Math 143
Fall 2012
Instructor : John Lott
The course grade will be based on 2/3 homework, 1/3 nal.
You are encouraged to work together on the homework but you should write
up your own solutions.
Homework is due at the beginning of class. No late homework
Math 143, Fall 2012. HW1 Solutions
(UAG) 1.5 Let k be eld of characteristic = 2. Let V be a three dimensional vector space
over k and Q : V V k a nondegenerate quadratic form on V ; let B be corresponding
symmetric bilinear form, so that
1
B (u , v ) = (Q
Math 143, Fall 2012. HW2 Solutions
(UAG) 1.8 Let P1 , . , P4 P2 be four points, no three of which are collinear. This means
k
that any three of these points correspond to three lines in k 3 which are linearly independent.
Let L1 , L2 , L3 k 3 be the three
Math 143, Fall 2012. HW3 Solutions
(UAG) 2.4 Consider the cubic curve
C = cfw_(X : Y : Z ) P2 | ZY 2 = X 3 + 4XZ 2
R
so that the only point with Z = 0 is the point at innity O = (0 : 1 : 0). This is the neutral
element in the simplied group law described
Math 143, Fall 2012. HW4 Solutions
(UCA) 0.7 Let A be a UFD, K its eld of fractions and f A[T ] a nonzero monic polynomial,
say
f = T n + an1 T n1 + . + a0 , a0 , . , an1 A.
Suppose that K is a root of f . Then, as A is a UFD we have that = p /q , where p
Math 143, Fall 2012. HW5 Solutions
(UCA) 1.1 Consider the ring A = Z and the ideals I = 2Z, J = 3Z. Then, 2, 3 I J but
2 + 3 = 5 I J . Hence, I J is not an ideal.
/
(UCA) 1.2 Let A = Z and consider the ideal I = J = 2Z. Then, I J = 2Z while IJ = 4Z.
(UCA)
Math 143, Fall 2012. HW6 Solutions
(UCA) 1.8 We have nil (A) = cfw_f A | f n = 0,
f n = g m = 0. Then,
n+m
(f + g )n+m =
i =0
some n > 0. Let f , g nil (A) with
n + m i n+mi
fg
= 0,
i
since g n+mi = 0, for i = 0, . , n, and f i = 0, for i = n + 1, . , n +