Download Homework 5 Solutions

Homework 5 Solutions
Problem 5.3.20. In the ring Z[i] of Gaussian integers let hpi be the ideal generated by a prime number. Show
that Z[i]/hpi has p2 elements, and has characteristic p.
Proof. Define a map φ : Zp [x] → Z[i]/hpi where φ(f (x)) = f (i) + hpi. One can verify that φ is a well-defined
surjective ring homomorphism with ker φ = hx2 + 1i. Thus Z[i]/hpi ∼
= Zp [x]/hx2 + 1i is a 2 dimensional vector
2
space over Zp and hence |Z[i]/hpi| = p .
Let n be the characteristic of Z[i]/hpi. Since p(1 + hpi) = p + hpi = hpi, n ≤ p. Also n(1 + hpi) = hpi so
n ∈ hpi and thus p | n. Thus n = p.
Problem 5.3.22. In the ring Z[i] of Gaussian integers show that the ideal h5 − ii is not a prime ideal.
Proof. Note that 26 = (5 − i)(5 + i) ∈ h5 − ii. To show that h5 − ii is not a prime ideal it suffices to show
that 2 6∈ h5 − ii and 13 6∈ h5 − ii. Note that if z ∈ h5 − ii then z = (5 − i)w for some w ∈ Z[i] and hence
|z|2 = |5 − i|2 |w|2 = 26|w|2 . Since z, w ∈ Z[i] then |z|2 , |w|2 ∈ Z and hence z ∈ h5 − ii implies that 26 | |z|2 .
Since 26 - |2|2 and 26 - |13|2 it follows that 2, 13 6∈ h5 − ii and therefore h5 − ii is not a prime ideal.
Problem 5.3.24. Let I be the smallest ideal of Z[i] that contains both 2 and x. Show that I is not a principal
ideal.
Proof. Suppose for a contradiction that I = ha(x)i for some a(x) ∈ Z[x]. Since 2 ∈ I = ha(x)i it follows that
a(x) | 2 in Z[x]. Also since x ∈ I = ha(x)i it follows that a(x) | x. The only polynomials dividing 2 and x in Z[x]
are ±1, i.e. a(x) = ±1. However I = h2, xi = {f (x) ∈ Z[x] | f (x) = 2p(x) + xq(x) with p(x), q(x) ∈ Z[x]} and
in particular every polynomial in h2, xi has constant term a multiple of 2. Thus ±1 6∈ h2, xi = I = ha(x)i which
contradicts a(x) = ±1.
Problem 5.4.1. Complete the proof in Lemma 5.4.3,to show that multiplication of equivalence classes in Q(D)
is well-defined.
Proof. Given [a, b], [c, d], [a0 , b0 ] and [c0 , d0 ] in Q(D) with [a, b] = [a0 , b0 ] and [c, d] = [c0 , d0 ] it’s required to show
that [ac, bd] = [a0 c0 , b0 d0 ]. Since [a, b] = [a0 , b0 ] and [c, d] = [c0 , d0 ] we then have ab0 = a0 b and cd0 = c0 d. Therefore
acb0 d0 = a0 c0 bd and hence [ac, bd] = [a0 c0 , b0 d0 ].
Problem 5.4.4. Let φ : D → Q(D) be the mapping φ(d) = [d, 1] defined in Theorem 5.4.4. Show that φ is an
isomorphism if and only if D is a field.
Proof. If φ is an isomorphism then φ is surjective so given d ∈ D\{0} there exists d0 ∈ D such that [d0 , 1] =
φ(d0 ) = [1, d], i.e. d0 d = 1. Thus D is a field.
Now suppose that D is a field. Since φ is not zero φ is injective by exercise 5.2.2. Given [d1 , d2 ] ∈ Q(D) then
−1
−1
d2 6= 0 so d−1
2 ∈ D and φ(d1 d2 ) = [d1 d2 , 1] = [d1 , d2 ], i.e. φ is surjective and hence an isomorphism.
Problem 5.4.5. In Theorem 5.4.6, verify that θ̂ is a one-to-one ring homomorphism
Proof. First we show θ̂ is a ring homomorphism. Suppose [a, b], [c, d] ∈ Q(D) then θ̂([a, b] + [c, d]) = θ̂([ad +
bc, bd]) = θ(ad + bc)θ(bd)−1 = θ(a)θ(b)−1 + θ(c)θ(d)−1 = θ̂([a, b]) + θ̂([c, d]). Also θ̂([a, b][c, d]) = θ̂([ac, bd]) =
θ(ac)θ(bd)−1 = θ(a)θ(b)−1 θ(c)θ(d)−1 = θ̂([a, b])θ̂([c, d]).
Since Q(D) is a field and θ̂ is not zero it follows from exercise 5.2.2 that θ̂ is one-to-one.
Problem 5.4.8. Let p be a prime number, and let D = {m/n | m, n ∈ Z and p - n}. Verify that D is an integral
domain and find Q(D).
Proof. As D ⊆ Q, to show that D is an integral domain it suffices to show that D is a subring of Q. If a/b, c/d ∈ D
then a/b − c/d = (ad − bc)/bd and a/b · c/d = (ac)/(bd) where p - b, d ⇒ p - bd and hence D is a subring of Q.
Since D is a subring of Q and every element of Q is a quotient of elements in D it follows from Corollary
5.4.7 that Q(D) ∼
= Q.
Problem 5.4.10. Considering Z[x] as a subring of Q[x], show that both rings have the same quotient field.
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Proof. Note that Q(Q[x]) = Q(x) = {p(x)/q(x) | p(x), q(x) ∈ Q[x]}. Also note that Z[x] is an integral domain
that is a subring of Q(x) and every element of Q(x) is a quotient p(x)/q(x) of elements in Q[x]. Moreover if d is
the product of denominators of coefficients of p(x) and q(x) then p(x)d and q(x)d have integer coefficients and
p(x)/q(x) = (p(x)d)/(q(x)d). Hence by Corollary 5.4.7 Q(Z[x]) ∼
= Q(x) ∼
= Q(Q[x]).
Problem
√ 6.1.1. Show that the following complex numbers are algebraic over Q.
(b) pn, for n ∈ Z+
√
(d) 2 + √3
(e) (−1 + 3i)/2
√
2
Proof. (b) Clearly n is a root
p of√the polynomial x − n ∈ Q[x].
(d) One can verify that
2 + 3 is a root of (x2 − 2)2 − 3 ∈ Q[x].
√
(e) Note that (−1 + 3i)/2 = e2πi/3 and thus is a root of x3 − 1 ∈ Q[x].
Problem 6.1.2. Let F be an extension field of K, and let u be a nonzero element of F that is algebraic over K.
Show that u−1 is also algebraic over K.
Proof. Since u ∈ F is algebraic over K there is a nonzero polynomial p(x) = a0 + a1 x + a2 x2 + · · · + an xn ∈ K[x]
such that p(u) = 0. Since u is nonzero it follows that an + an−1 u−1 + · · · + a0 u−n = u−n p(u) = 0. Thus u−1 is
algebraic over K.
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