Download Math 403 Assignment 1. Due Jan. 2013. Chapter 11. 1. (1.2) Show

Math 403 Assignment 1. Due Jan. 2013.
Chapter 11.
1. (1.2) Show that, for n 6= 0, cos(2π/n) is algebraic over Q.
Solution. Since e2πi/n and e−2πi/n are each algebraic (each satisfies the equation xn −1 = 0),
half their sum is algebraic too, i.e., cos(2π/n) = (1/2)(e2πi/n + e−2πi/n ) is algebraic.
√ √
2. (1.3) Let R = Q[ 2, 3] ⊂ R.
√ √
√
√
(a). Is Q[ 2, 3] = Q[ 2 + 3]?
√
√
√
√
Solution.√ Yes.√To begin, we have the fact that
is a field. Hence, ( √2 + √3)−1
√ Q[√ 2 +
√ 3] √
lies√in Q[√ 2 + 3]. Next,
that ( 2 + 3)( 2 − 3) = −1, and so, 2 − 3 =
√we observe
√
−( 2 + 3)−1 lies in Q[ 2 + 3].
√
√
√
√
√
√
√
√
3] lie in Q[ 2 + 3], so do their sum 2 2 and difference 2 3.
Since 2 +
√ 3√and Q[ √2 + √
Hence, Q[ 2, 3] = Q[ 2 + 3].
√ √
√
√
(b). Is Z[ 2, 3] = Z[ 2 + 3]?
√ √
√ √ √ √
√ √
Solution. Q[ 2 + 3] is a Q-vector space with basis {1, 2 + 3, ( 2 + 3)2 , ( 2 + 3)3 };
hence, each of the elements of the ring has a unique expression as a linear combination of
the basis.
√ 2
√
√
√
√
√
= 5√+ 2 6, we see that 6 = −5/2 + (1/2)( 2 + 3)2 , which is not
Since ( 2 + 3)√
an element of√Z[ 2 + 3] since the√coefficients
are not all√integers
(and
expression is
√
√
√ the √
unique). But 6 is an element of Z[ 2, 3]. Therefore, Z[ 2, 3] 6= Z[ 2 + 3].
3. (1.6) Is {a/b ∈ Q|(b, 3) = 1} a subring of Q?
Solution. Yes, a/b + c/d = (ad + bc)/bd and (a/b)(c/d) = ac/bd, where (bd, 3) = 1 when
(b, 3) = 1 and (d, 3) = 1 since 3 is prime.
4. (1.8) Determine the units in the following rings: Z/(12)Z, Z/8Z, and Z/nZ.
Solution. In a ring R, if a and b are nonzero elements of R such that ab = 0, then a and b
are nonunits. To see that, if a had a multiplicative inverse a−1 , then b = a−1 (ab) = a−1 0 = 0,
1
contrary to the hypothesis that b 6= 0. In Z/nZ, if (m, n) 6= 1, then m + nZ is a nonunit
since
(m + nZ)(n/(m, n) + nZ) = 0 + nZ.
m + nZ is a unit if (m, n) = 1, since by arithmetic, if (m, n) = 1, there are integers a and b
such that am + bn = 1. Hence,
1 + nZ = am + bn + nZ = am + nZ = (a + nZ)(m + nZ.
Units of Z/(12)Z: 1 + (12)Z, 5 + (12)Z, 7 + (12)Z, 11 + (12)Z.
Units of Z/8Z: 1 + 8Z, 3 + 8Z, 5 + 8Z, 7 + 8Z.
Units of Z/nZ: {m + nZ|0 ≤ m < n, (m, n) = 1}.
5. (1.9) Let R, +, · satisfy all the ring axioms except possibly the commutative rule for
+. Use the distributative rule to prove that + is in fact commutative. (Thus, we want a
distributative rule, we must base a ring on an abelian group R, +.)
Solution. We want to show that a + b = b + a, i.e., that a + b − (b + a) = 0.
First, we show that −c = (−1)c for each element c ∈ R. In fact, c + (−1)c = 1c + (−1)c,
since 1c = c. Hence,
c + (−1)c = 1c + (−1)c = (1 + (−1))c = 0c = 0,
using the distributative law. Hence,−c = (−1)c. Therefore, −(b + a) = (−1)(b + a) =
(−1)b + (−1)a = −b − a, using the distributative law.
Therefore, we have the comlutation a + b − (b + a) = a + b − b − a = 0, as we wished to show.
6. (2.1) For which positive integers n does x2 + x + 1 divide x4 + 3x3 + x2 + 7x + 5 in the
polynomial ring Z/nZ[x]?
Solution. For the division to take place, we need to solve the equation
(x2 + x + 1)(x2 + ax + b) = x4 + 3x3 + x2 + 7x + 5,
for a and b in Z/nZ.
Multiplying out the left-hand side, we get
x4 + (a + 1)x3 + (b + 1 + a)x2 + (a + b)x + b = x4 + 3x3 + x2 + 7x + 5,
2
in Z/nZ[x]. Comparing coefficients, we have a + 1 = 3, b + 1 + a = 1, a + b = 7, and b = 5 in
Z/nZ. By those equations, we see that a = 2 and b = 5; hence, by the equation b+ 1 + a = 1,
we see that 8 = 1, i.e., 7 = 0 in Z/nZ, i.e., n = 7.
7. (2.2) Let F be a field and let F [[t]] be the ring {a0 + a1 t + a2 t2 + · · · aj tj + · · · =
j
Σ∞
j=0 aj t |aj ∈ F ; j = 0, 1, 2, ...} where addition is gotten from the addition of polynomials and
multiplication is gotten from the distributative rule and the multiplication of polynomials.
Find the units in this ring.
j
Solution. For Σ∞
j=0 aj t to be a unit, we need to solve the equation
j
∞
j
(Σ∞
j=0 aj t )(Σj=0 bj t ) = 1,
for the coefficientsbj . The product equals Σj=0 (aj b0 + aj−1 b1 + · · · a1 bj−1 + a0 bj )tj . For that
product to equal 1, we need a0 b0 = 1 and aj b0 + aj−1 b1 + · · · a1 bj−1 + a0 bj = 0. So long as
a0 6= 0, we can solve successively for b0 , b1 ,...,bj ,.... Therefore, the units are the elements
j
Σ∞
j=0 aj t with a0 6= 0.
3