{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

103b_06w_1e

# 103b_06w_1e - a ∈ R is called a zero divisor if there are...

This preview shows page 1. Sign up to view the full content.

Math 103B First Hour Exam (40 points) 1 February 2006 Please put your name and ID number on your blue book. The exam is CLOSED BOOK except for one page of notes. Calculators are NOT allowed. In a multipart problem, you can do later parts without doing earlier ones. You must show your work to receive credit. 1. (10 pts.) Consider the ring R = Z 2 [ x ] / ( x 2 + 1 ) . (a) How many elements does R contain? (b) Show that R is not an integral domain. 2. (5 pts.) Find all the zero divisors of Z 3 Z 4 . (Recall that R S is the set of pairs ( r,s ) with addition and multiplication done componentwise.) 3. (10 pts.) For k Z n , define ϕ k : Z n Z n by ϕ k ( x ) = kx . (a) Prove that ϕ k is a ring homomorphism if and only if k 2 = k in Z n . Hint : ϕ k (1 · 1) = ϕ k (1) ϕ k (1). (b) What is the kernel of ϕ 4 : Z 6 Z 6 ? 4. (15 pts.) Let R be a ring. An element of R is called nilpotent if some power of it is zero. (In particular, 0 is nilpotent.) A nonzero a
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: a ∈ R is called a zero divisor if there are nonzero b and c such that ab = 0 and ca = 0. (The text only de±ned zero divisors for commutative rings.) (a) Prove that every nonzero nilpotent element of a ring is a zero divisor. Suppose R is a commutative ring and suppose a, b ∈ R satisfy a n = 0 and b k = 0. It can be shown that ( a − b ) n + k = 0. (You do NOT need to do this.) (b) Prove: If R is a commutative ring, then the nilpotent elements of R are an ideal. (c) It was shown in class that the only ideals in the ring M 2 ( R ) of 2 × 2 real matrices are the trivial ones { } and M 2 ( R ). Use this to show that “commutative” is necessary in (b). Hint : Look at a = p 0 1 0 0 P and b = p 1 P . END OF EXAM...
View Full Document

{[ snackBarMessage ]}