# Multiplication in Direct Sum of Algebras

Say I have a (non-unital) algebra $A$ which decomposes as a direct sum $A = V \oplus W$, where $V$ and $W$ are subalgebras. In an algebra, the multiplication is distributive over addition. Therefore, for two elements $v \in V$ and $w \in W$, we have that $$(v+ w)(v+w) = v^2 + vw + wv + w^2.$$

On the other hand, since $A = V \oplus W$, we have that the multiplication is component-wise, $$(v+w)(v+w) = v^2 + w^2.$$

Is there a way to see these two are compatible? I.e. that, in this setup, $vw + wv = 0$?

EDIT: Ok, so here's another way to look at my problem. Suppose $A$ is an algebra, so in particular it's a vector space. Let $V$ and $W$ be subspaces such that $A = V \oplus W$ where $\oplus$ is a vector space direct sum. The question is then, if $V$ and $W$ are also subalgebras of $A$, does this then extend to a direct sum of algebras?

• This can never happen (for $V, W \neq 0$) unless your notion of subalgebra does not include sharing the multiplicative unit. The closest thing that can happen is that $A$ is the direct product $V \times W$ of algebras, in which case $vw = wv = 0$ for all $v \in V, w \in W$. – Qiaochu Yuan Jul 27 '16 at 23:13
• A isn't necessarily unital, so neither are the subalgebras. – user353840 Jul 27 '16 at 23:17
• I'm not sure what you're asking for. You've just given a proof that if $V$ and $W$ are subalgebras of $A$ such that $V\oplus W\cong A$ as algebras by sending $(v,w)$ to $v+w$, then $vw+wv$ for all $v\in V$ and $w\in W$. What more do you want? – Eric Wofsey Jul 27 '16 at 23:19
• In the direct product $R\times S$ of rings $R$ and $S$, we abuse notation by referring to elements $(r,0)\in R\times S$ as just $r$ and elements $(0,s)\in R\times S$ as just $s$. Thus, $rs$ is shorthand for $(r,0)(0,s)$, which by definition equals $(r\cdot0,0\cdot s)=(0,0)$, i.e. the zero element in $R\times S$. (In the case our rings are algebras, they are in particular vector spaces, which may inspire us to use the symbol $\oplus$ instead of $\times$.) – arctic tern Jul 27 '16 at 23:25
• Show it? Um, if you're using $\oplus$ to denote direct sum of algebras, then it's part of what the notation means. It's part of the definition of $V\oplus W$ (interpreted as an internal algebra direct sum) that $v$ times $w$ equals $0$ for all $v\in V,w\in W$. If it weren't the case that all those products were $0$, then we wouldn't be able to claim that our algebra is an algebra direct sum of $V$ and $W$, and thus wouldn't have the right to write the notation $V\oplus W$ (again, interpreted as an internal algebra direct sum) in the first place. – arctic tern Jul 28 '16 at 0:11

Let's clear up some things from the comments. For categorical reasons, writing a direct product of rings $R\times S$ as a direct sum $R\oplus S$ might be misleading since $R\times S$ is not a coproduct, although in the case of algebras they are in particular vector spaces which we like writing $\oplus$ for, so in the context of algebras we might use $\oplus$ too. However we even use $\oplus$ for vector space decompositions of algebras (in order to signify grading), such as $\mathbb{C}[x]=\mathbb{C}\oplus\mathbb{C}x\oplus\mathbb{C}x^2\oplus\cdots$. In this answer I will decide to use $\times$ for direct products of rings in order to distinguish it from internal vector space direct sum. Also, let's distinguish internal from external perhaps.

If $R$ and $S$ are any two rings, the external direct product $R\times S$ is the the set of all ordered pairs $(r,s)$ with $r\in R$ and $s\in S$ with "componentwise" multiplication $(r_1,s_1)(r_2,s_2)=(r_1r_2,s_1s_2)$. If we're talking about unital rings $R$ and $S$, then $(1_R,1_S)$ is the unique identity for $R\times S$, and the elements $e_R=(1_R,0_S)$ and $e_S=(0_R,1_S)$ are not identity elements.

Instead they are idempotents (i.e. $e^2=e$), they are central (so $ex=xe$ for all $x\in R\times S$) and they are orthogonal (since $e_Re_S=0_{R\times S}$). I'll leave you to verify my parenthetical statements as an exercise.

There are (non-unital) ring embeddings $R\to R\times S$ and $S\to R\times S$, in which (by abuse of notation) we may write $r$ when we mean $(r,0_S)$ or write $s$ when we mean $(0_R,s)$. In this case, the product $rs$ is short for $(r,0_S)(0_R,s)=(r\cdot0_R,0_S\cdot s)=(0_R,0_S)=0_{R\times S}$. In particular, elements of $R$ and $S$ are identified with zero divisors in $R\times S$. With this abuse of notation, multiplication-by-$e_R$ is just the projection map $R\times S\to R$, and multiplication-by-$e_S$ is the projection map $R\times S\to S$.

Now let's switch gears. Suppose $A$ is any rings, and $R$ and $S$ are subrings for which every $a\in A$ may be uniquely written as $a=r+s$ for some $r\in R$, $s\in S$. If in addition we have $rs=0$ for all $r\in R$ and $s\in S$, then we write $A=R\times S$ and we call $A$ an internal direct product of $R$ and $S$. Note that the external direct product $R\times S$ is an internal direct product of $R\times\{0_S\}$ and $\{0_R\}\times S$ under these definitions.

They're both ways of talking about the same thing, the difference is we use the term "internal" when we're viewing $R$ and $S$ as subrings of an already-existing algebra $A$, and we use the term "external" when we're creating a new algebra out of already-existing rings $R$ and $S$.

If $A$ is unital and $A=R\times S$ is an internal direct product, then $1_A=e_R+e_S$ for some unique elements $e_R\in R$ and $e_S\in S$. As an exercise, check that $e_R$ and $e_S$ are orthogonal, central idempotents. Next, check that $R:=e_RA$ and $S:=e_SA$ are ideals, and in particular they are subrings, and that $A=R\times S$ is an internal direct product. Third, verify that if $e\in R$ is a central idempotent, then $e$ and $1-e$ are orthogonal central idempotents. Finally, prove that if $1_A\in A$ is a primitive idempotent then $A$ is not the internal direct product $A=R\times S$ of any two subrings $R\times S$. (Here let's not assume subrings share the same identity element.)

In conclusion, for unital rings, direct factors correspond to central, orthogonal idempotents. Sorry for the entire chapter, but I thought you might like this theory.

Anyway, if $A$ is an algebra, then it is possible for $A$ to be the internal vector space direct sum $V\oplus W$ of subalgebras $V$ and $W$ without being an internal direct product $V\times W$ of subrings $V$ and $W$. For instance, if we don't care about having identity elements, $A=\mathbb{R}\times\mathbb{R}[\varepsilon]/(\varepsilon^2)$ is a vector space direct sum of subalgebras $V=\mathbb{R}\times\mathbb{R}$ and $W=\{0\}\times\mathbb{R}\varepsilon$ but it is not a direct product of them as subrings, since $(0,1)$ is not orthogonal to $(0,\varepsilon)$. However it is a direct product of subrings $Y=\mathbb{R}$ (the first component) and $X=\mathbb{R}[\varepsilon]/(\varepsilon^2)$. Notice that $\dim V=\dim X=2$ and $\dim W=\dim Y=1$ but $X\not\cong V$ as algebras.

• This is a great answer thank you. Do you by any chance have the book 'Representations of Finite and Compact Groups' by Barry Simon? Because the reason I posted this question is that I'm trying to understand a proof in there. But now that I understand direct sums more, I understand the proof less. – user353840 Jul 29 '16 at 9:16
• @user353840 I have access to it. What's up? – arctic tern Jul 29 '16 at 9:50
• It's theorem III.7.1 on page 56. In the book it is previously shown that the group algebra is isomorphic as an algebra to a direct sum of matrix algebras. (theorem III.1.5 page 38) Now, in the proof of theorem III.7.1 they construct the group algebra as a direct sum of vectorspaces $M_\alpha$. Then they state that $M_\alpha$ is isomorphic to a matrix algebra, to which I agree, but only as a vector space. But then they use the fact that it's a direct sum of algebras to show that $p^2 =p$ iff $p_\alpha^2 = p_\alpha$. So I'm really confused about the use of direct sums there. – user353840 Jul 29 '16 at 10:31
• @user353840 The isomorphism $$\Bbb C[G]\cong \bigoplus_{{\rm Irr}(G)}{\rm End}(V)$$ (as $\Bbb C$-algebras and as $\Bbb C[G]$-bimodules in fact not just as vector spaces) is called the Artin-Wedderburn theorem or Peter-Weyl theorem. The slickest proof of this decomposition for finite groups that I know, which also explicitly solves for the idempotents/projections associated to irreps and derives the Schur orthogonality relations as corollary, I wrote another another account here. – arctic tern Jul 29 '16 at 16:43

If $V$ and $W$ are subalgebras and $A=V\oplus W$ as a vector space, it is not necessarily true that $A=V\oplus W$ as an algebra. For instance, consider a commutative algebra $A$ with basis $\{x,y,z\}$ with $xy=y^2=z$ and $x^2=xz=yz=z^2=0$. Take $V$ to be the span of $\{x\}$ and $W$ to be the span of $\{y,z\}$. Then $V$ and $W$ are subalgebras and $A=V\oplus W$ as a vector space. But for $v=x$ and $w=y$, $vw+wv=2z\neq 0$ (assuming our field does not have characteristic $2$).