When is a field extension galois

The field automorphisms of that fix do not fix any intermediate fields , i. Every irreducible polynomial over which has a root in factors into linear factors in.

Also, must be a separable extension. A field automorphism of the algebraic closure of for which must fix. That is to say that must be a field automorphism of fixing. A Galois extension has all of the above properties. For example, consider , the rationals adjoined by the imaginary number , over , which is a Galois extension.

Note that contains all of the roots of , and is generated by them, so it is the splitting field of. Of course, there are two distinct roots in so it is separable. The only nontrivial automorphism fixing is given by complex conjugation. The only irreducible polynomials with rational coefficients with roots of the form with are and. Both split into linear factors over. Can you see why this means that a number in a constructible field extension as defined above can be constructed using only an unmarked ruler and compass, and that only numbers in constructible field extensions can be made in this way?

This isn't very difficult to prove but requires some knowledge beyond what I'm assuming for this article. Here's the clever part.

It is easy to show that this has no rational roots, and so the roots are not constructible. You can use methods like this to prove other results about what shapes can or can't be constructed and so forth.

Main menu Search. An Introduction to Galois Theory. The two most important things to know about in order to understand the in depth part of the article are complex numbers and group theory. If you've not come across complex numbers before you can read An Introduction to Complex Numbers , which should be accessible to 15 or 16 year old students. If you haven't come across group theory before, don't worry. I introduce the idea of a group below, although it might be better to try and find a book or web site that goes into more detail.

Galois theory is a very big subject, and until you are quite immersed in mathematical study in a way which is unusual unless studying for a degree in maths, it can seem quite pointless. However, there are two problems which provide some motivation for studying Galois theory - the existence of polynomials which aren't soluble by radicals, and some results about classical Euclidean geometry, for example that you cannot trisect an angle using a ruler and compass, and that certain regular polygons cannot be constructed using a ruler and compass.

So, why is Galois theory called Galois theory? The answer is that it is named after a French mathematician Evariste Galois who did some very important work in this area. He had a very dramatic and difficult life, failing to get much of his work recognised due to his great difficulty in expressing himself clearly. For example, he wasn't admitted to the leading university in Paris, the Ecole Polytechnique , and had to make do with the Ecole Normale.

Asked 5 years, 4 months ago. Active 2 years, 6 months ago. Viewed 2k times. Eevee Trainer I have a valid question on a definition I don't understand I gave the definition and my intepretation in two possible ways for someone to clarify I don't know what else to do. Add a comment. Active Oldest Votes. Mathmo Mathmo However, the notions of being normal and separable make perfect sense for infinite extensions. Surb Surb Combine Lemmas 9. We will often use the notation. Theorem 9. It follows formally from these two observations that we obtain a bijective correspondence as in the theorem.

Then we obtain a short exact sequence.