Math 250: Higher Algebra
Problem Set #7 (12 November 2004):
Simple algebras, etc.
1. i) Fix a (commutative) field k. Is the (commutative, infinite-dimensional) algebra k[X], consid-
ered as a module over itself, semisimple?
ii) Now assume that kis not of characteristic 2. For each c∈k, let Acbe the (two-dimensional,
commutative) k-algebra k[X]/(X2−c). Prove that Ac, considered as a module over itself, is
simple unless cis a square in k, and semisimple unless c= 0.
iii) More generally, for any nonconstant polynomial f∈k[X], when is the commutative k-algebra
k[X]/(f) simple or semisimple as a module over itself?
2. Prove that if Mis a semisimple R-module then so is any submodule and any quotient module
of M. Give an example of a ring Rwith a module Mand a submodule M0such that M0and
M/M0are both semisimple but Mis not. [Hint: we’ve seen only a few examples of modules
that are not semisimple. . . ]
Linear algebra over skew fields. In our noncommutative setting, all modules, ideals, etc. will
be left modules/ideals/. .. unless otherwise specified. A module over a skew field Kis also called a
“vector space” over K. This generalizes the notion of a vector space over a commutative field. We
can use our generalities about semisimple modules to extend the basic notions of linear algebra to
this setting.
3. i) Show that any skew field K, considered as a left K-module, is simple, and that every simple
K-module is isomorphic to K.
Let Vbe a finitely generated1K-module, with generators v1, . . . , vn. We may assume that each
vi6= 0. Then Vis the sum of the simple modules Kvi, and is thus semisimple. Therefore it is the
direct sum of Kvjwith jrunning over some subset J⊆ {1,2, . . . , n}. If V=⊕j∈JKvj, we say that
{vj:j∈J}is a basis for V. Note that Vis isomorphic as a K-module to Km, where m= #(J)≤n.
If W⊆Vis any K-submodule (also called “[K-vector] subspace”), then V=W⊕W0for some
subspace W0⊆V, and W0can be chosen to be the direct sum of Kvjwith jrunning over some
subset of J. In particular, if {vj}is a basis of V, and {ui}is any independent subset (that is, a
subset such that the sum of Kuiin Vis direct), then {ui}can be completed to a basis.
ii) Prove that every linearly independent subset of Knhas cardinality at most n, and is a basis if
and only if its cardinality is exactly n.
Thus the basis cardinality is an invariant of finitely generated K-vector spaces. Naturally we call
this invariant the dimension of the space, and denote it by dimK.
iii) Let Vbe a K-vector space of finite dimension n, and TaK-linear map (that is, a K-module
homomorphism) from Vto some other K-vector space W. Prove that the dimensions of the
image and kernel of Tsum to n.
1We can extend the results of Problem 3 to arbitrary K-modules if we assume AC/Zorn.
