Chapter 18

Show that the kernel of an algebra homomorphism is an ideal.

Let \(A\) be a finite-dimensional algebra over \(F\) and let \(B\) be a subalgebra. Show that if \(b \in B\) is invertible, then \(b^{-1} \in B\).

If \(A\) is an algebra and \(S \subseteq A\) is nonempty, define the centralizer \(C_{A}(S)\) of \(S\) to be the set of elements of \(A\) that commute with all elements of \(S\). Prove that \(C_{A}(S)\) is a subalgebra of \(A\).

Show that \(\mathbb{Z}_{6}\) is not an algebra over any field.

Let \(G=\left\\{1=a_{0}, \ldots, a_{n}\right\\}\) be a finite group. For \(x \in F[G]\) of the form \(x=r_{1} a_{1}+\cdots+r_{n} a_{n}\) let \(T(x)=r_{1}+\cdots+r_{n}\). Prove that \(T: F[G] \rightarrow F\) is an algebra homomorphism, where \(F\) is an algebra over itself.

Prove the first isomorphism theorem of algebras: A homomorphism \(\sigma: A \rightarrow B\) of \(F\)-algebras induces an isomorphism \(\bar{\sigma}: A / \operatorname{ker}(\sigma) \approx \operatorname{im}(\sigma)\) defined by \(\bar{\sigma}(a \operatorname{ker}(\sigma))=\sigma a\).

Prove that the quaternion field is an \(F\)-algebra and a field. Hint: For $$ x=r_{0}+r_{1} i+r_{2} j+r_{3} k \neq 0 $$ \(\left(r_{0}=r_{0} 1\right)\) consider $$ \bar{x}=r_{0}-r_{1} i-r_{2} j-r_{3} k $$

Describe the left regular representation of the quaternions using the ordered basis \(\mathcal{B}=(1, i, j, k)\).

Let \(S_{n}\) be the group of permutations (bijective functions) of the ordered set \(X=\left(x_{1}, \ldots, x_{n}\right)\), under composition. Verify the following statements. Each \(\sigma \in S_{n}\) defines a linear isomorphism \(\tau_{\sigma}\) on the vector space \(V\) with basis \(X\) over a field \(F\). This defines an algebra homomorphism \(f: F\left[S_{n}\right] \rightarrow \mathcal{L}_{F}(V)\) with the property that \(f(\sigma)=\tau_{\sigma}\). What does the matrix representation of a \(\sigma \in S_{n}\) look like? Is the representation \(f\) faithful?

Prove that for \(n \geq 3\), the matrix algebras \(\mathcal{M}_{n}(F)\) are central and simple.

An element \(a \in A\) is left-invertible if there is a \(b \in A\) for which \(b a=1\), in which case \(b\) is called a left inverse of \(a\). Similarly, \(a \in A\) is rightinvertible if there is a \(b \in A\) for which \(a b=1\), in which case \(b\) is called a right inverse of \(a\). Left and right inverses are called one- sided inverses and an ordinary inverse is called a two-sided inverse. Let \(a \in A\) be algebraic over \(F\). a) Prove that \(a b=0\) for some \(b \neq 0\) if and only if \(c a=0\) for some \(c \neq 0\). Does \(c\) necessarily equal \(b\) ? b) Prove that if \(a\) has a one-sided inverse \(b\), then \(b\) is a two-sided inverse. Does this hold if \(a\) is not algebraic? Hint: Consider the algebra \(A=\mathcal{L}_{F}(F[x])\). c) Let \(a, b \in A\) be algebraic. Show that \(a b\) is invertible if and only if \(a\) and \(b\) are invertible, in which case \(b a\) is also invertible.