# Attaching an ideal whose square is zero: does this operation have a name and a notation?

I know I met the following construction somewhere, but I cannot remember where. Let $A$ be
a (unital associative) ring, and let $N$ be an $A$-$A$ bimodule. On the product set $A\times N$ we define multiplication by \begin{equation*} (a,m)(b,n) := (ab,\,an+mb)~. \end{equation*} The set $A\times N$ equipped with this multiplication is a (unital associative) ring with the
two-sided ideal $0\times N$ whose product with itself is $0$. The ideal $0\times N$ is the kernel of the surjective homomorphism of rings $A\times N\to A : (a,n)\mapsto a$. If $A$ is Jacobson-semisimple,
then $0\times N$ is the Jacobson radical of the ring $A\times N$.

Does this construction of the ring $A\times N$ from a ring $A$ and an $A$-$A$ bimodule $N$ have a name, and perhaps an established notation? The notation $A\times N$ is misleading since it suggests a direct product of rings, which it is not.

The comment by Dag Oskar Madsen to the answer by Jeremy Rickard got me thinking.

First consider the definition of the inner semidirect product of groups.

Let $G$ be a group with a subgroup $H$ and a normal subgroup $N$. The following statements are equivalent:

1. $G=NH$ ($=HN$) and $N\cap H=\{e\}$.

2. The natural embedding $H\to G$, composed with the natural projection $G\to G/N$,
is an isomorphism of groups.

3. There exist a group $H'$ and homomorphisms of groups $p\colon G\to H'$ and $j\colon H'\to G$
such that $p\circ j = \mathrm{id}_{H'}$ and $j(H')=H$, $\ker p = N$.

4. There exists an idempotent endomorphism $e$ of the group $G$ such that $e(G)=H$
and $\ker e = N$.

If one of these statements holds (and therefore all hold) we say that the group $G$ is the (inner) semidirect product of its normal subroup $N$ and its subgroup $H$, and write $G=N\rtimes H$.

Now compare this definition to an analogous situation with a ring in place of a group.

Let $R$ be a ring with a subring $A$ and a (two-sided) ideal $N$. The following statements are equivalent:

1. The underlying additive group of $R$ is a direct sum of the additive underlying groups
of $N$ and $A$, which we write $R=N\oplus A$.

2. The natural embedding $A\to R$, composed with the natural projection $R\to R/N$,
is an isomorphism of rings.

3. There exist a ring $A'$ and homomorphisms of rings $p\colon R\to A'$ and $j\colon A'\to R$
such that $p\circ j=\mathrm{id}_{A'}$ and $j(A')=A$, $\ker p = N$.

4. There exists an idempotent endomorphism $e$ of the ring $R$ such that $e(R)=A$
and $\ker e=N$.

I propose to say, whenever the situation described by any of the four cases above occurs,
that $R$ is the (inner) semidirect product of its ideal $N$ and its subring $A$, and write $R=N\rtimes A$.

The structure of the inner semidirect product of an ideal and a subring of a ring suggests the following definition of the outer semidirect product of a rng $N$ and a ring $A$ with respect to
a coherent biaction $\varphi$ of $A$ on $N$. (A rng is an additive group with an associative biadditive multiplication. A multiplicative identity is not required; if it is present, it is ignored.)
What we mean by a coherent biaction of $A$ on $N$: the rng $N$ is an $A$-$A$ (unital) bimodule,
where $(am)n=a(mn)$, $m(na)=(mn)a$ for all $m$, $n$ in $N$ and all $a$ in $A$. We define the multiplication on the set $B:=N\times A$ by \begin{equation*} (m,a)(n,b) \,:=\, (mn+an+mb,\,ab)~. \end{equation*} Then $B$ is a ring, $0\times A$ is a subring of $B$, $N\times 0$ is an ideal of $B$, and $B=(N\times 0)\oplus(0\times A)$. I propose to call the ring $B$ the (outer) semidirect product of a rng $N$ and a ring $A$ with respect to
a coherent biaction $\varphi$ of $A$ on $N$, and write $B = N\rtimes_\varphi A$.

I believe this is a natural transfer, by analogy, of the notion of a semidirect product from groups to rings. The outer semidirect product for rings is peculiar in that it constructs a ring from a rng and
a ring (thus it is an 'inter-species' construction) with the ring coherently biacting on the rng.

If we have just a plain $A$-$A$ bimodule $N$, with no multiplication on $N$, we equip $N$ with the
all-zero multiplication $($$mn=0 for all m, n in N\,$$)$, obtaining a legitimate rng, and so make $A$ coherently biacting on the rng $N$. I propose that in this special case we write $B=N\mathbin{{}_0\rtimes_\varphi} A$,
where $0$ stands for the all-zero multiplication on $N$ and $\varphi$ is the bimodule action of the ring $A$
on the additive group $N$.

I googled "semidirect product of rings" and got no exact matches. The approximate matches
were "crossed product of rings", "semidirect product of Hopf algebras", "semidirect product of Lie algebras/rings". There was also a 'mixed marriage' semidirect product $K[N]\rtimes_\varphi H$ $($isomorphic
to the group algebra $K[N\rtimes_\varphi H]$$)$ of the group algebra of a group $N$ with coefficients in
a commutative ring $K$, and a group $H$, with respect to an action $\varphi$ of the group $H$ on the group $N$
(by isomorphisms), which induces an action of the group $H$ on the group algebra $K[N]$
(by isomorphisms).

Mark the notion of a retract of a topological space: a subspace $A$ of a topological space $X$
is a retract of $X$ if there exists an idempotent continuous map $e\colon X\to X$ such that $e(X)=A$;
the restriction of $e$ to $r\colon X\to A$ (the codomain $X$ of $e$ is narrowed down to the codomain $A$ of $r$)
is a retraction.

• I'm not sure - is the second part of the question appropriate for meta.mathoverflow.net? – Anthony Quas May 16 '16 at 16:44
• @Anthony Quas - I think so. I would rather post it there, it is inappropriate here. But when you are in pain, you don't care for appropriate manners, you just cry out... How do I get there, to meta.mathoverflow.net? Below on this page there is only the link to Meta Stack Exchange. – chizhek May 16 '16 at 17:00
• @Anthony Quas - Stupid me: meta.mathoverflow.net gets me there. I am hurrying home right now. Tomorrow I will cut the second part of the question out and repost it on the meta site. Thanks. – chizhek May 16 '16 at 17:03
• This kind of construction has appeared in many papers on Banach algebras, many of which IMO are reinventing the wheel, even if I can't pin down where the wheel came from. For algebras over a field, I think it might be mentioned explicitly in Hochschild's papers from the 1940s, since it is precisely the extension of A by N that corresponds to the zero cocycle in H^2(A,N) – Yemon Choi May 16 '16 at 17:47
• Bourbaki treats this construction in A.II.1 Exercice 7, but does introduce neither name nor notation for it. – Fred Rohrer May 16 '16 at 21:26

In the representation theory of finite dimensional algebras, at least, it's called a "trivial extension algebra" (although that sometimes refers to the special case where $N$ is the vector space dual of $A$). Googling "trivial extension algebra" will find you several references.
• The notation $A \ltimes N$ is often used. – Dag Oskar Madsen May 16 '16 at 21:00
In the context of Hochschild- and André--Quillen (co-)homology, this would be called the trivial square-zero extension of $A$ by $N$. See e.g. Quillen's paper on the (co-)homology of commutative rings (1970).