“One Line” Algebra Qual Problems

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 F be a field of characteristic 0 and let f\in F[x] be an irreducible polynomial of degree >1 with splitting field E\subset F.  Define \Omega=\{\alpha\in E:f(\alpha)=0\}.

  • (a) Let \alpha\in \Omega and m a positive integer. If g is the minimal polynomial of \alpha^m over F, show that the roots of g are \{\beta^m:\beta\in \Omega\}.
  • (b) Now fix \alpha\in \Omega and assume r\alpha\in \Omega for some r\in F. Show that for all i\geq0, \beta r^i\in \Omega.
  • (c) If r is as in part (b), conclude r is a root of unity.
  • (d) If \alpha and r are as in part (b) and if m is the order of r, show that f(X)=g(X^m), where g is the minimal polynomial of \alpha^m over F.

Solution 1:

  • (a) We know \alpha satisfies g(X^m) so f divides g(X^m) and \{\beta^m:\beta\in \Omega\} is contained in the set of roots of g; further, g(X) divides \prod_{\beta\in\Omega}(X-\beta^m) as the latter polynomial is fixed by any Galois action and hence lies in F[X], which shows the reverse inclusion.
  • (b) The Galois action acts transitively on \Omega and preserves roots of f; a homomorphism mapping \alpha to \alpha r will map \alpha r to \alpha r^2 and a homomorphism mapping \alpha to \beta will send \alpha r^i to \beta r^i.
  • (c) Polynomials can only have finitely many roots so \{\alpha\cdot r^i:i\geq 0\} is finite.
  • (d) As in part (a), f divides g(X^m); both polynomials have the same degree as |\Omega|=m\cdot|\{\beta^m:\beta\in \Omega\}| since r is an mth root of unity.

 

Problem 2 [Atiyah MacDonald, Ch1 Ex 7]: Let A be a ring in which every element x satisfies x^n=1 for some n (depending on x). Show that every prime ideal in A is maximal.

Solution 2: If I is prime then A/I is an integral domain and we may apply the cancellation law on \overline{x}^n=\overline{x} to see that \overline{x} is invertible and hence A/I is a field, showing I is maximal.

 

Problem 3 [Multiple sources, none of which I can remember specifically]: Show that if F is a finite field, then |F|= p^k for some prime p.

Solution 3: The characteristic of F must be some prime p, so that F forms a vector space over \mathbb{F}_p which, after choosing a basis and counting, must have order p^k for some k\geq1.

 

Problem 4 [MSU Algebra Exam, Fall 2016]: If G is a finite p group and N\subset G is normal, show that N\cap Z(G)\neq \{1\}.

Solution 4: N is a union of conjugacy classes, so we may let G act on N via conjugation and examine orbits to see |N|=p^k=1+\sum_{conjugacy classes}|C_g|; each |C_g| is a power of p 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 G.

 

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

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

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