I will respond as swiftly as possible:
| Welcome to the web site for Math 123, Abstract Algebra II: Theory of Rings and Fields. This page will contain announcements, hints, extra explanations, and answers to independently posed questions. |
TERRIBLE JOKE OF THE DAY:
"Evariste Galois was not only a mathematical genius but also a
dedicated revolutionary. Ironically, he proved that many problems cannot
be solved by radicals."
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Professor Mike Nakamaye (nakamaye@math) Science Center 334; 5-5340 | |
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CA: Moon Duchin (mduchin@math) Dunster G-13; 3-2241 Section: Wednesdays 9pm, Sci. Ctr. 507 Office Hours: on request-- or electronic |
CA: Steve Wang (sswang@math) Dewolfe 10 #35; 3-0026 Section: Wednesdays 4pm, Sci. Ctr. 109 Office Hours: Thursdays 3-4 in the Math Lounge (4th floor) |
Class meets MWF 11 in Sever 214 (which hopefully you know) and homework is due Fridays in class.
For number 16, you need to at least mention that the field properties
about composition laws (distributivity, etc.) are naturally present in A
simply because the elements of A lie in the complex numbers.
[SOLUTION SET]
-SW
Also, think about what it means, if you know that alpha satisfies x^3-2,
to "choose" it as a real root for the purposes of showing, say, that a
primitive cube root of unity is not in Q(alpha). Why can you do this
without loss of generality?
[SOLUTION SET]
-MD
[SOLUTION SET]
-SW
[SOLUTION SET]
-MD
Problem Set Info
On number 4(d)&(e), though it may seem obvious, you were actually
expected to show that alpha is not in Q(beta)... say by writing
out a general
element in the latter and showing that it couldn't satisfy the
irreducible polynomial of the former.
Number (7) is not correct as posed; you need conditions on the
characteristic.
HINT: The induction in Lang 3.4 can be conducted on the number of
prime factors of m, so that when m is trivial, it reduces to the
case of 3(b).
Lots of people had trouble showing that the maximum abelian extension is
generated by commutator elements. The solution set does away with this
problem sort of slickly-- by doing it backwards: take the commutator
subgroup and show that its fixed field is a maximum abelian
subextension.
Food for Thought
Now here's something interesting: a student asked
me if, when alpha gives a degree n extension of F, it will always be true
for (m,n)=1 that F(alpha^m) gives the full extension F(alpha).
Recall that on the first problem set, you showed this was true for m=2.
If you're so inclined, think about this problem! The answer is "no":
see if you can find an example where F(alpha^m) is degenerate and another
where it is a genuine intermediate field. Finding the latter kind of
counterexample is somewhat tricky.
Links(!)
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Mail me any questions and I will respond as swiftly as possible: |
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| MD 3-5-97 |