Chapter 17

Algebra · 14 exercises

Problem 2

A ring is said to be Artinian if every descending sequence of left ideals \(J_{1} \supset J_{2} \supset \cdots\) with \(J_{i} \neq J_{i+1}\) is finite. (a) Show that a finite dimensional algebra over a field is Artinian. (b) If \(R\) is Artinian, show that every non-zero left ideal contains a simple left ideal. (c) If \(R\) is Artinian, show that every non-empty set of ideals contains a minimal ideal.

3 step solution

Problem 3

Let \(R\) be Artinian. Show that its radical is 0 if and only if \(R\) is semisimple. [Hint: Get an injection of \(R\) into a direct sum \(\oplus R / M_{i}\) where \(\left\\{M_{i}\right\\}\) is a finite set of maximal left ideals.]

3 step solution

Problem 4

Nakayama's lemma. Let \(R\) be any ring and \(M\) a finitely generated module. Let \(N\) be the radical of \(R .\) If \(N M=M\) show that \(M=0 .\) [Hint: Observe that the proof of Nakayama's lemma still holds. \(]\)

4 step solution

Problem 5

(a) Let \(J\) be a two-sided nilpotent ideal of \(R\). Show that \(J\) is contained in the radical. (b) Conversely, assume that \(R\) is Artinian. Show that its radical is nilpotent, i.e.. that there exists an integer \(r \geqslant 1\) such that \(N^{r}=0 .\) [Hint: Consider the descending sequence of powers \(N^{r}\), and apply Nakayama to a minimal finitely generated left ideal \(L \subset N^{*}\) such that \(N^{*} L+0\).

2 step solution

Problem 6

Let \(R\) be a semisimple commutative ring. Show that \(R\) is a direct product of fields.

6 step solution

Problem 7

Let \(R\) be a finite dimensional commutative algebra over a field \(k\). If \(R\) has no nilpotent element \(\neq 0\), show that \(R\) is semisimple.

4 step solution

Problem 8

(Kolchin) Let \(E\) be a finite-dimensional vector space over a field \(k\). Let \(G\) be a subgroup of \(G L(E)\) such that every element \(A \in G\) is of type \(I+N\) where \(N\) is nilpotent. Assume \(E \neq 0 .\) Show that there exists an element \(v \in E, v \neq 0\) such that \(A v=v\) for all \(A \in G .\) [Hint: First reduce the question to the case when \(k\) is algebraically closed by showing that the problem amounts to solving linear equations. Secondly, reduce it to the case when \(E\) is a simple \(k[G]\) -module. Combining Burnside's theorem with the fact that \(\operatorname{tr}(A)=\operatorname{tr}(I)\) for all \(A \in G\), show that if \(A_{0} \in G, A_{0}=I+N\), then \(\operatorname{tr}(N X)=0\) for all \(X \in\) End \(_{k}(E)\), and hence that \(\left.N=0, A_{0}=I .\right]\)

3 step solution

Problem 9

Let \(E\) be a finite dimensional vector space over a field \(k .\) Let \(R\) be a semisimple subalgebra of End \(_{4}(E)\). Let \(a, b \in R\). Assume that Ker \(b_{E} \supset\) Ker \(a_{E}\). where \(b_{E}\) is multiplication by \(b\) on \(E\) and similarly for \(a_{K}\). Show that there exists an element \(s \in R\) such that \(s a=b\). [Hint: Reduce to \(R\) simple. Then \(R=\) End \(_{D}\left(E_{0}\right)\) and \(E=E_{0}^{(\pi)} .\) Let \(v_{1}, \ldots, v, \in E\) be a \(D\) -basis for \(a E\). Define \(s\) by \(s\left(a v_{i}\right)=b v_{i}\) and extend s by D-linearity. Then \(s a_{E}=b_{\varepsilon}\), so \(\left.s a=b .\right]\)

5 step solution

Problem 10

Let \(E\) be a finite-dimensional vector space over a field \(k\). Let \(A \in\) End \(_{1}(E)\). 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 \(k\).

2 step solution

Problem 11

Let \(E\) be a finite-dimensional vector space over a field \(k\), and let \(S\) be a commutative set of endomorphisms of \(E\). Let \(R=k[S]\). Assume that \(R\) is semisimple. Show that every subset of \(S\) is semisimple.

2 step solution

Problem 12

Prove that an \(R\) -module \(E\) is a generator if and only if it is balanced, and finitely generated projective over \(R^{\prime}(E) .\) Show that Theorem \(5.4\) is a consequence of Theorem 7.1.

6 step solution

Problem 13

Let \(A\) be a principal ring with quotient field \(K .\) Let \(A^{n}\) be \(n\) -space over \(A\), and let $$ T=A^{*} \oplus A^{n} \oplus \cdots \oplus A^{n} $$ be the direct sum of \(A^{n}\) with itself \(r\) times. Then \(T\) is free of rank \(n r\) over \(A\). If we view elements of \(A^{n}\) as column vectors, then \(T\) is the space of \(n \times r\) matrices over \(A\). Let \(M=\) Mat \(_{n}(A)\) be the ring of \(n \times n\) matrices over \(A\), operating on the left of \(T\). By a lattice \(L\) in \(T\) we mean an \(A\) -submodule of rank \(n r\) over \(A\). Prove that any such lattice which is \(M\) -stable is \(M\) -isomorphic to \(T\) itself. Thus there is just one \(M\) -isomorphism class of lattices. [Hint: Let \(g \in M\) be the matrix with 1 in the upper left corner and 0 everywhere else, so \(g\) is a projection of \(A^{\prime \prime}\) on a 1 -dimensional subspace. Then multiplication on the left \(g: T \rightarrow A_{r}\) maps \(T\) on the space of \(n \times r\) matrices with arbitrary first row and 0 everywhere else. Furthermore, for any lattice \(L\) in \(T\) the image \(g L\) is a lattice in \(A_{r}\), that is a free \(A\) -submodule of rank \(r\). By elementary divisors there exists an \(r \times r\) matrix \(Q\) such that \(g L=A, Q \quad\) (multiplication on the right). Then show that \(T Q=L\) and that multiplication by \(Q\) on the right is an \(M\) -isomorphism of \(T\) with \(L\) ]

6 step solution

Problem 14

Let \(F\) be a field. Let \(\mathrm{n}=\mathrm{n}(F)\) be the vector space of strictly upper triangular \(n \times n\) matrices over \(F\). Show that \(n\) is actually an algebra, and all elements of \(n\) are nilpotent (some positive integral power is 0 ).

5 step solution

Problem 15

Conjugation representation. Let \(A\) be the multiplicative group of diagonal matrices in \(F\) with non-zero diagonal components. For \(a \in A\), the conjugation action of \(a\) on \(\operatorname{Mat}_{n}(F)\) is denoted by \(\mathrm{c}(a)\), so \(\mathrm{c}(a) M=a M a^{-1}\) for \(M \in \mathrm{Mat}_{n}(F) .\) (a) Show that it is stable under this action. (b) Show that \(n\) is semisimple under this action. More precisely, for \(1 \leqq i

3 step solution

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