Math 901
Solutions to Homework # 6
1. Suppose A1 × · · · × Ak ∼
= B1 × · · · × B` is a ring isomorphism where Ai and Bj are
simple rings for all i and j. Prove that k = ` and (after reordering) Ai ∼
= Bi for all i.
Solution: For ease of notation, we identify Ai and Bj with their canonical images (as
ideals) in the corresponding products. Let φ be the given isomorphism. Then φ(A1 ) is
an ideal of B1 ×· · ·×B` . Hence, φ(A1 ) = I1 ×· · ·×I` where Ij is an ideal of Bj for all j.
Note that I1 ×· · ·×I` is a simple ring (as A1 is). Hence, Ij 6= 0 for precisely one value of
j, say (without loss of generality), j = 1. As B1 is simple, I1 = B1 . Thus, φ(A1 ) = B1 ,
so A1 ∼
= B1 . Modding out by A1 and B1 , we get that A2 × · · · × Ak ∼
= B2 × · · · × B` .
∼
By induction, k = ` and (after rearranging), Ai = Bi for all i.
2. P
Suppose R = R1 × · · · × Rk where each Ri is a semisimple ring. Prove that λR (R) =
∼
i λRi (Ri ). Use this to deduce that if R = Mn1 (D1 ) × · · · Mnk (Dk ) where D1 , . . . , Dk
are division rings, then λ(R) = n1 + · · · + nk .
Solution: Let I be a left ideal of Ri . Then Iˆi = 0 × · · · × I × · · · × 0 is a left ideal
b
of R. Further, I is simple as a left Ri -module
Pni if and only if I is simple as a left Rmodule. As each Ri is semisimple, Ri = j=1 Iij where Iij are simple left ideals and
P
P P i c
Iij . Since each Ic
ni = λRi (Ri ). Then R = ki=1 nj=1
ij is simple, λR (R) =
i λRi (Ri ).
For the second statement, let R = Mn (D) where D is a division ring. Note that Dn
is a simple R-module (where the elements of Dn are viewed as column matrices) and
R∼
= ⊕ni=1 Dn . Hence, λR (R) = n. The second statement now follows from the first.
3. Let M be a finitely generated semisimple R-module. Prove that EndR M is a semisimple ring.
Solution: First, note that (as we proved in class for simple left ideals), if φ : N1 → N2
is an R-homomorphism of simple R-modules, then φ is either zero or an isomorphism.
Now, as M is a finitely generated semisimple module, M is the direct sum of finitely
many simple submodules. Let N1 , . . . , Nt be the distinct (up to isomorphism) simple
submodules appearing in such a sum. For each i = 1, . . . , t, let Bi be the sum of all
the submodules of M isomorphic to Ni . Then M = B1 ⊕ · · · ⊕ Bt . Let ψ : M → M
be an R-endormorphism of M . Then (as we showed in class), ψ can be represented
by a matrix (ψij ) where ψij : Bi → Bj . We claim that ψij = 0 for all i 6= j. Suppose
ψij 6= 0 for some i 6= j. Let Bi = A1 ⊕ · · · ⊕ Ar and Bj = C1 ⊕ · · · ⊕ Cs where Ak
and C` are simple. Then ψij ρk (Ak ) is nonzero for some k, where ρk : Ak → Bi is the
natural inclusion. Hence, π` ψij ρk (Ak ) is nonzero for some `, where π` : Bj → C` is
the projection map. But, by the remark above, this means Ak ∼
= C` , a contradiction.
Thus, EndR (M ) = EndR (B1 ) × · · · × EndR (Bt ). It suffices to show that each EndR (Bi )
is semisimple; that is, we may assume M is a direct sum of finitely many simple
modules which are all isomorphic. Say, M ∼
= E n , where E is a simple R-module. Then
EndR (M ) ∼
= EndR (E n ) ∼
= Mn (EndR (E)), which is semisimple as EndR (E) is a division
ring.
In each of the problems #4-#7, let F be a field of characteristic zero and A the (first) Weyl
algebra of F .
4. Prove that A is a domain.
Pn
i
Solution:
Suppose
f,
g
∈
A
\
{0}
with
f
g
=
0.
Let
f
=
i=1 fn (x)d and g =
Pm
j
j=1 gj (x)d where fn (x) and gm (x) are nonzero. Then, using dx = xd + 1 repeatedly,
one obtains
n+m−1
X
f (x)g(x) = fn (x)gm (x)dn+m +
ci (x)di
i=0
i j
for some ci (x) ∈ A. Since {x d | i, j ≥ 0} is a k-basis for A, we see that f g = 0 implies
fn (x)gm (x) = 0, a contradiction.
5. Prove that A is neither left nor right Artinian.
Solution: Consider the sequence of left ideals
A ⊇ Ad ⊇ Ad2 ⊇ · · · .
Note that for any i ≥ 0, di 6∈ Adi+1 . Otherwise, there exists an equation of the form
X
di = (
ck` xk d` )di+1 ,
k,`
with ck` ∈ F , contradicting that the set {xi dj } is an F -basis for A. This gives a
descending chain of left ideals which doesn’t stabilize. Hence, A is not left Artinian.
Similarly, one can consider the chain of right ideals
A ⊇ xA ⊇ x2 A ⊇ · · · .
Since xi 6∈ xi+1 A for all i (by the same argument as above), we see that A is not right
Artinian.
6. Recall that A is a subring of EndF (F [x]). Hence, F [x] is a left A-module. Prove that
F [x] is a simple A-module (and hence finitely generated).
Solution: It suffices to show that Af = F [x] for every nonzero f ∈ F [x]. Suppose
deg f = n. Then dn f is a nonzero constant, say c ∈ F . Hence, for g(x) ∈ F [x], let
h = c−1 g(x)dn ∈ A. Then g = hf ∈ Af , so Af = F [x].
7. Let A0 = EndA (F [x]), which is a division ring by Problem #6. Prove that F [x] is not
a finitely generated A0 -module. (Hint: A is not semisimple.)
Solution: Suppose F [x] is a finitely generated A0 -module. Then by a corollary to the
Jacobson Density Theorem, the natural map λ : A → EndA0 (F [x]) is surjective. Also,
as A is simple, ker λ = 0 so λ is injective. So λ is an isomorphism. However, as A0 is
a division ring and F [x] is a f.g. A0 -vector space, EndA0 (F [x]) is semisimple. But A is
not semisimple (or even left Artinian), a contradiction.