August 20, 2007
This is a closed book exam. To use a result, you can cite it by name (e.g., say "by the Main Theorem on Finitely Generated Modules over PIDs") or, if the result does not have a common name, you can just restate it (e.g., say "We know from class that the polynomial ring over a UFD is a UFD").
Make sure that I can understand what you write. It will help if you use complete sentences to communicate your ideas. I do not engage in reading minds. You really have to tell me what is going on. Make sure that it is impossible to misunderstand your write-up. I can be somewhat dense, and that will work to your disadvantage.
If you think a problem needs clarification, please ask. I will respond to the class.
There is a total of 50 points on this exam.
This exam has 9 problems and 10 pages. Make sure that none are missing in your copy.
Problems are not sorted according to difficulty.
Prove or disprove: the alternating group has a generating set that consists entirely of elements of order 2007.
Let
be a Galois extension of degree 12. The following is the lattice of
intermediate extensions.
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Here, vertices of the same hight may correspond to intermediate extension of different degree. The top vertex corresponds to and the bottom vertex corresponds to . Show that the Galois group of is .
The aim of this problem is to show that finite non-abelian simple groups do not arise as subgroups of . You may use the Feit-Thompson theorem that says that finite non-abelian simple groups have even order.
Show that there is exactly one element of order 2 in and determine that matrix.
Show that any finite non-abelian simple subgroup of is contained in .
Deduce that does not contain any finite non-abelian simple subgroups.
Let be the rational function field, and consider the two -automorphisms and of defined by and .
Determine the group generated by and .
Let be the fixed field of . Show that and that generates as a field extension over .
Let be an square matrix with rational coefficients. Suppose that has order 5 and assume that the equation is only satisfied for . Show that is divisible by 4 .
Show that there is some so that is not a PID.
Show that
is irreducible in .
Let be the -submodule of generated by
Find a basis for and determine the structure of as an Abelian group.
For the following, no reasoning is required:
True or false: in a PID, irreducible elements are prime.
True or false: in a domain, irreducible elements are prime.
True or false: in a PID, prime elements are irreducible.
True or false: in a domain, prime elements are irreducible.
True or false: a solvable-by-solvable group is solvable.
Compute .
State the definition of a prime ideal.
State Hilbertβs basis theorem for polynomial rings.
State the definition of a solvable group.
State the definition of a Euclidean ring.