Download MATH 819 HOMEWORK 8 SOLUTIONS 1. Let A1,...,Am be

MATH 819
HOMEWORK 8
SOLUTIONS
1. Let A1 , . . . , Am be commuting n × n matrices over an algebraically closed
field k. Using Burnside’s theorem, prove that A1 , . . . , Am have a common
eigenvector.
Proof. Let V = k n . We may view A1 , . . . , Am ∈ Endk (V ). We prove the
theorem by induction on n = dim V . For n = 1, the result is trivial (since
every vector in V is a multiple of every other (nonzero) vector in that case).
Suppose that n > 1. Let R be the k-subalgebra of Endk (V ) generated by
A1 , . . . , Am . Note that Endk (V ) ∼
= Matn (k) is not a commutative ring
(since n > 1). Since A1 , . . . , Am are commuting matrices, obviously R is a
commutative ring. So we must have R 6= Endk (V ). By Burnside’s theorem,
V cannot be a simple R-module (since this would imply that R = Endk (V )).
Thus, there is a nontrivial proper R-submodule V 0 ⊂ V . In other words,
Ai V 0 ⊂ V 0 for i = 1, . . . , m and we can view A1 , . . . , Am ∈ Endk (V 0 ).
Since 0 < dim V 0 < n, by induction A1 , . . . , Am have a common eigenvector
v ∈ V 0 . Viewing v ∈ V , we see that A1 , . . . , Am have a common eigenvector
(in V ).
2. (Lang, Ch. XVII, Ex. 1)
(a) Let R be a ring. We define the Jacobson radical of R to be the left ideal
N which is the intersection of all maximal left ideals of R. Show that
N E = 0 for every simple R-module E. Show that N is a two-sided
ideal.
Solution. Let E be a simple R-module. Let v ∈ E, v 6= 0. The kernel of the R-module homomorphism R → E, x 7→ xv is a maximal
left ideal Mv , as noted at the beginning of section 6. Then clearly
Ann(E) = ∩v∈E,v6=0 Mv . Since N is contained in every maximal left
ideal Mv , we have N ⊂ Ann(E) and N E = 0.
Consider ∩ Ann(R/M ), where M runs over the maximal left ideals
of R. Since each R/M is a simple R-module, from the above, N ⊂
∩ Ann(R/M ). On the other hand, Ann(R/M ) ⊂ M , so ∩ Ann(R/M ) ⊂
∩M = N . It follows that ∩ Ann(R/M ) = N . But for any left Rmodule E, Ann(E) is a two-sided ideal. So N is an intersection of
two-sided ideals and is therefore a two-sided ideal.
(b) Show that the Jacobson radical of R/N is 0.
Solution. Let M and M0 be the set of maximal left ideals of R/N and
R, respectively. Let φ : R → R/N be the canonical map. For any
M ∈ M, φ−1 (M ) is a maximal left ideal of R and conversely, for any
maximal left ideal M 0 of R, φ(M 0 ) = M 0 /N is a maximal left ideal
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MATH 819 HOMEWORK 8 SOLUTIONS
of R/N (since N ⊂ M 0 ). So the maximal left ideals of R/N and R
correspond under the canonical map φ. It follows that
∩M ∈M M = ∩M 0 ∈M0 φ(M 0 ) = φ(∩M 0 ∈M0 M 0 ) = φ(N ) = 0.
3. (Lang, Ch. XVII, Ex. 2) A ring is said to be (left) Artinian if every descending sequence of left ideals J1 ⊃ J2 ⊃ · · · with Ji 6= Ji+1 is finite.
(a) Show that a finite dimensional algebra over a field is Artinian.
Solution. Let R be a finite dimensional algebra over a field k. Let
J1 ⊃ J2 ⊃ · · · be a descending sequence of left ideals with Ji 6= Ji+1
for all i. Each ideal Ji is a vector space over k. If Ji 6= Ji+1 then
dimk Ji+1 < dimk Ji . Since R is finite dimensional, dimk J1 = n is
finite. This implies that the sequence must be finite. In fact, it must
have length ≤ n + 1.
(b) If R is Artinian, show that every non-zero left ideal contains a simple
left ideal.
Solution. Let J = J1 be a nonzero left ideal. If J is simple, there’s
nothing more to prove. Otherwise, J contains a proper nonzero left
ideal J2 . If J2 is simple, then we’re done. Otherwise, J2 contains
a proper nonzero left ideal J3 . Continuing in this way, we obtain a
sequence J1 ⊃ J2 ⊃ · · · which must be finite since R is Artinian. The
last left ideal in the sequence will be a simple ideal.
(c) If R is Artinian, show that every non-empty set of left ideals contains
a minimal element.
Solution. Let S be a non-empty set of left ideals. Let J ∈ S. If
J = J1 is not a minimal element of S then J properly contains a left
ideal J2 ∈ S. If J2 is not a minimal element of S then J2 properly
contains a left ideal J3 ∈ S. Continuing in this way we obtain a
sequence J1 ⊃ J2 ⊃ · · · which must be finite since R is Artinian. The
last left ideal in the sequence must be a minimal element of S.
4. (Lang, Ch. XVII, Ex. 3) Let R be Artinian. Show that its Jacobson radical
is 0 if and only if R is semisimple. (see hint in Lang)
Solution. Suppose that R is semisimple. Let N be the Jacobson radical of
R. Since R is semisimple, every R-module is a (direct) sum of simple left
R-modules. But then Exercise 1(a) implies that N E = 0 for any R-module
E (not necessarily simple). In particular, N R = 0. Obviously, since 1 ∈ R,
this can only happen if N = 0.
Now suppose that R is Artinian and its Jacobson radical N is 0. Let M1
be a maximal left ideal of R. If M1 6= 0, since N = 0, there exists a maximal
left ideal M2 with M1 ∩ M2 6= M1 . If M1 ∩ M2 6= 0, then again there exists
a maximal left ideal M3 with M1 ∩ M2 ∩ M3 6= M2 . Continuing in this
way, we obtain a sequence of left ideals M1 ⊃ M1 ∩ M2 ⊃ · · · with each
containment proper. Since R is Artinian, this sequence must be finite, and
MATH 819
HOMEWORK 8
SOLUTIONS
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we find that for some maximal left ideals M1 , . . . , Mn , M1 ∩ · · · ∩ Mn = 0.
Then we have an R-module homomorphism
R → ⊕ni=1 R/Mi
x 7→ (x + M1 , . . . , x + Mn ).
The kernel is clearly ∩ni=1 Mi = 0, so the map is injective. Since each
Mi is a maximal left ideal, R/Mi is a simple R-module. So ⊕ni=1 R/Mi is
semisimple. Since R is a submodule of this module, R is semisimple.
5. (Lang, Ch. XVII, Ex. 5)
(a) Let J be a two-sided nilpotent ideal of R. Show that J is contained
in the Jacobson radical.
Solution. It suffices to show that J is contained in every maximal left
ideal M of R. Let M be a maximal left ideal of R. Then M + J is a
left ideal of R containing M . If M + J = M then J ⊂ M . Otherwise,
we must have M + J = R. So we can write 1 = m + j for some m ∈ M
and j ∈ J. Since J is nilpotent, j n = 0 for some positive integer n.
Then
(1 + j + · · · + j n−1 )m = (1 + j + · · · + j n−1 )(1 − j) = 1 − j n = 1.
So 1 ∈ M , a contradiction, and we must have J ⊂ M . Then J is
contained in the radical.
(b) Conversely, assume that R is Artinian. Show that its Jacobson radical
N is nilpotent, i.e., that there exists an integer r ≥ 1 such that N r = 0.
(see hint in Lang and assume Nakayama’s lemma)
Solution. We’ll follow the hint. Note that we have a descending chain
of (two-sided) ideals N ⊃ N 2 ⊃ N 3 ⊃ · · · . Since R is Artinian, this
must stabilize at some point to an ideal N ∞ (= N r for some positive
integer r). Suppose that N ∞ 6= 0. Let S be the set
S = {left ideal L | N ∞ L 6= 0}.
Clearly R ∈ S, so S is nonempty. Then since R is Artinian, S has a
minimal element L. Since N ∞ L 6= 0, let x ∈ L be such that N ∞ x 6= 0.
Then N ∞ x is a left ideal of R and N ∞ N ∞ x = N ∞ x 6= 0. So N ∞ x ∈
S. Since L is minimal and N ∞ x ⊂ L, we must have N ∞ x = L. In
particular, L is finitely generated (in fact, generated by x), and we
have N ∞ L = L. Then by Nakayama’s lemma, we have L = 0, a
contradiction. So N ∞ = 0, that is, N r = 0 for some positive integer
r.
6. (Lang, Ch. XVII, Ex. 6) Let R be a semisimple commutative ring. Show
that R is a direct product of fields.
Qs
Solution. By Theorem 4.3, R = i=1 Ri where each Ri is a simple ring.
Since R is commutative, each Ri is commutative. Thus, it suffices to show
that a commutative simple ring Ri is a field. This is immediate from the
fact that Ri contains no ideals other than 0 and Ri (if α ∈ Ri , α 6= 0,
then (α) = (1), so there exists β ∈ Ri with αβ = 1 and α is invertible).
Alternatively, we know that Ri ∼
= Matn (D) for some division ring D. As is
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MATH 819 HOMEWORK 8 SOLUTIONS
easy to see, the only commutative such ring is when D = k is a field and
n = 1.
7. (Lang, Ch. XVII, Ex. 7) Let R be a finite dimensional commutative algebra
over a field k. If R has no nilpotent element 6= 0, show that R is semisimple.
Solution. By Exercise 2, R is Artinian. By Exercise 5, the radical N is
nilpotent. This implies that every element of N is nilpotent. Then by
assumption, we must have N = 0. Now by Exercise 3, R is semisimple. 8. (Lang, Ch. XVII, Ex. 10) Let E be a finite-dimensional vector space over a
field k. Let A ∈ Endk (E). We say that A is semisimple if E is a semisimple
A-space, or equivalently, let R be the k-algebra generated by A, then E is
semisimple over R. Show that A is semisimple if and only if its minimal
polynomial has no factors of multiplicity > 1 over k.
Add: If k is algebraically closed, then A is semisimple if and only if A is
diagonalizable.
Solution. By Proposition 4.7, E being semisimple over R is equivalent to
R being semisimple. We have a ring homomorphism
k[t] → k[A]
t 7→ A.
By definition, the kernel of this map is the ideal (qA (t)), where qA is the
minimal polynomial of A. So
∼ k[t]/(qA (t)).
R = k[A] =
Qn
ji
Let qA (t) = i=1 pi (t) , where pi is an irreducible polynomial over k and
the ji are positive integers. By the Chinese remainder theorem, we have an
isomorphism
n
Y
k[t]/ pi (t)ji .
k[t]/(qA (t)) ∼
=
i=1
If ji = 1 for all i, then each of the rings k[t]/ pi (t)ji = k[t]/ (pi (t)) is a
field. Then R is isomorphic to a (finite) direct product of fields, which is
semisimple. Conversely, from Exercise 6, since R is commutative, it must
be a direct product of fields. Note that a direct product
Qn of fields contains
no nonzero nilpotent elements. If some ji > 1, then i=1 pi (A) is clearly a
nonzero nilpotent element of k[A] (raise it to the power maxi ji ), and so R
cannot be semisimple in this case.
Finally, note that if k is algebraically closed, then A is diagonalizable if
and only if its minimal polynomial has no factors of multiplicity > 1 over
k (look at the Jordan canonical form of A).