12 January 2015
Let R be a ring with 1.
1. Show that (1)2 = 1 in R.
(1)2 = (1)(1) = (1 1 1)(1) = 1 + (1)2 + (1)2 ,
after subtracting (1)2 and adding 1 to both sides, we obtain 1 = (1)2 .
11. Prove that
23 February 2015
Chapter 10.3: R is a ring with 1 and M is an R-module.
10.3.4 An R-module M is called a torsion module if for each m M there is a
nonzero element r R such that rm = 0, where r may depend on m (i.e.,
20 September 2015
Let R be a commutative ring with 1.
7.2.2 Let p(x) = an xn + an1 xn1 + + a1 x + a0 be an element of the polynomial
ring R[x]. We prove that p(x) is a zero divisor in R[x] if and only there is a
2 March 2015
12.1.2 Let M be a module over the integral domain R.
(a) Suppose that M has rank n and that x1 , x2 , . . . , xn is any maximal set
of linearly independent elements of M . Let N = Rx1 + + Rxn be
16 February 2015
9.4.1 Determine whether the following polynomials are irreducible in the rings indicated. For those that are reducible, determine their factorization into irreducibles. The notation Fp denotes the n
9 February 2015
8.3.8 Let be a the quadratic integer ring Z[ 5]. Prove that 2, 3, 1 + 5 and
1 5 are irreducibles in R, two of which are associate in R, and that
6 = 2 3 = (1 + 5) (1 5) are two distinct factoriz
26 January 2015
7.4.7 Let R be a commutative ring with 1. Prove that the principal ideal generated
by x in the polynomial ring R[x] is a prime ideal if and only if R is an integral
domain. Prove that (x) is a maxima
2 February 2015
8.2.3 Prove that the quotient of a P.I.D. by a prime ideal is again a P.I.D.
Proof. Let R be a P.I.D. and let I be a prime ideal. Then I is a maximal ideal
by a proposition. Then R/I is a eld by anot