# Algebraic Structures, Fall 2014 Homework 10 Solutions Clinton Conley

```Algebraic Structures, Fall 2014
Homework 10 Solutions
Clinton Conley
Problem 1 (Herstein 3.2.4). If every x ∈ R satisfies x2 = x, prove that R
must be commutative.
We first show that every element is its own additive inverse (or, phrased
another way, that the ring R has characteristic 2).
Claim. For all x ∈ R, x + x = 0.
Proof of the claim. We compute x + x = (x + x)2 = x(x + x) + x(x + x) =
x2 + x2 + x2 + x2 = x + x + x + x. Then adding −(x + x) to both sides yields
0 = x + x as desired.
Now we fix x, y ∈ R with the goal of showing xy = yx. We first compute
x + y = (x + y)2 = x(x + y) + y(x + y) = x2 + xy + yx + y 2 = x + xy + yx + y.
Then adding −(x + y) to both sides yields 0 = xy + yx. Finally, we add xy to
both sides, yielding xy = xy + xy + yx = yx since by the claim xy + xy = 0.
Problem 2 (Herstein 3.4.19). Let R be a ring in which x3 = x for every
x ∈ R. Prove that R is a commutative ring.
We’ll try to break this down into manageable claims. Let’s also adopt for
n ∈ N and x ∈ R the notation nx = x
| +x+
{z· · · + x}.
n times
Claim. For all x ∈ R, 6x = 0.
Proof of the claim. We calculate (x + x)3 in two different ways. First, we
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3
3
3
have (x + x)3 = x(x + x)2 + x(x + x)2 = · · · = x
| +x +
{z · · · + x} = 8x = 8x,
8 times
since x3 = x. On the other hand, (x + x)3 = x + x. So 8x = 2x, and
subtracting 2x from both sides yields 6x = 0.
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Claim. For all x ∈ R, 3x2 + 3x = 0.
Proof of the claim. This time we expand (x2 +x)3 in two different ways. The
long way yields (x2 + x)3 = · · · = x6 + 3x5 + 3x4 + x3 = x4 + 3x3 + 3x2 + x =
x2 + 3x + 3x2 + x = 4x2 + 4x. The short way yields (x2 + x)3 = x2 + x. So
4x2 + 4x = x2 + x, so subtracting (x2 + x) from both sides yields 3x2 + 3x = 0
as desired.
Claim. For all x, y ∈ R, 3xy + 3yx = 0.
Proof of the claim. Now we expand out 3(x + y)2 + 3(x + y), which by the
previous claim we know equals 0. We then compute 3(x + y)2 + 3(x + y) =
3(x2 +xy+yx+y 2 )+3(x+y) = 3x2 +3x+3y 2 +3y+3xy+3yx = 0+0+3xy+3yx,
using again the previous claim to conclude that 3x2 + 3x = 3y 2 + 3y = 0. All
in all, we have shown 3xy + 3yx = 0 as desired.
Claim. For all x, y ∈ R, 2xy = 2yx.
Proof of the claim. We expand out (x + y)3 − (x + y), which by hypothesis
we know equals 0. This yields
(x + y)3 − (x + y) = x3 + x2 y + xyx + yx2 + xy 2 + yxy + y 2 x + y 3 − x − y
= x2 y + xyx + yx2 + xy 2 + yxy + y 2 x
= 0.
An analogous expansion of (x − y)3 − (x − y) yields
(x − y)3 − (x − y) = x3 − x2 y − xyx − yx2 + xy 2 + yxy + y 2 x − y 3 − x + y
= −x2 y − xyx − yx2 + xy 2 + yxy + y 2 x
= 0.
Summing the results of these two calculations gives 2xy 2 + 2yxy + 2y 2 x = 0.
Multiplying this last equation on the left by y gives 2yxy 2 + 2y 2 xy + 2y 3 x =
y0 = 0. Multiplying it on the right by y gives 2xy 3 + 2yxy 2 + 2y 2 xy = 0.
Finally, subtracting the first equation from the second yields
0 = 2xy 3 + 2yxy 2 + 2y 2 xy − (2yxy 2 + 2y 2 xy + 2y 3 x)
= 2xy 3 − 2y 3 x
= 2xy − 2yx
as desired. Phew!
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Claim. For all x, y ∈ R, xy = yx.
Proof. We start with the equation 3xy + 3yx = 0. This can be rewritten as
xy + 2xy + 3yx = 0. Using the previous claim, 2xy = 2yx, so we can rewrite
this equation as xy + 5yx = 0. Next, we add yx to both sides, obtaining
xy + 6yx = yx. Finally, recall by the first claim that 6yx = 0, so we have
xy = yx as desired!
Problem 3 (Herstein 3.4.20). If R is a ring with unit element 1 and ϕ is a
homomorphism of R onto R0 , prove that ϕ(1) is the unit element of R0 .
I’m going to relabel the ring R0 to S to make this easier to read. Fix
a surjective homomorphism ϕ : R → S and some y ∈ S. Our goal is to
show that ϕ(1R )y = yϕ(1R ) = y, since then ϕ(1R ) = 1S by definition of the
multiplicative identity.
Towards that end, by surjectivity of ϕ, fix x ∈ R with ϕ(x) = y. Then
y = ϕ(x) = ϕ(1R x) = ϕ(1R )ϕ(x) = ϕ(1R )y. Analogously, y = ϕ(x1R ) =
yϕ(1R ), and we’ve shown what we want.
Problem 4 (Herstein 3.4.21*). If R is a ring with unit element 1R and ϕ
is a homomorphism of R into an integral domain R0 such that ker(ϕ) 6= R,
prove that ϕ(1) is the unit element of R0 .
Let’s again rename the ring R0 to S. Since ker(ϕ) 6= R, we can fix
some a ∈ R with ϕ(a) 6= 0. Our goal is to show that for all x ∈ S we have
ϕ(1R )x = x, which is what it means for ϕ(1R ) to be the multiplicative identity
element of S. Note that since S is an ID multiplication is commutative by
assumption, so we don’t need to waste time also showing that xϕ(1R ) = x.
Towards that end, fix x ∈ S. Note that ϕ(a)ϕ(1R )x = ϕ(a1R )x = ϕ(a)x.
We can pull everything over to one side, yielding ϕ(a)ϕ(1R )x − ϕ(a)x = 0,
or ϕ(a)(ϕ(1R )x − x) = 0. Since S is an ID and ϕ(a) 6= 0 by hypothesis, we
conclude that ϕ(1R )x − x = 0 and hence ϕ(1R )x = x as desired.
Problem 5 (Herstein 3.6.1). Prove that if [a, b] = [a0 , b0 ] and [c, d] = [c0 , d0 ]
then [a, b][c, d] = [a0 , b0 ][c0 , d0 ].
The book doesn’t make this explicit in the problem, but remember that
the field of fractions is defined over an integral domain so in particular multiplication is commutative. Recall that by definition of multiplication in the
field of fractions, we have [a, b][c, d] = [ac, bd] and [a0 , b0 ][c0 , d0 ] = [a0 c0 , b0 d0 ].
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To show that these are equal equivalence classes, we need to show that
the cross products agree, namely that acb0 d0 = a0 c0 bd. Note that our hypotheses can be restated as ab0 = a0 b and cd0 = c0 d. All in all, we have
acb0 d0 = (ab0 )(cd0 ) = (a0 b)(c0 d) = a0 c0 bd as required.
Problem 6 (Herstein 3.6.6). Let D be an integral domain, a, b ∈ D. Suppose
that an = bn and am = bm for two relatively prime integers m and n. Prove
that a = b.
Note that if either of a or b is zero, so too must be the other (since
x = 0 iff x = 0 in an ID), so we may assume both a and b are nonzero. We
first digress and prove a lemma about fields. The key point is that negative
exponents make sense in fields, and they will simplify our analysis.
n
Lemma 6.1. Suppose that F is a field, m and n are relatively prime integers,
and c ∈ F such that cn = cm = 1. Then c = 1.
Proof. Fix i, j ∈ Z such that im + jn = 1. Then c = cim+jn = (cm )i (cn )j = 1
as desired.
Now work in the field of fractions F associated with the integral domain
D, and consider the element c = [a, b]. Note that the conditions an = bn and
am = bm translate in F into saying that cm = cn = 1, where as usual we
write 1 in place of [x, x] for any nonzero x. Then by the lemma we conclude
that c = 1, which is the same as saying that a = b.
Problem 7 (Herstein 3.7.6). Prove that the units in a commutative ring with
a unit element form an abelian group.
The overuse of the word “unit” with two distinct meanings in this question
hurts my brain. Let U denote the set of units of a commutative ring R
with multiplicative identity 1; we wish to show that U is an abelain group
under the multiplication inherited from R. To do this, we need to show that
multiplication is a commutative, associative binary operation on U with an
identity element and inverses.
• Binary operation: This means we need to show that whenever u and
v are units, so too is uv. Fixing inverses u−1 and v −1 respectively, the
usual calculation shows that v −1 u−1 uv = 1, so uv is also a unit.
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• Associativity: We get this for free since R is a ring.
• Identity element: We get this for free since R has a 1.
• Inverses: This is the definition of unit.
Problem 8 (Herstein 3.7.7). Given two elements a, b in the Euclidean domain R their least common multiple c ∈ R is an element such that a|c and
b|c and such that whenever a|x and b|x for x ∈ R, then c|x. Prove that any
two elements in the Euclidean domain R have a least common multiple in R.
We prove this in the greater generality of principal ideal domains by
reformulating this in terms of ideals. First we need a warm-up lemma.
Lemma 8.1. Suppose that R is a ring and I, J are two ideals of R. Then
I ∩ J = {x ∈ R : x ∈ I and x ∈ J} is also an ideal of R.
Proof. We check the laundry list of criteria. Fortunately, though, way back
in HW3 Problem 1 (Herstein 2.5.1) we showed that the intersection of two
subgroups of a group is again a subgroup. So since (I, +) and (J, +) are
subgroups of (R, +), we get for free that (I ∩ J, +) ≤ (R, +). Thus to show
that I ∩ J is an ideal we really only need to show that it is closed under left
and right multiplication by arbtirary elements of r.
Towards that end, arbitrarily pick r ∈ R and x ∈ I ∩ J. Then x ∈ I so
rx ∈ I and xr ∈ I. Similarly, x ∈ J, so rx ∈ J and xr ∈ J. So we conclude
rx ∈ I ∩ J and xr ∈ I ∩ J as desired.
Now suppose that R is a PID, and a, b ∈ R. Consider the ideal (a) ∩ (b);
since it is an ideal and R is a PID, there is some c ∈ R with (c) = (a) ∩ (b).
Note in particular that c ∈ (c) since R has a 1, so we have that c ∈ (a) and
c ∈ (b). This is just another way of saying that a|c and b|c. We’ll show that
this c is indeed a least common multiple of a and b.
Claim. If x ∈ R such that a|x and b|x, then c|x.
Proof of the claim. As we know, a|x is just another way of saying x ∈ (a);
similarly b|x means x ∈ (b). So x ∈ (a) ∩ (b) = (c), so c|x.
The claim proves that c is a least common multiple of a and b as desired.
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