrt(2)) = \pm \sqrt(2)$ and $\theta(\sqrt(3)) = \pm \sqrt(3)$. It then goes on to define the automorphisms, which I agree taking the above for granted.

However, what I can't figure out is why $\theta(\sqrt(2)) = \sqrt(3)$ is not a valid permutation? It doesn't violate the definition given in the book. And it doesn't prove anywhere that the automorphisms only permute to other roots in the same irreducible factor. And I can't think of why, just based on the fact of $\theta$ being an isoomorphism that I can elinimate such possibilites as well. Am I missing something obvious?

2. Originally Posted by mpetnuch
However, what I can't figure out is why $\theta(\sqrt(2)) = \sqrt(3)$ is not a valid permutation?
Do you know we are in the same class in set theory? Unbelievable, I know.

Because $\theta( \sqrt{2} )$ needs to a zero of the minimal polynomial of $\sqrt{2}$, i.e. either $\pm \sqrt{2}$. While it is true that $\pm \sqrt{2},\pm \sqrt{3}$ both solve $(x^2-2)(x^3-3)$ it need not to be true because if we restrict out consideration to minimal polynomials.

I want to add, the theorem: "automorphisms permute zeros of a polynomial" is necessary but not sufficient. Meaning if you have a zero then it must be mapped into its permuted zero. The other way around: that if you map a zero into a permuted zero, does not mean that you have an automorphism.

3. Thanks. Didn't check my email so I missed this reply.

I also want to elaborate on your response. It is actually quite funny I kept on reading the theorm in my textbook over and over and then I relaized that it didn't say F was a splitting field of f(x), it said extension field of K! So in my problem I found the splitting field $F = K(u_{1}, u_{2})$ of my polynomial $p(x)$ over the field $K$. As you have said the elements in the Galois group must permute the root of $p(x)$, but this not sufficient. What I failed to initally see was that the minimal polynomials of $u_1, u_2$ are in $K[x]$ and so their roots are in F as well and hence the automorphisms in the Galois group must permute those root as well. Now life is good Hah.

BTW, I dropped the Set Theory course. I felt bad too because Prof. Brinkman is really nice guy and a decent lecturer, but I just found the material too boring. I the middle of the semester I wound up taking a job which needed me at those hours and I chose to make money over staying in the class. I would have liked him to cover some of the more interesting (what I consider more interesting at least) topics like Axoim of choice, Alexander basis thorem, things of that nature earlier in the coures. Perhaps then I would not have taken the job.

4. Originally Posted by mpetnuch
BTW, I dropped the Set Theory course.
(I am the one who wears black pants, white shirt, briefcase , sits in the front and distrubts the class for most of the time (If you were wondering who I am)).

But we where just getting to some new stuff. Ordinal numbers, transfinite induction and recursion.

You know what Cantor said about the Transfinite?
Cantor believed that God communicated to him the knowledge of the transfinite, so he can teach it to the world.

Another nice fact.
Cantor's work was attacked by Christian theologians which said Cantor is blashpheming against the uniqueness of the infinitude of God (Cantor's work on cardinals).

I felt bad too because Prof. Brinkman is really nice guy and a decent lecturer
Brinkmann. I like him too. I wanted to tell him, "Professor, you should just quit math and become a comedian", but maybe that is impolite.
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About your question. The way you find all the automorphism is by using the fact that $| \text{Gal}(F/K) | = [F:K]$ (given the proper conditions on $F/K$). Thus, if you know you have a degree 4 extension then the Galois group is order 4. Which will mean if you can get at least 4 mappings on the roots then you have found all 4 automorphisms. For example, given $F=\mathbb{Q}(\sqrt{2},\sqrt{3})$ as in the above example. The mappings CAN be $\sqrt{2} \mapsto \pm \sqrt{2}$ and $\sqrt{3}\mapsto \pm \sqrt{3}$. This means there are at least four automorphism but the degree extension is 4 which means these actually have to be the automorphism (overwise you cannot get 4).
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Which book are you using? Beachy and Blair? I love that book. Eventhough I knew algebra before that book I strengthed my algebra skills with that book by a lot.
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I am going to be doing independent study over the summer in Field and Galois theory. Maybe you want to join?

5. I would love to join. Who are you doing it with? My algebra skills are only so-so, although I try really hard :-)