tut5

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Department of Mathematics National University of Singapore 2014/15 Sem II MA3201 Algebra II Tutorial 5 (1) Let R be a subring of an integral domain S , and suppose R is a PID. Let a 2 R be a greatest common divisor of r 1 and r 2 in R (r 1 ,r 2 2 R, not both zero). Prove that if b 2 S is a common divisor of r 1 and r 2 in S , then b | a in S . Show by a counter-example that the conclusion is invalid if R is a UFD, but not a PID. (2) Let R be a commutative ring with 1, and let R[X, Y ] denote the set of all polynomials in two variables X and Y with coecients in R. (a) Prove that R[X, Y ]= S [Y ] where S = R[X ]. (b) Suppose R is a UFD. Show that R[X, Y ] is a UFD. (c) If R is a field, is R[X, Y ] a PID/Euclidean domain? (3) Factor x 3 - y 3 into irreducibles in Q[x, y] and prove that each of the factors is irre- ducible. (4) Let F be a field. Prove that if f (x) 2 F [x] and f (a) = 0 for some a 2 F , then f (x) is reducible if deg(f ) > 1. (5) Factorize x 3 +3x 2 + 3 in Z 5 [x] and x 4 - x 2 + 1 in Z 7 [x] into product of irreducibles. (6) Prove that the following polynomials in Z[X ] are irreducible in Q[X ]. (a) pX 2 + aX + q, where p and q are prime integers, and a 6= ±(p + q), ±(1 + pq). (b) aX 2 + bX + c, where a, b and c are odd integers. (c) X 4 - 5X 2 + X + 1. (7) Let R be a UFD, and let f (X ) 2 R[X ] be a primitive polynomial. Prove that if g(X ) | f (X ) in R[X ], then g(X ) is primitive. (8) Let R be a ring. Prove that if R[X ] is a UFD, then so is R. (9) Let R be a UFD, with field of fraction F , and let f (X )= a 0 +a 1 X +··· +a n X n 2 R[X ] with a n 6= 0. Let x 2 F be a root of f (X ). Show that x = r s for some r, s 2 R, s 6=0 satisfying r | a 0 and s | a n in R. (10) Prove that the following polynomials are irreducible in R[X ]: (a) f (X )= P p-1 i=0 (-1) i X i where p is a prime integer, R = Z. (b) f (X )= X 6 + X 3 + 1, R = Q. (Hint: Try Y = X - 1.) (c) f (X )= X 3 - 6X 2 +4iX + (1 + 3i), R = Z[i]. 1

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Page 1: tut5

Department of Mathematics

National University of Singapore

2014/15 Sem II MA3201 Algebra II Tutorial 5

(1) Let R be a subring of an integral domain S, and suppose R is a PID. Let a 2 R

be a greatest common divisor of r1 and r2 in R (r1, r2 2 R, not both zero). Prove

that if b 2 S is a common divisor of r1 and r2 in S, then b | a in S. Show by a

counter-example that the conclusion is invalid if R is a UFD, but not a PID.

(2) Let R be a commutative ring with 1, and let R[X, Y ] denote the set of all polynomials

in two variables X and Y with coe�cients in R.

(a) Prove that R[X, Y ] = S[Y ] where S = R[X].

(b) Suppose R is a UFD. Show that R[X, Y ] is a UFD.

(c) If R is a field, is R[X, Y ] a PID/Euclidean domain?

(3) Factor x3 � y

3 into irreducibles in Q[x, y] and prove that each of the factors is irre-

ducible.

(4) Let F be a field. Prove that if f(x) 2 F [x] and f(a) = 0 for some a 2 F , then f(x)

is reducible if deg(f) > 1.

(5) Factorize x

3 + 3x2 + 3 in Z5[x] and x

4 � x

2 + 1 in Z7[x] into product of irreducibles.

(6) Prove that the following polynomials in Z[X] are irreducible in Q[X].

(a) pX

2 + aX + q, where p and q are prime integers, and a 6= ±(p+ q),±(1 + pq).

(b) aX

2 + bX + c, where a, b and c are odd integers.

(c) X

4 � 5X2 +X + 1.

(7) Let R be a UFD, and let f(X) 2 R[X] be a primitive polynomial. Prove that if

g(X) | f(X) in R[X], then g(X) is primitive.

(8) Let R be a ring. Prove that if R[X] is a UFD, then so is R.

(9) Let R be a UFD, with field of fraction F , and let f(X) = a0+a1X+· · ·+anXn 2 R[X]

with an 6= 0. Let x 2 F be a root of f(X). Show that x = rs for some r, s 2 R, s 6= 0

satisfying r | a0 and s | an in R.

(10) Prove that the following polynomials are irreducible in R[X]:

(a) f(X) =Pp�1

i=0 (�1)iX i where p is a prime integer, R = Z.(b) f(X) = X

6 +X

3 + 1, R = Q. (Hint: Try Y = X � 1.)

(c) f(X) = X

3 � 6X2 + 4iX + (1 + 3i), R = Z[i].1

Yong Xien
z[x] is a ufd that is not a pid2, x z[x] z[x/2]
Yong Xien
Yong Xien
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4
Yong Xien
6(c) is there any better way than consider cases?
Yong Xien
why need PID
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Challenging Problem

Let R be a commutative ring. Prove that R has ACCP if and only if every non-empty

collection of principal ideals of R has a maximal element.

Prove further that if R is an integral domain and has ACCP, then R[X] has ACCP.

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