Math 901
Solutions to Homework # 5
Throughout R denotes a ring with identity.
1. Let R1 , . . . , Rn be rings and R = R1 × · · · × Rn . Prove that R is left semisimple if and
only if Ri is left semisimple for i = 1, . . . , n.
Solution: For i = 1, . . . , k let ρi : Ri → R be the natural inclusion map into the ith
component of R. Let Ri0 = ρi (Ri ). Then Ri0 is an ideal of R and Ri0 Rj0 = 0 for all i 6= j.
Identifying Ri with Ri0 , we can assume each Ri is an ideal of R. Let I be a left ideal of
R. Then I = RI = (R1 + · · · + Rn )I = R1 I + · · · + Rn I. Furthermore, each Ri I is a
left ideal of Ri and also a left ideal of R. Now, let I be a simple left ideal of R. Then
Ri I is a left R-subideal of I for all i. Hence for each i, Ri I = 0 or Ri I = I. If Ri I = 0
for all i, the I = RI = (R1 + · · · + Rk )I = 0, a contradiction. Therefore, Ri I = I for
some i. Hence, I is a simple left ideal of Ri . Conversely, any simple left ideal of Ri is
also a simple left ideal of R.
Now suppose R is semisimple. Then R is a sum of (finitely many) simple left ideals,
say R = I1 + · · · + Ir . Then Ri = Ri R = Ri I1 + · · · + Ri Ir . By above, each Ri Ij
is either zero (if Ij 6⊂ Ri ) or equal to Ij (if Ij ⊆ Ri ) and hence a simple left ideal of
Ri . This implies each Ri is semisimple. Conversely, if each Ri is semisimple, the each
Ri is the sum of simple left ideals of Ri , and hence of simple left ideals of R. Then
R = R1 + · · · + Rn is a sum of simple left ideals of R and is thus semisimple.
2. Prove that a commutative ring is semisimple if and only if R is a finite product of
fields.
Solution: Clearly, any field is semisimple. Thus, by problem #1, a finite product
of fields is semisimple. Conversely, suppose R is commutative and semisimple. By
a theorem from class, R is the direct product of finitely many simple rings, each of
which is simple and commutative. Since a simple commutative ring is a field, the result
follows.
3. Let R be a commutative k-algebra, where k is a field, and suppose R/m ∼
= k for every
maximal ideal m of R. Let M be an R-module. Prove that λR (M ) = dimk M . Use
this to find λR (R) where R = k[x, y, z]/(x3 , yz, y 2 , xz 2 , z 3 ).
Solution: Clearly, any R-module is also a k-vector space. Let L be a simple Rmodule. Then L ∼
= R/m where m is a maximal ideal of R. Since R/m ∼
= k, we see
that dimk L = 1. Thus, L is a simple k-module (vector space). Suppose first that
λR (M ) < ∞. Then M has a composition series as an R-module. This series is also a
composition series for M as a k-vector space, since the factor modules will be simple
as both R and k-modules. As the length of any composition series of a k-vector space
is its dimension as a k-vector space, we get that λR (M ) = dimk M .
Suppose now that λR (M ) = ∞. If dimk M is finite, then any descending or ascending
chain of k-subspaces of M (and hence any chain of R-submodules of M ) stabilizes,
which implies that M is both Noetherian and Artinian as an R-module. But this
means that λR (M ) < ∞, a contradiction. Hence, dimk M = ∞ = λR (M ).
For the example, we use that λR (R) = dimk R. A k-basis for R is the set of nonzero
monomials in R: {1, x, y, z, xy, xz, x2 , z 2 , x2 y, x2 z}. Hence, λR (R) = 10.
4. Let R be a commutative ring containing an algebraically closed field k such that R is
finitely generated as a ring over k; i.e., R = k[u1 , . . . , un ] for some u1 , . . . , un ∈ R. Let
M be an R-module. Prove that λR (M ) = dimk M . (Hint: You may use the following
version of the Nullstellensatz: If E/F is a field extension and E is finitely generated
as a ring over F then E/F is algebraic.)
Solution: Let m be a maximal ideal of R. Then R/m is a finitely generated field
extension of k. (Note that the composition k → R → R/m is injective, else m contains
a unit from k. And R/m is finitely generated as a ring over k by the images of
u1 , . . . , un .) By the version of the Nullstellensatz stated above, R/m is algebraic over
k. As k is algebraically closed, we must have R/m = k; i.e., the map k → R/m is an
isomorphism. The result now follows by Problem #3.
5. Suppose R contains a simple left ideal I.
(a) Let y ∈ R. Prove that Iy = 0 or Iy ∼
= I.
Solution: Suppose Iy 6= 0. Consider the R-linear map φ : I → Iy given by
φ(i) = iy for all i ∈ I. Clearly, φ is surjective. Since I is simple and ker φ 6= R
(as Iy 6= 0), we have ker φ = 0. Thus, φ is an isomorphism.
P
(b) Let B(I) =
J where the sum is over all left ideals J which are isomorphic to
I. Prove that B(I) is a two-sided ideal.
Solution: Clearly, B(I), being the sum of left ideals of R, is a left ideal. Let
a ∈ B(I) and y ∈ R. Then a = a1 + · · · + ak where ai ∈ Ji and Ji is a left ideal
isomorphic to I. Clearly, ai y ∈ Ji y. By part (a), or Ji y = 0 or Ji y ∼
=I ⊆
= Ji ∼
B(I). Thus, ai y ∈ B(I) for all i and hence ay ∈ B(I). This shows that B(I) is
an ideal of R.
(c) Suppose R is simple. Prove that R is left semisimple.
Solution: As R is simple and B(I) is a nonzero ideal of R, we have R = B(I).
Thus, R is the sum of simple left ideals. Hence, R is left semisimple.