Jacobson radical of non-unital rings This question come from a colleague who does operator theory, and hence works with non-unital rings, which I know little about. It's well-known that any non-unital ring $R$ can be embedded in a unital ring $R_1= \mathbf{Z} \oplus R$. Is it true that the Jacobson radical of $R$ is the same as the radical of $R_1$? In this situation, $R$ is also a two-sided ideal in $R_1$. Is the radical of $R$ considered as an $R_1$-module the same as its radical as a ring?
 A: I'll write elements of $R_1$ as $(m,x)$, where $m\in\mathbb{Z}$ and $x\in R$. Addition is componentwise and multiplication is
$$(m,x)(n,y)=(mn,nx+my+xy).$$

An element $x$ in a ring $R$ is called right quasi-regular if there exists $y$ in $R$ such that $x+y+xy=0$.
In a unital ring this amounts to saying that $1+x$ is right invertible, because then $(1+x)(1+y)=1+x+y+xy=1$.
Just like in the unital case, one can show that $x\in J(R)$ (the Jacobson radical) if and only if $xy$ is right quasi-regular for all $y\in R$. This is the same, in the unital case, as $1+xy$ being right invertible for all $y\in R$.
Notice that the Jacobson radical is defined as the intersection of all maximal regular right ideals. A right ideal $I$ is regular if there exists $e\in R$ such that $ex-x\in I$ for all $x\in I$.
Now, let's see what elements in $R_1$ are in the Jacobson radical. If $(m,x)$ is in the Jacobson radical, then $(1,0)+(m,x)$ must be right invertible:
$$
(1,0)=(m+1,x)(n,y)=((m+1)n,nx+(m+1)x+xy)
$$
so $(m+1)n=1$. This implies $m+1=1$ or $m+1=-1$. But also $(1,0)+(-m,-x)$ must be right invertible and the same computation shows $1-m=1$ or $1-m=-1$. The only possibility is then $m=0$.
Thus elements in the Jacobson radical of $R_1$ must be of the form $(0,x)$. Saying that $(1,0)+(0,x)$ is right invertible, using the computation above, means a right inverse is of the form $(1,y)$ with $x+y+xy=0$: that is, $x$ is right quasi-regular, and conversely.
It's easy to conclude now that an element $(m,x)$ belongs to the Jacobson radical of $R_1$ if and only if $m=0$ and $x\in J(R)$.
