Galois theory

I’ve been quite exhausted lately with work and other annoying life things. So sadly, I haven’t been able to think about math much, let alone write about it. However, this morning on the public transit I was able to indulge a bit by reviewing in my head some essentials behind Galois theory, in particular how its fundamental theorem is proved.

The first part of it states that there is the inclusion reversing relation between the fixed fields and subgroups of the full Galois group and moreover, the degree of the field extension is equal to the index of corresponding subgroup. This equivalence can be easily proved using the primitive element theorem, which I will state and prove.

Primitive element theorem: Let F be a field. F(\alpha)(\beta), the field from adjoining elements \alpha, \beta to F can be represented as F(\gamma) for some single element \gamma. This extends inductively to that any field extension can be represented by some adjoining some primitive element.

Proof: Let \gamma = \alpha + c\beta for some c \in F. We will show that there is such a c such that \beta is contained in F(\gamma). Let f, g be minimal polynomials for \alpha and \beta respectively. Let h(x) = f(\gamma - cx). The minimal polynomial of \beta in F(\gamma) must divide both h and g. Suppose it has degree at least 2. Then there is some \beta' \neq \beta which induces \alpha' = \gamma - c\beta' that is another root of f. With \gamma = \alpha + c\beta = \alpha' + c\beta', there is only a finite number of c such that \beta is not in F(\gamma). QED.

The degree of a field extension corresponds to the degree of the minimal polynomial of its primitive element. That primitive element can be in an automorphism mapped to any one of the roots of the minimal polynomial, thereby determining the same number of cosets.

The second major part of this fundamental theorem states that normality subgroup wise is equivalent to there being a normal extension field wise. To see this, remember that if a field extension is normal, a map that preserves multiplication and addition cannot take an element in the extended field outside it as that would imply that its minimal polynomial has a root outside the extended field, thereby violating normality. Any g in the full Galois group thus in a normal extension escapes not the extended field (which is fixed by the subgroup H we’re trying to prove is normal). Thus for all h \in H, g^{-1}hg also fixes the extended field, meaning it’s in H.


Galois group of x^10+x^5+1

This was a problem from an old qualifying exam, that I solved today, with a few pointers. First of all, is it reducible? It actually is. Note that x^{15} - 1 = (x^5-1)(x^{10}+x^5+1) = (x^3-1)(x^{12} + x^9 + x^6 + x^3 + 1). 1 + x + x^2, as a prime element of \mathbb{Q}[x] that divides not x^5-1 must divide the polynomial, the Galois group of which we are looking for. The other factor of it corresponds to the multiplicative group of \mathbb{F}_{15}, which has 8 elements. Seeing that it has 3 elements of order 2 and 4 elements of order 4 and is abelian, it must be C_2 \times C_4. Thus, the answer is C_2 \times C_2 \times C_4.