Abstract Nonsense

Crushing one theorem at a time

Direct Sum of Algebra

Point of post: In this post we shall discuss a natural way to build new algebras out of a collection of algebras. This is the direct sum of algebras which, unsurprisingly, mimics the construction of the direct sum of groups or direct sum of vector spaces.


Just as is the case for vector spaces, groups, modules etc. one can define the direct sum of algebras.

Direct Sum of Algebras

Namely, if \left\{\mathscr{A}_j\right\}_{j\in\mathcal{J}} is a collection of F-algebras and \displaystyle f\in\prod_{j\in\mathcal{J}}\mathscr{A}_j we define the support of f to be the set


We then define the direct sum of \left\{\mathscr{A}_j\right\}_{j\in\mathcal{J}}, denoted \displaystyle \bigoplus_{j\in\mathcal{J}}\mathscr{A}_j,  to be the set of all elements of \displaystyle \prod_{j\in\mathcal{J}}\mathscr{A}_j with finite support with coordinate wise operations. More explicitly if we denote \displaystyle f,g in \displaystyle \bigoplus_{j\in\mathcal{J}}\mathscr{A}_j by (a_j),(b_j)  respectively (where we evidently mean f(j)=a_j and g(j)=b_j) then we define addition in \displaystyle \bigoplus_{j\in\mathcal{J}}\mathscr{A}_j by (a_j)+(b_j)=(a_j+b_j), multiplication by (a_j)(b_j)=(a_jb_j), and scalar multiplication c(a_j)=(ca_j) for c\in F. In other words, we just give \displaystyle \bigoplus_{j\in\mathcal{J}}\mathscr{A}_j the ring structure of the direct sum of rings and just multiply scalars coordinate wise. It’s evident that this does, in fact, define an F-algebra on \displaystyle \bigoplus_{j\in\mathscr{J}}\mathscr{A}_j with unity equal to (1_j).

It’s clear that if \mathscr{J} is in fact finite that this reduces to defining coordinate wise operations on the set \displaystyle \prod_{j\in\mathcal{J}}\mathscr{A}_j.

For now it will suffice to know just the definition of the direct sum of algebras, but just because they are easy we prove a fundamental results. Namely,

Theorem: Let F be a field and \left\{\mathscr{A}_j\right\}_{j\in\mathcal{J}} a collection of F-algebras then

\text{ }

\displaystyle \mathcal{Z}\left(\bigoplus_{j\in\mathcal{J}}\mathscr{A}_j\right)=\bigoplus_{j\in\mathcal{J}}\mathcal{Z}\left(\mathscr{A}_j\right)

\text{ }

Proof: Let (a_j) be an element of \displaystyle \bigoplus_{j\in\mathcal{J}}\mathcal{Z}\left(\mathscr{A}_j\right) then evidently for any (b_j) in \displaystyle \bigoplus_{j\in\mathcal{J}}\mathscr{A}_j we have that (a_j)(b_j)=(a_jb_j)=(b_ja_j)=(b_j)(a_j) so that (a_j) is a member of \displaystyle \mathcal{Z}\left(\bigoplus_{j\in\mathcal{J}}\mathscr{A}_j\right). Conversely, if (a_j) is an element of \displaystyle \mathcal{Z}\left(\bigoplus_{j\in\mathcal{J}}\mathscr{A}_j\right) then we evidently have for each k\in\mathcal{J} and b\in\mathscr{A}_k that \widetilde{b}=(b_j) where b_j=\bold{o}_j for j\ne k and b_k=b is an element of \displaystyle \bigoplus_{j\in\mathcal{J}}\mathscr{A}_j and thus (a_jb_j)=(a_j)\widetilde{b}=\widetilde{b}(a_j)=(b_ja_j so that a_jb_j=b_ja_j for every j\in\mathcal{J} and in particular a_kb=ba_k. Since b\in\mathscr{A}_k was arbitrary it follows that a_k is an element of \mathcal{Z}\left(\mathscr{A}_k\right) and thus since k was arbitrary we have that (a_j) is an element of \displaystyle \prod_{j\in\mathcal{J}}\mathscr{A}_j, but since all but since \text{supp}((a_j)) is finite we must have that it, in fact, an element of \displaystyle \bigoplus_{j\in\mathcal{J}}\mathcal{Z}\left(\mathscr{A}_j\right). The conclusion follows. \blacksquare


Roman, Steven. Advanced Linear Algebra. New York: Springer, 2005. Print.


April 6, 2011 - Posted by | Algebra, Linear Algebra | , , ,


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