h NRA
1. For each statement below, CIRCLE true or false. You do not need to show your work.
(a) (b) (C) (d) (6)
True False True False True False True False True False
(a) There exist integers a and I) such that a2 + if)2 = 32003. (Note that 32003 is prime
HOMEWORK 12 (DUE WEDNESDAY, MAY 6)
(1) Chapter 16, 6.2.
Solution [K : Q] = 8. I’ll omit the details, but the idea is to build a. tower of ﬁeld
extensions
Q g Q(x/§) g QM, ﬂ) 9 (110/5, f3, V5),
and show that each extension is a degree two extension. At eac
. . 2. WORKSHEET 1 SOLUTIONS: JAN 23
Prove the following
(1) Any ring homorphism satisﬁes f (0) = 0.
Solution: For any 7“ E R we have
f(T)=f(0+T)=f(0)+f(r) H) : _'
Subtracting f (r) from both sides, we see that f (0) = O. \f/ {tr} 0"“
(2) There is a uniq
w .xw m. w,

Math 542 Midterm 1 SOLUTIONS Spring 2015
I
l
1. (20 points) (a) Every prime element is irreducible in Z[\/5 . True.
(b) There exists a ring homorphism (,1) : Z/6 —> Z/2 x Z/4 that sends an element
ﬁ = n + (6) E Z/6 to the pair (ﬁﬁz‘) = (n
HOMEWORK 1: SELECTED SOLUTIONS
(1) Carefully read the webpage. Email Daniel Erman any questions that you have. Write
(on the "homework that you hand in) a statement acknowledging that you have read
the website in full and that you understand the course po
“.3
HOMEWORK 10 (PARTIAL ANSWERS)
(1) Let p, g be distinct prime integers. Prove that ﬁ+ ﬂ is a primitive element for the
extension Q g Q(/f), Solution: Here is an outline of a solution:
0 Note that ([p' ¢ Q but is a root of $2 — p and hence has basis {1,
PRACTICE PROBLEMS FOR FINAL EXAM
(1) Determine all of the ideals in the ring Q[:z:] /(:z:2 — 4). Give an example of a ring with
exactly 5 ideals.
(2) Fix the ring R :2 Q[a:, y] and the ideal I := (m2 — y) g R. The correspondence
theorem says that there is
HW due Mar 5
Tao Ju
March 5, 2014
Problem 20
Let at be transcendental over Zg. Let F = Z2 (a) and let p(a:) = $2 oz.
(a) Prove that p is irreducible over F.
(b) Prove that if ,8 is a root of p in some extension eld, then p(:r) :2 (a: ,8)?
(0) Su
(iii/.
rim
HW due Apr 4
Tao Ju
April 4, 2014
Problem 25
Prove that [Fe : Q} is innite. FC is the eld of constructible reals (straight edge and compass).
Proof:
It is clear 27x72 6 FC for any n, which is a root of (102n 2 = 0. Let f(.13) = $2 2,