Download Solutions - Dartmouth Math Home

Math 31 Problem Set # 6 Solutions
3. If R is a finite integral domain, show that R is a field.
To show that R is a field, we must show that every nonzero element of R is a unit. Let r ∈ R
be nonzero. Consider the infinite sequence of ring elements r1 , r2 , r3 , . . . . Because R is finite, the
elements of this sequence cannot all be distinct. Thus there exist postive integers n > m with
rn = rm . Using the ring axioms, we have
0 = rn − rm = rk rm − rm = (rk − 1)rm ,
where we set k = n − m.
Because R is an integral domain, we conclude that either rk − 1 = 0 or rm = 0. If rm = 0, then it
follows (since R is an integral domain) that r = 0. But this is impossible as r was assumed to be
nonzero. Therefore we must have rk − 1 = 0, that is, rk = 1. This implies r · rk−1 = 1. This shows
that r is a unit R, since rk−1 is the multiplicative inverse of r.
4. Let R be a ring. If a ∈ R, prove that a cannot be both a zero divisor and a unit.
Suppose a ∈ R is a unit. Then there exists b ∈ R with ab = 1. We must show that a is not a
zero divisor. Suppose that ca = 0 for some c ∈ R. If we can show c = 0, then it follows that a is
not a zero divisor. But multiplying the equation ca = 0 on the right by b, we have (ca)b = 0b = 0,
and so
0 = (ca)b = c(ab) = c1 = c.
6. If R and S are commutative rings (with multiplicative identity), you can make R × S
into a ring by defining
(r1 , s1 ) + (r2 , s2 ) = (r1 + r2 , s1 + s2 )
and
(r1 , s1 ) · (r2 , s2 ) = (r1 · r2 , s1 · s2 ),
where r1 , r2 ∈ R and s1 , s2 ∈ S. It is straightforward to check that the ring axioms hold
for R × S.
(a) Express the unit group (R × S)× in terms of the unit groups R× and S × .
Let 1R and 1S denote the multiplicative identities of R and S. Then the multiplicative identity
of R × S is (1R , 1S ). We have
(r, s) ∈ (R × S)× ⇐⇒ (r, s) · (u, v) = (1R , 1S ) for some (u, v) ∈ R × S
⇐⇒ (ru, sv) = (1R , 1S ) for some (u, v) ∈ R × S
⇐⇒ ru = 1R for some u ∈ R and sv = 1S for some v ∈ S
⇐⇒ r ∈ R× and s ∈ S ×
⇐⇒ (r, s) ∈ R× × S × .
1
Therefore (R × S)× = R× × S × .
(b) If both R and S are integral domains, is R × S also an integral domain? Prove this
or find a counterexample.
The product R × S is never an integral domain. For example, we have (1R , 0) · (0, 1S ) =
(1R · 0, 0 · 1S ) = (0, 0), but neither (1R , 0) nor (0, 1S ) are zero. This shows that both (1R , 0)
and (0, 1S ) are zero divisors.
7. Consider the ring Z[i] = {a + bi : a, b ∈ Z}, where i is the usual 4th root of unity. We
can view Z[i] as a subring of C.
(a) If α ∈ Z[i], let α denote the complex conjugate of α. If α = a + bi, then α = a − bi,
and so Z[i] is closed under complex conjugation. Consider the map
N : Z[i] → Z[i]
by N(α) = αα.
Show that for any α ∈ Z[i], we have N(α) ∈ Z. That is, N is actually a map Z[i] → Z.
Let α = a + bi for a, b ∈ Z. Then N(α) = (a + bi)(a − bi) = a2 + b2 , an integer.
(b) Prove that N is respects multiplication. That is, show that for any α, β ∈ Z[i], we
have N(αβ) = N(α)N(β).
The nicest way to prove this is to first show that αβ = αβ, that is, complex conjugation respects
multiplication. To see this, let α = x + iy and let β = u + iv. Then
αβ = (x + iy)(u + iv) = xu − yv + i(yu + xv) = (xu−yv)−i(yu+xv) = (x−iy)(u−iv) = αβ.
With this fact in mind, we have
N(αβ) = αβαβ = αβαβ = ααββ = N(α)N(β).
(c) Prove that α ∈ Z[i]× if and only if N(α) ∈ Z× .
Suppose α ∈ Z[i]× . Then there exists β ∈ Z[i] with αβ = 1. Applying the N map to
this equality, using the fact that N respects multiplication, and using N(1) = 1, we have
N(α)N(β) = N(αβ) = N(1) = 1. But both N(α) and N(β) are integers, and their product is
the multiplicative identity 1 of Z. Therefore N(α) ∈ Z× .
Conversely, suppose N(α) ∈ Z× . By definition, there exists u ∈ Z with N(α)u = 1. In other
words, ααu = 1. We can view this as an equation in the larger ring Z[i], in which Z sits as a
subring. Then α(αu) = 1. Because α, u ∈ Z[i], so is their product. This shows that α is a unit
in Z[i], as its multiplicative inverse is αu ∈ Z[i].
(d) Find the unit group Z[i]× . Is this group cyclic?
We can use the result of part (c) as a criterion for α to belong to Z[i]× . Let α = a + bi ∈ Z[i].
By part (c), α is a unit if and only if N(α) ∈ Z× . We know N(α) = a2 + b2 , and we also have
Z× = {±1}. Therefore a2 + b2 = ±1. The only integers a and b which satisfy this equation
are a = ±1, b = 0 and a = 0, b = ±1. The corresponding values of α are α = ±1 and α = ±i.
Hence Z[i]× = {1, i, −1, −i} = θ4 . We’ve previously studied this group. It is cyclic of order 4.
2
(e) Is Z[i] an integral domain? Prove or disprove.
We know that Z[i] is a subring of C. Because C is a field, C is necessarily an integral domain.
If αβ = 0 for some α, β ∈ Z[i], we can also consider this equality in C. Since C is an integral
domain, either α = 0 or β = 0. This shows that Z[i] is an integral domain.
8. Recall that any fraction can be written in lowest terms. In other words, if a ∈ Q,
we can always write a = m/n where gcd(m, n) = 1. Consider the set
nm
o
R=
∈ Q : gcd(m, n) = 1 and n is odd .
n
That is, R is the set of all rationals which, when written in lowest terms, have an odd
denominator.
(a) Prove that R is a ring under the usual addition and multiplication of rational
numbers.
To verify that R is a ring, we can use the subring criterion. We must check the following: R is
closed under +; R contains 0; R is closed under additive inversion; R is closed under ·; and R
contains 1. Let a/b and c/d be elements of R. We can assume both b and d are odd. Then
a c
ad + bc
+ =
b d
bd
and
a c
ac
· = .
b d
bd
These two fractions (ad + bc)/bd and ac/bd may no longer be reduced, but they still have odd
denominator. After dividing out by some common divisor to put these fractions into lowest
terms, the denominators will remain odd. This shows that R is closed under + and ·. It is
clear the 0 = 0/1 ∈ R and 1 = 1/1 ∈ R. Lastly, if a/b ∈ R, then −(a/b) = (−a)/b ∈ R, which
shows that R is closed under additive inversion.
(b) Find R× .
Because R is a subring of Q, and we know how to take multiplicative inverses of nonzero
elements in Q, we just need to figure out when the multiplicative inverse is contained in R. If
a/b ∈ Q is nonzero, then (a/b)−1 = b/a. Therefore R× = {a/b ∈ R : b/a ∈ R}. Writing a/b
in reduced form, we must have that both a and b are odd. Therefore R× is the multiplicative
group of fractions whose numerators and denominators are both odd when written in lowest
terms.
9. Let R be a commutative ring (with multiplicative identity). The characteristic of
R is the smallest positive number n such that
· · + 1} = 0
|1 + ·{z
n summands
if such an n exists, and 0 otherwise. For example, Z has characteristic 0, while Z4 has
characteristic 4, and Zn has characteristic n.
3
(a) Let p be a prime number
and let 0 < k < p be an integer. Prove that p divides the
p
binomial coefficient k .
p!
We have the formula kp = k!(p−k)!
. Rewriting this, we have p! = kp k!(p − k)!. We see
that p divides the left-hand side of this equality, so p must also divide the right-hand side. If
1 ≤ k ≤ p − 1, then p divides neither k! nor (p − k)!. But because
p is prime, p must divide
p
one of the terms on the right-hand side. Therefore p divides k .
(b) Let R be a ring of characteristic p, a prime number. Prove the child’s binomial
theorem:
(x + y)p = xp + y p
for any x, y ∈ R.
By the binomial theorem,
p p−1
p p−2 2
p
(x + y) = x +
x y+
x y + ··· +
xy p−1 + y p .
1
2
p−1
p
p
Because R is a ring of characteristic p, we have pr = 0r = 0 for any ring element r. Every term
in the above sum is of the form kp xp−k y k . If 1 ≤ k ≤ p − 1, because p divides kp , this term
equals 0 in the ring R. We see that all except for the first and last terms drop away. We are
left with (x + y)p = xp + y p .
(c) Consider the Frobenius map
by F (x) = xp .
F :R→R
Prove that F is a ring homomorphism.
By part (b), we have F (x + y) = (x + y)p = xp + y p = F (x) + F (y), which shows that F
preserves addition. Because R is commutative, we have F (xy) = (xy)p = xp y p = F (x)F (y).
That is, F preserves multiplication. Thus F is a ring homomorphism.
4