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MATH 113 - HOMEWORK 6
SOLUTIONS
18.12. Let Q[
√
2] = {a + b√2 | a, b ∈ Q}. Note that Q[√2] ⊂ R, and the operations of
+ and · on Q[√2] are the usual + and · of real numbers. Not only is Q[√2] closed under +
and ·, but Q[√2] is a field (a subfield of R).
• Q[√2] is closed under addition. (a+ b√2) + (c+ d√2) = (a+ c) + (b+ d)√2.
• Q[√2] is closed under multiplication. (a + b√2)(c + d√2) = ac + ad√2 + bc√2 +
bd
√
2
2
= (ac+ 2bd) + (ad+ bc)
√
2.
• Addition is associative and commutative on Q[√2], since it is associative and com-
mutative in R.
• Identity for addition: 0 = 0 + 0√2 ∈ Q[√2].
• Inverses for addition: The inverse of a+ b√2 is −(a+ b√2) = −a+−b√2 ∈ Q[√2].
• Multiplication is associative and commutative on Q[√2], since it is associative and
commutative in R.
• Distributivity holds in Q[√2], since it holds in R.
• Identity for multiplication: 1 = 1 + 0√2 ∈ Q[√2].
• Inverses for multiplication: This is the interesting one. Given a, b ∈ Q with a+b√2 6=
0, (either a 6= 0 or b 6= 0), we need to find c, d ∈ Q such that (a+ b√2)(c+d√2) = 1.
Solution 1: (a+b
√
2)(c+d
√
2) = (ac+2bd)+(ad+bc)
√
2, so we need ac+2bd = 1
and ad+ bc = 0. Let’s do some linear algebra.(
a 2b
b a
)(
c
d
)
=
(
1
0
)
The matrix on the left has determinant a2 − 2b2, so it’s invertible unless a2 = 2b2.
Fortunately, if a2 = 2b2, then a = ±√2b, but a, b ∈ Q, so the only way this could
happen is if a = b = 0.
Multiplying by the inverse matrix
1
a2 − 2b2
(
a −2b
−b a
)
,
we have(
c
d
)
=
1
a2 − 2b2
(
a −2b
−b a
)(
1
0
)
=
1
a2 − 2b2
(
a
−b
)
,
so the inverse of a+ b
√
2 is
a
a2 − 2b2 −
b
a2 − 2b2
√
2 ∈ Q[
√
2].
Solution 2: We can invert a+ b
√
2 in R:
1
a+ b
√
2
. Rationalizing the denominator,
1
a+ b
√
2
=
a− b√2
(a+ b
√
2)(a− b√2) =
a− b√2
a2 − 2b2 =
a
a2 − 2b2 −
b
a2 − 2b2
√
2 ∈ Q[
√
2].
18.18. (a, b, c) is a unit in the product ring if and only if each of a, b, and c are units in the
factor rings. Indeed, if (a, b, c) has an inverse (d, e, f), then (a, b, c)(d, e, f) = (ad, be, cf) =
(1, 1, 1), and (d, e, f)(a, b, c) = (da, eb, fc) = (1, 1, 1), so a−1 = d, b−1 = e, and c−1 = f .
Conversely, if a, b, and c are units, then (a−1, b−1, c−1) is the inverse of (a, b, c) in the
product ring: (a, b, c)(a−1, b−1, c−1) = (aa−1, bb−1, cc−1) = (1, 1, 1) = (a−1a, b−1b, c−1c) =
(a−1, b−1, c−1)(a, b, c).
So U(Z×Q× Z) = U(Z)× U(Q)× U(Z) = {(a, b, c) | a = ±1, c = ±1, b 6= 0}.
18.38. Suppose R is commutative. Then for all a, b ∈ R, (a+b)(a−b) = a(a−b)+b(a−b) =
aa− ab+ ba− bb = a2 − ab+ ab− b2 = a2 − b2.
Conversely, suppose that for all a, b ∈ R, a2 − b2 = (a + b)(a − b). Then a2 − b2 =
(a + b)(a − b) = a2 − ab + ba − b2 by the computation above. Subtracting a2 from both
sides and adding b2 to both sides, we have 0 = −ab+ ba. Adding ab to both sides, we have
ab = ba. So R is commutative.
18.44.a. Let a, b ∈ R be idempotent. Then (ab)2 = abab = aabb = a2b2 = ab. So ab is
idempotent.
b. (a, b) is idempotent in the product ring if and only if each of a and b are idempotent in
the factor rings. Indeed, (a, b) is idempotent if and only if (a, b) = (a, b)2 = (a2, b2), if and
only if a2 = a and b2 = b.
In Z6, by direct inspection, the idempotent elements are 0, 1, and 3.
In Z12, the idempotent elements are 0, 1, 4, and 9.
So the idempotent elements of Z6 × Z12 are (0, 0), (1, 0), (3, 0), (0, 1), (1, 1), (3, 1), (0, 4),
(1, 4), (3, 4), (0, 9), (1, 9), (3, 9),
19.7. Z3 × 3Z has characteristic 0. Indeed, n · (0, 3) = (0, 3n) 6= (0, 0) for any n ∈ Z+. So
no positive integer n kills every element of the ring.
19.8. Z3×Z3 has characteristic 3. For all (n,m) ∈ Z3×Z3, 3 · (n,m) = (3n, 3m) = (0, 0).
19.29. Let D be an integral domain, and suppose for contradiction that D has of charac-
teristic c, with c a positive composite number. We can write c = nm with n,m < c. Now
(n·1) 6= 0, since n < char(D), and similarly (m·1) 6= 0. But (n·1)(m·1) = (1+· · ·+1)(m·1) =
(1(m · 1) + · · ·+ 1(m · 1)) = n · (m · 1) = nm · 1 = c · 1 = 0. So (n · 1) and (m · 1) are a pair
of zero divisors in D, contradiction.
20.10. The numbers between 1 and 23 relatively prime to 24 are 1, 5, 7, 11, 13, 17, 19, 23,
so φ(24) = 8.
Now 7 is relatively prime to 24, so by Euler’s theorem, 71000 ≡ (78)125 ≡ 1125 ≡ 1 (mod 24).
20.29. To show that n37 − n is divisible by 383838, we’ll show that it is divisible by each
of the primes 37, 19, 13, 7, 3, and 2. This suffices, since 383838 = (37)(19)(13)(7)(3)(2).
Note that p | n37 − n if and only if n37 − n ≡ 0 (mod p) if and only if n37 ≡ n (mod p).
37: If 37 | n, then automatically 37 | n37 − n. If not, then by Fermat’s Little Theorem,
n37 ≡ n (mod 37), so by the equivalence noted above, 37 | n37 − n.
19: If 19 | n, then automatically 19 | n37 − n. If not, then by Fermat’s Little Theorem,
n37 ≡ (n19)(n18) ≡ (n)(1) ≡ n (mod 19), as desired.
13: If 13 | n, then automatically 13 | n37 − n. If not, then by Fermat’s Little Theorem,
n37 ≡ (n12)3n ≡ 13n ≡ n (mod 13), as desired.
7: If 7 | n, then automatically 7 | n37 − n. If not, then by Fermat’s Little Theorem,
n37 ≡ (n6)6n ≡ 16n ≡ n (mod 7), as desired.
3: If 3 | n, then automatically 3 | n37 − n. If not, then by Fermat’s Little Theorem,
n37 ≡ (n2)18n ≡ 118n ≡ n (mod 3), as desired.
2: If 2 | n, then automatically 2 | n37 − n. If not, then by Fermat’s Little Theorem, n37 ≡
(n1)36n ≡ 136n ≡ n (mod 2), as desired. Alternatively, n is odd, so n37 ≡ 1 ≡ n (mod 2).
Fraleigh, Section 21: Prove that the field of fractions of an integral domain really is a field.
To do this, refer to “Step 3” of the construction of the field of fractions in Section 21 of
Fraleigh, page 193. Parts 1 and 9 of Step 3 are done for you. You should read the proof of
Part 1, complete Parts 2-8, and finish by reading Part 9.
You can copy down or work out for yourself the proofs of Parts 1 and 9 if you want to
have the whole proof in one place, but you don’t have to.
Solution:
(1) Addition in F is commutative. Let [(a, b)], [(c, d)] ∈ F .
[(a, b)] + [(c, d)] = [(ad+ bc, bd)]
= [(cb+ da, db)]
= [(c, d)] + [(a, b)].
(2) Addition is associative. Let [(a, b)], [(c, d)], [(e, f)] ∈ F .
([(a, b)] + [(c, d)]) + [(e, f)] = [(ad+ bc, bd)] + [(e, f)]
= [((ad+ bc)f + (bd)e, (bd)f)]
= [(adf + bcf + bde, bdf)]
= [(a(df) + b(cf + de), b(df))]
= [(a, b)] + [(cf + de, df)]
= [(a, b)] + ([(c, d)] + [(e, f)]) .
(3) [(0, 1)] is an identity element for addition in F . Let [(a, b)] ∈ F .
[(a, b)] + [(0, 1)] = [(a1 + b0, b1)] = [(a, b)].
We only need to check the sum in this order, since by (1) addition is commutative.
(4) [(−a, b)] is an additive inverse for [(a, b)] in F .
[(−a, b)] + [(a, b)] = [((−a)b+ ba, b2)] = [(−ab+ ab, b2)] = [(0, b2)]
Now I claim that [(0, b2)] = [(0, 1)]. Indeed, 0 · 1 = b2 · 0 = 0, so (0, b2) ∼ (0, 1).
We only need to check the sum in this order, since by (1) addition is commutative.
(5) Multiplication in F is associative. Let [(a, b)], [(c, d)], [(e, f)] ∈ F .
([(a, b)][(c, d)]) [(e, f)] = [(ac, bd)][(e, f)]
= [((ac)e, (bd)f)]
= [(a(ce), b(df))]
= [(a, b)][(ce, df)]
= [(a, b)] ([(c, d)][(e, f)]) .
(6) Multiplication in F is commutative. Let [(a, b)], [(c, d)] ∈ F .
[(a, b)][(c, d)] = [(ac, bd)]
= [(ca, db)]
= [(c, d)][(a, b)].
(7) The distributive laws hold in F . Let [(a, b)], [(c, d)], [(e, f)] ∈ F .
[(a, b)] ([(c, d)] + [(e, f)]) = [(a, b)][(cf + de, df)]
= [(a(cf + de), b(df))]
= [(acf + ade, bdf)].
On the other hand,
[(a, b)][(c, d)] + [(a, b)][(e, f)] = [(ac, bd)] + [(ae, bf)]
= [((ac)(bf) + (bd)(ae), (bd)(bf))]
= [(abcf + abde, b2df)].
Now I claim that [(acf+ade, bdf)] = [(abcf+abde, b2df)]. Indeed, (acf+ade)(b2df) =
(bdf)b(acf + ade) = (bdf)(abcf + abde), so (acf + ade, bdf) ∼ (abcf +abde, b2df).
We only need to check distributivity on the left, since by (6) multiplication is
commutative. (If distributivity on the left holds and · is commutative, then (y+z)x =
x(y + z) = xy + xz = yx+ zx, which is distributivity on the right.)
(8) [(1, 1)] is a multiplicative identity element in F . Let [(a, b)] ∈ F .
[(a, b)][(1, 1)] = [(a1, b1)] = [(a, b)].
We only need to check the product in this order, since by (6) multiplication is
commutative.
(9) If [(a, b)] ∈ F is not the additive identity element, then a 6= 0 in D and [(b, a)] is a
multiplicative inverse for [(a, b)].
For the first statement, we’ll prove the contrapositive: If a = 0 in D, then [(a, b)] =
[(0, 1)]. Indeed, if a = 0, then a1 = b0 = 0, so (a, b) ∼ (0, 1).
Now we know that if [(a, b)] 6= [(0, 1)], then a 6= 0, so [(b, a)] is an element of F .
Let’s check that it is [(a, b)]−1. We have [(a, b)][(b, a)] = [(ab, ba)] = [(ab, ab)]. The
claim is that [(ab, ab)] = [(1, 1)]. Indeed, ab1 = ab1, so (ab, ab) ∼ (1, 1).
We only need to check the product in this order, since by (6) multiplication is
commutative.
