Today I’m going to try to distill some solutions to previous algebra qual problems into one sentence answers. This will likely mean that the solutions are missing details, but my hope is to be able to give a decent overview of how the solution would go. My goal in doing this is to develop my intuition for the main ideas used in these solutions, and also to work on being able to talk about mathematics succinctly (there is also a “puzzle” element to finding the shortest way to express the solution which I have enjoyed). Apologies in advance if this ends up being more confusing than helpful.

**Problem 1** [Wisconsin Algebra Qual, Jan 2007]: Let be a field of characteristic and let be an irreducible polynomial of degree with splitting field . Define .

- (a) Let and a positive integer. If is the minimal polynomial of over , show that the roots of are .
- (b) Now fix and assume for some . Show that for all , .
- (c) If is as in part (b), conclude is a root of unity.
- (d) If and are as in part (b) and if is the order of , show that , where is the minimal polynomial of over .

Solution 1:

- (a) We know satisfies so divides and is contained in the set of roots of ; further, divides as the latter polynomial is fixed by any Galois action and hence lies in , which shows the reverse inclusion.
- (b) The Galois action acts transitively on and preserves roots of ; a homomorphism mapping to will map to and a homomorphism mapping to will send to .
- (c) Polynomials can only have finitely many roots so is finite.
- (d) As in part (a), divides ; both polynomials have the same degree as since is an th root of unity.

**Problem 2** [Atiyah MacDonald, Ch1 Ex 7]: Let be a ring in which every element satisfies for some (depending on ). Show that every prime ideal in is maximal.

Solution 2: If is prime then is an integral domain and we may apply the cancellation law on to see that is invertible and hence is a field, showing is maximal.

**Problem 3** [Multiple sources, none of which I can remember specifically]: Show that if is a finite field, then for some prime .

Solution 3: The characteristic of must be some prime , so that forms a vector space over which, after choosing a basis and counting, must have order for some .

**Problem 4** [MSU Algebra Exam, Fall 2016]: If is a finite group and is normal, show that .

Solution 4: is a union of conjugacy classes, so we may let act on via conjugation and examine orbits to see ; each is a power of and as the identity element has a trivial conjugacy class we must have other elements with trivial conjugacy classes which are precisely elements in the center of .

**Problem 5** [MSU Algebra Exam, Fall 2016]: Show that there are only finitely many simple groups containing a subgroup of index .

Solution 5: If is not normal and has index , then the action of on the cosets of gives a nontrivial map from to which cannot be injective if is too big.