Download Problem Score 1 2 3 4 or 5 Total - Mathematics

Johns Hopkins University, Department of Mathematics
110.402 Abstract Algebra - Spring 2013
Midterm Exam–Solution
Instructions: This exam has 6 pages. No calculators, books or notes allowed. You must answer the first
3 questions, and then answer one of question 4 or 5. Do not answer both. No extra points will be
rewarded. Place an “X” through the question you are not going to answer. Be sure to show all work for all
problems. No credit will be given for answers without work shown.
If you do not have enough room in the space provided you may use additional paper provided by the
instructor. Be sure to clearly label each problem and attach them to the exam.
You have 75 MINUTES.
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Problem
1
2
3
4 or 5
Total
Score
1
1. (25 points) Consider the ring of Gaussian integers Z[ i ] = {a + bi | a, b ∈ Z}
(i =
√
−1).
(a) Is 2 + 3i a prime element in Z[ i ]? (Prove it or disprove it)
(b) Compute the cardinality of the set Z[ i ]/(2 + 3i). What group is it?
(c) Find G.C.D.(1 + 3i, 5 + i) (up-to associates).
Solution: (a) N r(2 + 3i) = 22 + 32 = 13 is prime in Z, so 2 + 3i is an irreducible element in Z[ i ]. Since
Z[ i ] is a Euclidean domain, every irreducible element is a prime element. Therefore 2 + 3i is a prime
element in Z[ i ].
(b) Note that N r(2 + 3i) = 13. Consider then the ring homomorphism ϕ: Z[ i ] → Z/13Z, ϕ(a + bi) =
(a + 8b) mod 13. By applying the division algorithm to a + bi and 2 + 3i (Z[ i ] is an Euclidean domain)
we can write a + bi = q(2 + 3i) + r with N r(r) < N r(2 + 3i) for some q, r ∈ Z[ i ]. Hence every coset
a + bi + I (I = (2 + 3i)) in Z[ i ]/(2 + 3i) has a representative with norm less than 13, and since such
representatives have integer real and imaginary parts, there are finitely many distinct cosets.
The definition of ϕ is motivated by the following facts: ϕ should be a ring homomorphism, so necessarily
ϕ(1) = 1 (mod 13), also we must have ϕ(2+3i) = 2+3ϕ(i) = 0 (mod 13), with 2+3ϕ(i) = 2−10ϕ(i) =
2(1 − 5ϕ(i)) = 2(1 + 8ϕ(i)) = 0 (mod 13). Then one sees that ϕ(i) = 8 (mod 13) is the good choice.
Also ϕ is evidently surjective. We have: (2 + 3i) ⊆ Ker(ϕ), moreover Z[ i ] is a P.I.D. and since we
know from (a) that (2 + 3i) is a prime ideal it is also maximal, finally Ker(ϕ) is a maximal ideal as the
image of ϕ is the field Z/13Z. Thus Z[ i ]/Ker(ϕ) = Z[ i ]/(2 + 3i) ' Z/13Z and it has cardinality 13.
(c) Apply the division algorithm for Z[ i ] :
4 7
5+i
= − i
1 + 3i
5 5
we choose the approximate quotient 1 − i, to get
5 + i − (1 − i)(1 + 3i) = 1 − i.
Therefore
5 + i = (1 − i)(1 + 3i) + 1 − i
where N r(1 − i) = 2 < N r(1 + 3i) = 10. Now we repeat the process with 1 + 3i and 1 − i:
1 + 3i
= −1 + 2i
1−i
and so
1 + 3i = (−1 + 2i)(1 − i)
and the division algorithm ends. The algorithm tells us that 1 − i is the GCD of 5 + i and 1 + 3i (up-to
associates).
2
2. (25 points) Give a proof or disprove the following statement:
√
Z[ −5] is an Euclidean domain.
√
Solution: We turn the attention to the possibility of finding an element √
of Z[ −5] with non-unique
factorization. We search for possible candidates among elements of Z[ −5] with small norm,
the
√
norm itself providing a means to discover possible factorizations. By trying out N r(a + b −5) =
a2 + 5b2 √
for different
we soon find that 6 = 12 + 5 · 12 =√ 2 · 3. So
√ small integer values of a and b, √
6 = (1 + −5)(1 − −5) = 2 · 3. Notice that 2, 3, (1 ± −5) are irreducible elements in Z[ −5] (this
statement
√ has been proven several times in class), thus the above two different factorizations of 6 show
that Z[ −5] is not U.F.D. thus it cannot be an Euclidean domain.
3
3. (25 points) Let I be a proper ideal of a commutative ring with identity R. Prove or disprove:
R/I free R-module =⇒ I = (0)
Solution: Since I is a proper ideal, R/I has non-zero elements. Suppose R/I is a free R-module, then
R/I has a non-empty basis, and in particular it contains at least one linearly independent element
x + I, x ∈ R. Let i ∈ I. Then i(x + I) = ix + I = I since I is an ideal so ix ∈ I. I is the zero element
in R/I and since x + I is R-linearly independent we must have i = 0. Therefore I = (0).
4
4. (25 points) (ANSWER THIS QUESTION OR NUMBER 5)
Prove or disprove (X = indeterminate):
(a) Z[X]/(X 2 + 1) and Z ⊕ Z are isomorphic as Z-modules and as rings.
(b) Q[X]/(X 2 − X − 1) and Q[X]/(X 2 − 1) are isomorphic as rings and Q-vector spaces.
Solution: (a) Z[X]/(X 2 + 1) ' Z[ i ] and Z ⊕ Z are isomorphic as abelian groups (i.e. as Z-modules)
in fact ϕ : Z[ i ] → Z ⊕ Z, ϕ(a + bi) = (a, b) is a group isomorphism. On the other hand, they are not
isomorphic as rings: the former one being integral but not the latter one.
(b) Q[X]/(X 2 − X − 1) and Q[X]/(X 2 − 1) are isomorphic as Q-vector spaces (of rank 2): a Q-vector
space isomorphism is given by: 1 + (X 2 − X − 1) 7→ 1 + (X 2 − 1) and X + (X 2 − X − 1) 7→ X + (X 2 −√1).
However they are not isomorphic as rings since the former one is (isomorphic to) the field Q( 5)
((X 2 −X −1) is irreducible over Q) whereas the latter one is not an integral domain (i.e. X ±1+(X 2 −1)
are zero-divisors).
5
5. (25 points) (ANSWER THIS QUESTION OR NUMBER 4)
√
(a) Decompose 22 into irreducible elements of Q( −7).
√
(b) Does each factor in the decomposition determine a prime ideal in Z[ −7]? Why?
√
√
Solution: (a) The element 22 = 11 · 2 splits in Z[ 1+ 2 −7 ] ⊂ Q( −7) as follows:
22 = (2 +
√
−7)(2 −
√
−7)
1+
√
2
−7
1−
√
−7
2
√
√
with Nr(2 ± −7) = 11 and Nr( 1± 2 −7 ) = 2. All the factors are irreducible since they have prime
norms.
√
√
ideals in Z[ −7] since√Z[X]/(X√2 +7, 2±X) ' Z/11Z is a field. Moreover, the
(b) (2± √ −7) are prime
√
ideal ( 1+ 2 −7 ) in Z[ 1+ 2 −7 ] is prime since Z[ 1+ 2 −7 ]/( 1+ 2 −7 ) ' Z[X]/(X 2 − X + 2, X) ' Z/2Z a field.
√
√
√
Then, if ι : Z[ −7] → Z[ 1+ 2 −7 ] is the natural immersion, ι−1 (( 1+ 2 −7 )) is a prime ideal since the map
√
√
√
√
Z[ −7]/ι−1 ( 1+ 2 −7 ) → Z[ 1+ 2 −7 ]/( 1+ 2 −7 ) is an injective ring homomorphism (a similar argument
√
√
shows that ( 1− 2 −7 ) determines (by contraction) a prime ideal in Z[ −7]).
6