Download Math 210A: Algebra, Homework 8

Math 210A: Algebra, Homework 8
Ian Coley
December 1, 2013
Problem 1.
Show that if 1 = 0 in a ring R, then R is the zero ring.
Solution.
Let x ∈ R be an element. Then x = 1 · x = 0 · x = 0. Therefore R is the zero ring.
Problem 2.
(a) For any ring R, define a ring structure on the abelian group R̃ = R × Z such that (0, 1)
is the identity of R̃ and the inclusion map R ,→ R̃, r 7→ (r, 0) is a rng homomorphism.
(b) Let C be the category of rings without identity. Show that the functor F : C → Rings
such that F (R) = R̃ is a left adjoint to the forgetful functor G : Rings → C taking a
ring with identity R to R considered as an object of C.
Solution.
(a) Since R has an abelian group structure (i.e. a Z-module structure), multiplication on
R by Z is well-defined. We define multiplication on R × Z by
(r, m)(s, n) = (rs + (n · r) + (m · s), mn).
We need to show this satisfies the distributive property and that (0R , 1) is indeed the
identity. First,
(r, m)(0R , 1) = (r · 0R + 1 · r + m · 0R , m · 1) = (r, m)
and it is trivial to see that (0R , 1)(r, m) is the same. Therefore (0R , 1) is the identity.
Now on the one hand,
(a, b) [(r, m) + (s, n)] = (a, b)(r + s, m + n)
= (a(r + s) + (m + n) · a + b · (r + s), b(m + n))
and on the other
= (a, b)(r, m) + (a, b)(s, n)
= (ar + m · a + b · r, bm) + (as + n · a + b · s, bn)
= (ar + as + m · a + n · a + b · r + b · s, bm + bn).
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which are equal by the distribution of the action of Z on R. Therefore this is well
defined multiplication.
Now let i : R → R̃ be the inclusion map r 7→ (r, 0). Clearly i(r+s) = (r+s, 0) = (r, 0)+
(s, 0) and i(0) = (0, 0) is the identity. Therefore we need only check multiplication:
i(r)i(s) = (r, 0)(s, 0) = (rs + 0 · r + 0 · s, 0 · 0) = (rs, 0) = i(rs).
Therefore i is a ring homomorphism.
(b) We need to show that, for every ring without identity R and ring with identity S we
have
HomRings (F (R), S) ∼
= HomC (R, G(S)).
Let i be the inclusion R ,→ R̃ = F (R). Suppose that f : F (R) → S is a ring
homomorphism. Then consider the map f ◦ i : R → S. Since i does not map identity
to identity (since R does not have an identity), we forget the identity of S so we
properly have a map f ◦ i : R → G(S).
Now suppose we have a map g : R → G(S) not respecting the identity of S. Then we
define a map g 0 : F (R) → S by (0, 1) 7→ 1S so that (r, m) 7→ g(r) + m · 1S . Then we
need to check this respects multiplication. In particular,
g 0 (r, m)g 0 (s, n)) = (g(r) + m · 1S )(g(s) + n · 1S )
= g(r)g(s) + n · g(r) + m · g(s) + mn · 1S = g 0 ((r, m)(s, n)).
Now we need only show we can transform a map twice back to itself. Starting with a
map f : F (R) → S and the associated map f ◦ i : R → S not respecting identity, we
claim the extended map (f ◦ i)0 = f . Indeed, suppose (f ◦ i)(r) = f (r, 0) = s. Then
f (r, m) = s+f (0, m) = s+m·f (0, 1). But (f ◦i)0 (r, m) = (f ◦i)(r)+m·1S = s+m·f (0, 1)
precisely. Therefore we have an isomorphism of Hom-sets, so F is the left adjoint of
G.
Problem 3.
Prove that a finite nonzero ring with no zero divisors is a division ring and a finite integral
domain is a field.
Solution.
Let R be a finite nonzero ring with no zero divisors. Let x ∈ R be a nonzero element. Since
R is finite, there are only finitely many distinct powers of x. Suppose that xm = xn for some
m > n. Then
0 = xm − xn = xn (xm−n − 1).
Since R has no zero divisors, one of xn and xm−n − 1 must be zero. If xn = 0, then x is
zero divisor, which is a contradiction. Therefore xm−n − 1 = 0, i.e. xm−n = x · xm−n−1 = 1.
Therefore x has an inverse, and since this holds for all nonzero x, R is a division ring.
If R is a finite integral domain, then it is a commutative division ring, therefore a field.
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Problem 4.
Let R be the set of all 2 × 2 matrices over C of the form
u v
.
−v u
Show that R is a subring with identity in M2 (C) that is isomorphic to the ring of real
quaternions H.
Solution.
First, the usual I2 ∈ M2 (C) is in R, so R has an identity. Now given two elements of R,
uy + vx
ux − vy
u v
x y
ux − vy uy + vx
=
=
.
−v u
−x y
−uy − vx ux − vy
−(uy + vx) (ux − vy)
Therefore R is closed under multiplication, so R is a subring. Let A ∈ R be a matrix defined
by u, v ∈ C, where u = a+bi and v = c+di. Then we claim the map A 7→ a+bi+(c+di)j ∈ H
is an isomorphism. It is clear that this map is bijective and respects the abelian group
structure of R, so we need only show it respects multiplication. Indeed, using the above
calculation, let x = α + βi and y = γ + δi define B. Then AB is defined by the elements
ux − vy = (aα − bβ − cγ − dδ) + (aβ + bα + cδ − dγ)i
uy + vx = (aγ − bδ + cα + dβ) + (aδ + bγ − cβ + dα)i.
Multiplying the elements in H gives us
(a + bi + cj + dk)(α + βi + γj + δk) = (aα − bβ − cγ − dδ) + (aβ + bα + cδ − dγ)i
+ (aγ − bδ + cα + dβ)j + (aδ + bγ − cβ + dα)k.
These correspond perfectly, and it is also clear that I2 7→ 1+0i+(0+0i)j = 1 ∈ H. Therefore
these rings are isomorphic.
Problem 5.
Show that the subset Q = {±1, ±i, ±j, ±k} in the real quaternion ring H is a group (with
respect to the multiplication) isomorphic to the quaternion group Q8 .
Solution.
First, it is clear that Q is closed under multiplication. Second, since i2 = j 2 = k 2 = −1, we
have i−1 = −i and similarly for the other elements, and −1−1 = −1. 1 is still the identity,
so Q is a group. Let G be the usual quaternion group, defined by
G = ha, b, c : ab = cba, a2 = b2 = c, c2 = ei
We claim that the map i 7→ a and j 7→ b (and by extension ij = k 7→ ab) is an isomorphism.
We necessarily have −1 7→ c, and ij = −1 · ji just as ab = cba. Also under the relations of
the quaternion group, we have
abab = abcba = ab(b2 )ba = a(b2 )(b2 )a = ac2 a = a2 = c,
just as k 2 = −1. Therefore Q ∼
= G.
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Problem 6.
(a) Prove that if a nonzero matrix a ∈ Mn (F ), where F is a field, is a zero divisor if and
only if det(a) = 0.
(b) Prove that a nonzero matrix a ∈ Mn (R), where R is a commutative ring, is a zero
divisor if det(a) = 0.
Solution.
(a) We use the fact that the determinant det : Mn (F ) → F is a ring homomorphism.
Suppose that a ∈ Mn (F ) is a zero divisor, and let ab = 0 for a nonzero matrix b.
Then det(ab) = det(a) det(b) = 0. Since F is a field, either det(a) = 0 or det(b) = 0.
Suppose that det(a) 6= 0. Then a is invertible (since we are working over a field), so
a−1 ab = b = 0, so b is the zero matrix. But this is a contradiction, so we must have
det(a) = 0.
Conversely, suppose that det(a) = 0. Then by linear algebra, there exists some x ∈ F n
such that a · x = 0. Then consider the matrix b = [x x · · · x], where each column of the
matrix is the vector x. Then ab = 0, so a is a zero divisor.
(b) As above, the equation a · x = 0 has a solution for some nontrivial x ∈ Rn . The same
matrix b = [x x · · · x] gives ab = 0, so a is a zero divisor.
Problem 7.
Let S = Mn (R), where R is a ring with identity. Show that for any ideal J ⊂ S there is a
unique ideal I ⊂ R such that J = Mn (I).
Solution.
Let I = {a11 : a ∈ J}. Then we claim I is an ideal. Indeed, if a, b ∈ J, then (a + b)11 =
a11 + b11 ∈ I. Further, for the diagonal matrix c = r · I, where r ∈ R, (c · a)11 = r · a11 ∈ I,
so I is an ideal.
We claim that J = Mn (I). Let eij be the matrix with 1 in the (i, j) position and 0
elsewhere. Then for every a = (aij ) ∈ Mn (R), we have eij aek` = ajk ek` . In particular,
for a ∈ J, aij e11 = e1i aej1 ∈ J, so we have aij ∈ I for every i, j ∈ {1, . . . , n}. Therefore
J ⊂ Mn (I). Conversely, let r ∈ I. Then let c ∈ J such that r = c11 . Further, reij =
ei1 ce1jP∈ J for all i, j ∈ {1, . . . , n}. Therefore any a ∈ Mn (I) can be written in the form
a = i,j aij eij ∈ J, so Mn (I) ⊂ J. Since we have shown double inclusion, we are done.
Further, by this construction, I is unique.
Problem 8.
(a) Let f : R → S be a ring homomorphism, I an ideal in R, J an ideal in S. Show that
f −1 (J) is an ideal in R that contains ker f .
(b) If f is surjective, then f (I) is an ideal in S. If f is not surjective, f (I) need not be an
ideal in S.
Solution.
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(a) We need to show that f −1 (J) is closed under left multiplication by R. Let r, x ∈ R
such that f (x) ∈ J. Then f (rx) = f (r)f (x) ∈ J, so rx ∈ f −1 (J). Therefore f −1 (J) is
an ideal.
(b) Let y ∈ f (I) and s ∈ S. Since f is surjective, there exists some r such that f (r) = s.
Then f −1 (sy) = rf −1 (y) ∈ I so sy ∈ f (I). Therefore f (I) is an ideal.
Consider the ideal 2Z ⊂ Z and let f : Z → Z[t] be the inclusion map. Then 2Z ⊂ Z[t]
is not an ideal since 2t ∈
/ 2Z (for example). Therefore f (I) need not be an ideal if f is
not surjective.
Problem 9.
(a) An element a of a ring R is called nilpotent if an = 0 for some n ∈ N. Show that if
R is a commutative ring, then the set Nil R of all nilpotent elements of R is an ideal.
Prove that the factor ring R/ Nil R has no nonzero nilpotent elements.
(b) Prove that a polynomial f (X) = a0 + a1 X + . . . + an X n ∈ R[X] is nilpotent if and
only if all ai are nilpotent in R.
Solution.
(a) Let x ∈ Nil R and r ∈ R. Then suppose xn = 0. Using commutativity,
(rx)n = rn xn = 0
so rx ∈ Nil R. Therefore Nil R is an ideal. Now let R = R/ Nil R and suppose that
x ∈ R is a nilpotent element. Then xn ∈ Nil R for some n, so (xn )m = xnm = 0 for
some m. Therefore x ∈ Nil R, so x = 0 in R. Hence R/ Nil R has no nonzero nilpotent
elements.
(b) Suppose that each ai is nilpotent. Then since R[X] is commutative
and Nil R ⊂
P
i
i
Nil R[X] is an ideal, we have ai X ∈ Nil R[X] for all i and f =
ai X ∈ Nil R[X].
Therefore f is nilpotent.
Now suppose that f is nilpotent. Let ai be the first non-nilpotent coefficient. Then
consider R[X] → (R/ Nil R)[X]. Letting¯denote the image of this map, we have
!
n
n−i
X
X
f (X) =
aj X j = X i
aj X j
j=0
j=i
Pn−i
j
which is still nilpotent. Therefore g(X) =
j=0 aj X must be nilpotent as well.
Suppose g m = 0. Then we see the constant term is am
i = 0. But R/ Nil R has no
nonzero nilpotent elements, so we must have ai = 0. Therefore every coefficient of
f (X) must be nilpotent, so we are done.
Problem 10.
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(a) Prove that if a is a nilpotent element of a ring R, then the element 1 + a is invertible.
(b) Prove that a polynomial f (X) = a0 + a1 X + . . . + an X n ∈ R[X] is invertible in R[X]
if and only if a0 is invertible and all ai are nilpotent in R for i ≥ 1.
Solution.
(a) Since a ∈ Nil R, let an = 0. Then
1 = 1 + an = (1 + a)(1 − a + a2 + . . . + (−1)n−1 an−1 )
so 1 + a has an inverse.
(b) We can generalise the above construction: assuming R is commutative, if u is an
invertible element and a is nilpotent, then u + a is invertible. Since the nilradical is
−1
an ideal, au−1 ∈ Nil R, so 1 + au−1 is invertible, hence
) = u + a is too.
Pu(1 + au
i
By Problem 9(b), if all
for i ≥ 1, then i≥1 ai X is nilpotent. If a0 is
P ai are nilpotent
i
invertible, then a0 + i≥1 ai X is invertible.
P
j
Conversely, suppose f has an inverse g, where g = m
j=0 bj X and we may assume
bm 6= 0. Since the constant term of f g is a0 b0 = 1, it is clear that a0 is invertible.
To see that ai are nilpotent, first assume that R is an integral domain. Then f g = 1
implies that the leading coefficient of f (x)g(x) = an bm = 0, which means that an = 0
(since bm 6= 0). Therefore f (X) must have a zero coefficient on all X terms, i.e. ai = 0
for all i ≥ 1, which are certainly nilpotent.
Now if R is not a domain, let p ⊂ R be any prime ideal. Then let R = R/p and ai the
images of ai in R and f (X), g(X) the images of the polynomials in R[X]. Then
f (X)g(X) = 1 =⇒ ai = 0 for all i ≥ 1
as above. This implies that ai ∈ p for all i ≥ 1. By T
a thus-far unproven lemma (but
it should be before this assignment is due), Nil R = p⊂R p. By above, ai ∈ p for all
primes, so ai ∈ Nil R for all i ≥ 1. This completes the proof.
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