Math 312 Assignment 4 Answers
1. Find all ring-homomorphisms from Z ⊕ Z into Z.
(1, 0)2
[4]
(φ(1, 0))2
Answer: Let φ be a homomorphism. Since
= (1, 0), then
φ(1, 0) = 0 or 1. Similarly φ(0, 1) = 0 or 1. There are four cases:
(i) φ(1, 0) = φ(0, 1) = 0;
= φ(1, 0). Therefore
(ii) φ(1, 0) = 1, φ(0, 1) = 0;
(iii) φ(1, 0) = 0, φ(0, 1) = 1;
(iv) φ(1, 0) = φ(0, 1) = 1.
This leads to (i) φ1 (m, n) = 0, (ii) φ2 (m, n) = m, (iii) φ3 (m, n) = n. It is straightforward to
verify that each is indeed a ring-homomorphism.
The fourth case in not possible. You catch early as it implies that the idempotent (1, 1) is
mapped to 2 which is not idempotent, or you can catch it later as it implies that φ(m, n) = m+n
which you can verify that it is not a ring-homomorphism.
2. Find the GCD g of 8 − 6i and 5 − 15i in the ring of Gaussian integers by using the Euclidean [4]
Algorithm and express g in the form (8 − 6i)z + (5 − 15i)w for z, w ∈ Z[i].
ANSWER: 5−15i
8−6i = 1.3 − 0.9i, so for a quotient you can take 1 − i (note 2 − i works too, but
not standard). So (5 − 15i) = (8 − 6i)(−1 + i) + (3 − i).
Next: 8 − 6i = (3 − i)(3 − i). No Remainder.
So the GCD is 3 − i (or any of its 4 associates) and 3 − i = (8 − 6i)(−1 + i) + (5 − 15i).
3. (a) If I is a nonzero proper ideal in Z[i], prove that Z[i]/I is a finite ring.
[3]
ANSWER: Every ideal in Z[i] is principal, so I =< a + ib >. Every coset in the factor ring
equals z + I for z ∈ Z[i], but by the division algorithm, z = (a + bi)(q1 + iq2 ) + (r + is) for
integers q1 , q2 , r, s with r2 + s2 < a2 + b2 . So z + I = ri s + I. There are only finitely many
values of r, s that satisfy the above inequality, so the number of cosets is finite.
(b) Find the cardinality and the characteristic of the field Z[i]/ < 1 + 2i >.
[4]
ANSWER If I is the ideal < 1 − 2i >, then 5(a + ib + I) = (1 − 2i)(1 + 2i)(a + bi) + I = 0 + I.
So the characteristic divides 5, so it is 5.
Z[i]/ < 1 + 2i >= {r + is + I : r2 + s2 < 12 + 22 = 5}. There are 13 remainders but only
five distinct cosets since for example 1 + I = −2i + I = −1 + i + I. The following is a possible
listing of the cosets: −2 + 1, −1 + I, 0 + I, 1 + I, 2 + I. The field has 5 elements.
4. Let a is a zero of f (x) = x3 + x + 1 in an extension field K of Z2 .
(a) Prove that a2 is also a zero of the same polynomial.
ANSWER: Since
a3
[3]
+ a + 1 = 0,
a3 = a + 1,
a4 = a2 + a,
a5 = a3 + a2 = a2 + a + 1,
a6 = a3 + a2 + a = a2 + 1,
f (a2 ) = a6 + a2 + 1 = (a2 + 1) + a2 + 1 = 0.
A shorter proof: (x3 + x + 1)2 = x6 + x2 + 1. Now evaluate at a.
(b) Factor f (x) into a product of linear factors over K. (Short Answer.)
f (x) = (x − a)(x −
a2 )(x
−a−
a2 )
1
[2]
p
√ 5. (a) Find a polynomial p(x) ∈ Q[x] such that Q[x]/ < p(x) > is isomorphic to Q
1 + 5 . [4]
p
√
√
ANSWER: If a = 1 + 5, then a2 = 1+ 5 and (a2 −1)2 = 5, so a is a zero of the polynomial
p(x) = x4 − 2x2 − 4. To prove minimality, it suffices to prove that the polynomial is irreducible
over Q or equivalently over Z.
First check for linear factors. If p(x) = (x − m)g(x), then m|4 and m is a zero of p(x). By
straightforward calculation, p(m) 6= 0 for m ∈ {±10 ± 2, ±4}.
Next consider p(x) = (x2 + ax + b)(x2 + cx + d for integers a, b, c, d. Comparing coefficients,
you get
a + c = 0, ac + b + d = −2, ad + bc = 0, bd = −4
So c = −a. If a 6= 0, then ad + bc = 0 gives b = d and bd = −4 implies that b2 = −4 which
is impossible. If a = 0, then c = 0 and the equations above reduce to b + d = −2, bd = −4
which imply that b2 − 2b − 4 = 0 and this is easily seen to have integer solution b. This ends
the proof of irreducibility and hence minimality.
NOTE: It may be temptingpto think that the form (two nested square roots) will always have
√
degree 4, but this is false. 6 + 2 5 al algebraic of degree 2, not 4.
p
√ √
1 + 5 over each of Q and Q( 5). NOT MARKED
(b) Write down a basis for the field Q
p
√ √
ANSWER: Let K = Q
1 + 5 . Using Theorem 21.5 for Q ⊂ QQ( 5) ⊂ K, you get
√
√
[K : Q( 5)] = 2 since [Q( 5) : Q] = 2 and [K : Q] = 4.
p
√
√
√
√
A basis for K over Q( 5) is {1, α} where α = 1 + 5 and for Q( 5) over Q is {1, 5}.
You√get a basis
√ for K over Q by “multiplying” these two bases as in Theorem 21.5. You get
{1, 5, α, α 5}
Another basis is {1, α, α2 , α3 }.
√ √
6. Find [Q( 3 2, 4 2) : Q]
√
√ √
√
ANSWER: Let E = Q( 3 2) and K = Q( 3 2, 4 2) = E( 4 2) and consider Q ⊂ E ⊂ K.
[4]
So [K : Q] = [K : E][E : Q] = 3[K : E]. At this point we don’t know that [K : E] = 4. The
adjoined element is a zero of x4 − 2 and we don’t yet know that this polynomial is irreducible
of E. But we can conclude that [K : E] ≤ 4. Therfore [K : Q] ≤ 12 and is divisible by 3.
√
A similar calculation, using Q( 3 2) as the intermediae field instead of E implies that [K : Q]i
is divisible by 4. So [K : Q] = 12
√
Another approcah starts by provibg that K = Q(( 12 2).
√
7. prove that Z[ −6] is not a unique factorization domain.
[3]
ANSWER There are many many ways to do this. One is:
√
√
10 = (2)(5) = (2 + −6)(2 − −6). Each of the four factors can be shown to be irreducible.
√
√
√
√
For 2 + −6 assume that you have a factorization 2 + −6 = (a + b −6)(c + d −6) for
integers a, b, c, d. Multiply the latest equation by its complex adjoint (or use the so-called
norm of the textbook) to get that 4 + 6 = (a2 + 6b2 )(c2 + cd2 ). There are no integers a, b such
that a2 + 6b2 = 2√or 5. So a2 + 6b2 = 1 or 10. If a2 + 6b2 = 1, then b = 0 and a = ±1, and
2
2
2
2
the factor
√ (a + b −6) is a unit.
√ Similarly, if a + 6b = 10 then c + 6d = 1 and the factor
(c + c −6) is a unit. So 2 + −6 does not factor nontrivially.
Other factors are done in exactly the same way.
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