Problem Sets 150B: Modern Algebra Winter 2009

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Click here for the study list you put your contact info, to find collaborators on HW. (updated Sept 27) Also, on MyUCDavis, we have a discussion page. Feel free to post questions and answers.

Problem sets

 
Homework 1: , due January 12, 2009
Solutions by Jeff: HW1     Optional problems:

Homework 2: , due January 21, 2009
Solutions by Jeff: HW2    

Homework 3: , due January 26, 2009     (but it's short and straightforward)

Solutions by Jeff: HW3

Homework 4: , due February 2, 2009      
Solutions by Jeff: HW4     7.4.16  


See study aid for midterm at vocab .
The solutions for the midterm are here (barring typos).
A rough curve might look like
95-79 A;   78-64 B;   63-55 C;   54-41 D;  
It may be a little harsh in the C/D/F range, but those people should take this as a bit of a warning/push.


Homework 5: , due February 9, 2009      
Solutions by Jeff: HW5

Homework 6: , due February 18, 2009  
Solutions by Jeff: HW6

Homework 7: , due February 23, 2009  
Solutions by Jeff: HW7
    Optional problems:     3.2.7 [There is more detail here than is needed.]
  3.2.12 [There are shortcuts one can take here; if one does Q well, you need not repeat the row reducing for Fp every time.]  
Here are two alternate solutions to 3.2.15 one [ Justifying a quadratic poly has only 2 roots is lacking here. It's only true for fields. Not for rings, such as Z/8Z. Find a counter example! To justify it, one good argument is that the polynomial (x 2 -1) factors as (x-1)(x+1) and it can have no other roots A else (x-A) would also divide it (which we can see by the Division Alogorithm it does not). Recall the Division Algorithm needs us to work over a field. ]   And 3.2.15 two [This one provides justification, BUT it is not as convincing an argument as the one given above as it uses the fairly BIG result that (Fp)x is cyclic. However, it's still a correct proof, so I have included it.]

Homework 8: , due Mar 2, 2009
Solutions by Jeff: HW8 version 1 HW8 version 2 has minor modifications, but i'm leaving version 1 up for now...     On 10.4.3b, note another approach is: Z[i]/(2+i) = (Z[x]/(x^2+1))/(2 + bar x) = Z[x]/(x^2+1, 2+x) = (Z[x]/(2+x)) / (bar x ^2 +1) = (Z[x]/(2+x)) / ( (-2)^2 + 1) = (Z[x]/(2+x)) / (5) = Z/(5).
    Optional problems:     10.3.19     10.4.7     10.4.8a (revised) [The proof IJ = I intersect J here could still use a bit more, so see 10.4.7 for that. ]     10.7.5

You had a quiz in section Mar 4th. Here are the Solutions by Jeff: quiz1

Homework 9: , due Mar 9, 2009
Solutions by Jeff: HW9 Please note on problem 10, we had to amend it to assume R is an integral domain (or the proof won't work.)  
Optional problems:     10.6.7


Homework 10: , due Mar 16, 2009   make a copy for your records before you turn it in if you want to study from it
Solutions by Jeff: HW10
Optional problems:     11.1.1b     11.5.6


Homework 11: , due Mar 21, 2009   (you knew it was coming...)
Solutions by Professor Vazirani: final exam solutions