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While working through some lecture notes on semigroups, it seemed to me like a semigroup doesn't buy you much generality over a monoid. But I wondered whether the situation is different for non-unital versus (unital) rings. Then I worked through the following reasons against non-unital rings, which made the point that all naturally occurring non-unital (sub)rings actually have important additional structures, giving ideals and R-algebras as examples. But the author of that paper later later agreed to most of the reasons for non-unital rings.

Let $A$ be a ring (possibly non-unital) and $\tilde{A}=\mathbb Z\oplus A$ as abelian group. I wondered what Martin Brandenburg meant by the "obvious" multiplication so that $A\subseteq \tilde{A}$ is an ideal and $1\in\mathbb Z$ is the identity. After some trying, I came up with $$ (r,a)\cdot(s,b)=(rs,rb+sa+ab)$$ I haven't checked associativity, but at least the above two conditions are satisfied. Now I wonder what would happen if I replace $\mathbb Z$ by an arbitrary commutative ring $R$ with identity $1$, and assume that I'm also given a scalar multiplication $R \times A \mapsto A$ denoted by $(r, a) \mapsto ra$. Would the above multiplication turn $R\oplus A$ into an (unital) ring, if $A$ is an $R$-algebra? And would conversely $A$ be an $R$-algebra, whenever $R\oplus A$ happens to be an (unital) ring under the above multiplication?

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If you write symbolically $r+a$ for $(r,a)$, you get $$ (r+a)\cdot(s+b)=rs+rb+sa+ab.$$ What you did looks correct to me. These kind of things can be stated in term of adjoint functors, I presume. –  Pierre-Yves Gaillard Jan 4 '12 at 15:51
    
@Pierre-YvesGaillard Thanks, now I see why it's "obvious". I guess this also helps with checking associativity. –  Thomas Klimpel Jan 4 '12 at 15:56
    
The answer to most of your questions is yes, though you want to be careful with some logic. In your very last question, if A is not an R-algebra, then what does "rb" mean? As in Leon's questions, I think you'll get farther if you just forget about rings, and only consider R-algebras. If you want them to have unity, then require them to by calling them unital R-algebras. Rings are not related to R. R-algebras are. –  Jack Schmidt Jan 4 '12 at 15:58
    
I think you'll find very interesting the paper by Burgess and Stewart reference in my MO post here. –  Bill Dubuque Jan 4 '12 at 16:30
    
@BillDubuque: thanks, I added a summary of Burgess–Stewart, and another recent paper on Dorroh extensions that paints them in a more favorable light. –  Jack Schmidt Jan 4 '12 at 20:25

2 Answers 2

up vote 2 down vote accepted

Let R be a unital, associative, possibly non-commutative ring. A possibly non-unital, non-associative, non-commutative R-algebra is simply an R-R bimodule M and an R-R bimodule homomorphism f from $M\otimes_R M$ to M, where $m\cdot n = f(m\otimes n)$. The result is a distributive multiplication on M where elements of R act in normal fashions: $mr\cdot sn=m\cdot rsn$, etc.

Then the R-R bimodule $S=R\oplus M$ becomes a unital $R$-algebra using the homomorphism $(r\oplus m) \otimes (s\oplus n) \mapsto (rs \oplus (rn+ms+m\cdot n))$. If the multiplication on M is associative, then S is an associative, unital R-algebra. If both R and M are commutative, then S is commutative.


If the multiplication on M is the zero map (the only multilication you can guarantee in general), then you can view this as a (non-unital) R-subalgebra of the triangular R-algebra defined by R, R, and M.

$$S = \left\{\begin{bmatrix} r & m \\ 0 & r \end{bmatrix} : r \in R, m \in M \right\} \qquad \begin{bmatrix} r & m \\ 0 & r \end{bmatrix}\cdot \begin{bmatrix} s & n \\ 0 & s \end{bmatrix} = \begin{bmatrix} rs & rn+ms \\ 0 & rs \end{bmatrix}$$

Triangular rings are very good to produce counterexamples by taking a module property of M and making it a ring property of S. Lam's Lectures on Modules and Rings has a few good examples. The "D+M" construction is also a popular use of triangular rings, as in the ring of polynomials with rational coefficients in general, but integral constant coefficients: it behaves like a mix of the ring Z[x] and the Z-module Q[x].


In your question, you have a scalar multiplication $R\times A\to A$. This multiplication needs to satisfy a few axioms. Those axioms are precisely that A is an R-module under this multiplication, and that the multiplication map $f:A\otimes A \to A:a\otimes b \mapsto ab$ is an R-R bimodule homomorphism.


A description of this trivial $R\oplus M$ extension was originally given in Dorroh (1932), which is clearly written and in modern enough language. This preserves basic structural features such as distributivity, associativity, commutativity, and adjoins a multiplicative unity. However, the elements of M are much closer to an ideal than a large subring, and so properties such as being an integral domain need not hold. For instance if M had a 1, then $$(1⊕0 - 0⊕1)(0⊕m) = (0⊕(m+0+0)) - (0⊕(0+0+m)) = 0,$$ and every element of M is a zero-divisor, even if M began life as a nice field.

A few different embeddings are given in Burgess–Stewart (1989) that address some of these problems. First they find a subring K of End(M) that is as large as possible while still being similar to the “integers” inside M, and then embed both M and K inside End(M) and let S be the ring they both generate. In this case, elements of S are all of the form km, and so M is "dense" inside S, automatically giving most of the good properties (in the non-commutative world) one would expect from a localization. For instance, the non-unital ring $2\mathbb{Z}$ is embedded in $\mathbb{Z}$, where the latter is really the ring of endomorphisms of the additive group of $2\mathbb{Z}$. This works well as long as M has only one multiplicative 0. In the case where this is not true, they take an approach fairly similar to Dorroh, but using a ring like K instead of $\mathbb{Z}$.

Sometimes (my version of) Dorroh's extension works fairly well. The version I describe here is also described in Mesyan (2010), where it is shown that these extensions behave well with regards to semi-primeness (similar to a direct product of integral domains), and a similar sounding result is given for primeness (similar to integral domain), though the actual impact is not as great.

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Now I understand what you mean by "I think you'll get farther if you just forget about rings, and only consider R-algebras." Your explanation answers indeed one question I had, but I feel more comfortable with a (possibly non-associative) R-algebra over a commutative ring R. Are the quaternions an algebra over $\mathbb C$ according to your definition? And what is the "D+M" construction? –  Thomas Klimpel Jan 4 '12 at 20:54
    
@ThomasKlimpel: You can view the quaternions as an algebra over C, because the quaternions are a C-C bimodule. However, the two actions on the left and right are not the same (ij ≠ ji), and so you'll find most commutative ring ideas don't work well. Considering the quaternions as an algebra over the reals is much more pleasant (where the left and right module structures agree; in other words, where the image of the coefficient ring lies in the center). D+M is described in Fuchs–Salce books.google.com/books?id=4NzTiOuirN4C&pg=PA7&lpg=PA7 –  Jack Schmidt Jan 4 '12 at 21:05

A counterexample to a claim I erroneously thought someone had made: if $A$ is a left $R$-module, but not necessarily an $R$-algebra, must the Dorroh construction produce an associative algebra?

Let $A=\mathbb{Q}[i]$ and $R=\mathbb{Q}[\sqrt{2}]$, and define an $R$-module structure on $A$ by defining $(r_1+r_2\sqrt{2})\cdot(a_1+a_2i) = (r_1 a_1 + 2 r_2 a_2) + (r_2a_1 + r_1a_2)i$, where $r_1, r_2, a_1, a_2 \in \mathbb{Q}$ are rational numbers with the standard multiplication.

This defines an $R$-module structure on $A$, namely $A$ is a free $R$-module of rank 1, so that ${}_RR \cong {}_RA$ as left $R$-modules, but not as rings. This also defines both $R$ and $A$ as $\mathbb{Q}$-algebras, since the multiplication when $r_2=0$ or $a_2=0$ is the expected one.

Now we use the formulation in the question to define a multiplication on $R \oplus A$ via $$(r,a)\cdot(s,b) = (rs, rb+sa+ab)$$

Now consider the multiplication $$(0,i) \cdot (0,i) \cdot (\sqrt{2},0)$$

On the one hand, it is supposed to be: $$((0,i)\cdot(0,i)) \cdot (\sqrt{2},0) = (0,-1) \cdot (\sqrt{2},0) = (0,0+(\sqrt{2}(-1)) + 0 )$$ but the multiplication of $R$ on $A$ defines $\sqrt{2}(-1) = -i$, so we get $$((0,i)\cdot(0,i)) \cdot (\sqrt{2},0) = (0,-1) \cdot (\sqrt{2},0) = (0,-i)$$

On the other hand, it is supposed to be: $$(0,i)\cdot((0,i)) \cdot (\sqrt{2},0)) = (0,i) \cdot (0,0+\sqrt{2}(i)+0)$$ but the multiplication of $R$ on $A$ defines $\sqrt{2}(i) = 2$, so we get: $$(0,i)\cdot((0,i)) \cdot (\sqrt{2},0)) = (0,i) \cdot (0,2) = (0,2i)$$

In particular, the resulting multiplication is not associative.


I misread the question (sorry!). Suppose $A$ is a left $R$-module so that one does manage to get an associative unital ring structure, must $A$ be an $R$-algebra?

The counterexample is fairly general. I'll reorder it a little to avoid assuming $R$ is central: $$(r,0)\cdot((0,a)\cdot(0,b)) = (r,0) \cdot(0,0+0+ab) = (r,0)\cdot(0,ab) = (0,r(ab)+0+0) = (0,r(ab))$$ but $$((r,0)\cdot(0,a))\cdot(0,b) = (0,ra+0+0) \cdot(0,b) = (0,ra)\cdot(0,b) = (0,0+0+(ra)b) = (0,(ra)b)$$ so for $R\oplus A$ to be associative, $A$ must be an $R$-algebra, as in, $r(ab) = (ra)b$.

I believe your multiplication will also force $r(ab) = a(rb)$, whereas my multiplication only requires $(ar)b = a(rb)$, so that associative properties follow from associative properties, rather than commutative properties.

Notice that your multiplication can be used to give $A$ an R-R bimodule structure since $m\cdot r$ can be defined as $(0,m)\cdot (r,0) = (0,m\cdot r)$, but then of course $rm=m\cdot r$ and we have the situation common in commutative algebra where the left and right module structures are required to coincide.

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Abelian groups (the additive subgroups of rings) have a unique unital Z-module structure, and so Dorroh does not worry about sides or compatibility issues. These only matter for larger rings, but for instance Z/2Z x Z/2Z or Z[i] as R will also provide counterexamples. –  Jack Schmidt Jan 4 '12 at 23:15

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