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.

[Added a day later.]

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.

  • 1
    $\begingroup$ I'm not sure - is the second part of the question appropriate for meta.mathoverflow.net? $\endgroup$ May 16, 2016 at 16:44
  • 1
    $\begingroup$ @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. $\endgroup$
    – chizhek
    May 16, 2016 at 17:00
  • 2
    $\begingroup$ @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. $\endgroup$
    – chizhek
    May 16, 2016 at 17:03
  • $\begingroup$ 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) $\endgroup$
    – Yemon Choi
    May 16, 2016 at 17:47
  • $\begingroup$ Bourbaki treats this construction in A.II.1 Exercice 7, but does introduce neither name nor notation for it. $\endgroup$ May 16, 2016 at 21:26

2 Answers 2


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.

  • 1
    $\begingroup$ The notation $A \ltimes N$ is often used. $\endgroup$ May 16, 2016 at 21:00
  • 1
    $\begingroup$ @Jeremy Rickard - Thanks. I googled "trivial extension algebra" and found some references which, however, all consider only special instances of the construction described in the question. Whatever, I got what I asked for: the construction is no longer nameles and I have a sensible notation for it. About the notation, borrowed from semidirect products of groups: this is more than just a distant analogy, see the continuation of my question. $\endgroup$
    – chizhek
    May 17, 2016 at 13:16

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).


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