Transcribed Text
Denition 0.1. Let R be a ring. A non-zero element a 2 R is called a zero-divisor if there
exists a non-zero element b 2 R such that a b = 0. A ring with no zero-divisors is called
an integral domain. An element a 2 R is called a unit if it has a multiplicative inverse,
i.e., if there exists b 2 R such that a b = 1.
(1) Find all units and zero-divisors in the following ring: Z=18Z
(2) Describe all units and zero-divisors in the ring (Z=4Z)[x], i.e., in the normal polyno-
mial ring with coecients in the ring Z=4Z.
(3) Show that every eld is in integral domain.
(4) Show that 2n 1 is prime only if n is prime. Is the converse true?
(5) Let F be a eld. Exhibit innitely many irreducible elements in F[x], none of which
is a constant multiple of another.
(6) Find minimal polynomials over Q for the following algebraic numbers:
* 3 +
p
5
*
p
2 +
p
3
(7) Consider the following polynomial f(x) = x42x37x+14. Note that f can be con-
sidered as an element of F[x], where F is one of the following elds Q;R;C. Factor f into
irreducibles as an elements of F[x] in each of the three cases, i.e., when F = Q; F = R
and F = C.
(8) Do the same for the following polynomial x4 + 2x3 + 2x 1, but take F to be
Q; Q(
p
2); R; C.
(9) Consider := 2 + 4
p
7 2 Q(
p
7). Find the inverse of and write it so that it is
clear that it belongs to Q(
p
7).
(10) Consider := 1 + 1
2
3 p
2 2 Q( 3
p
2). Find the inverse of and write it so that it
is clear that it belongs to Q( 3
p
2).
(11) Find a polynomial f(x) of degree 2 such that x4 + x2 + 2x 1 f(x) (mod
x3 1).
(12) Recall that since x2 2 is an irreducible element of Q[x], the ring of congruence
classes Q[x]=(x2 2)Q[x] is a eld. Find the multiplicative inverse of the following ele-
ments: x + (x2 2)Q[x]; 2 + (x2 2)Q[x]; 3x 1 + Q[x]:
(13) Consider the following two polynomials f(x) = x2 5 and g(x) = x2 4x 1.
The roots of f are obviously
p
5 and
p
5. Let ; be the two roots of g. Consider
the following 4 elds (why are they elds?): Q(
p
5;Q(
p
5);Q();Q(). Are they all the
same subeld of C? Prove your answer.
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