Definition of Jacobson radical This may be a rather silly question, but I wonder why the definition of the Jacobson radical always is
 $$\{x\in R\mid 1-xy \text{ is a unit for all } y\in R\}$$
and not
 $$\{x\in R\mid 1+xy \text{ is a unit for all } y\in R\}$$
Clearly, if $y\in R$ then so is $-y$ and we might as well write $1+xy$. So why do we keep the minus instead of a plus?
 A: If we believe that the first set is an additive subgroup of $R$, then they are equal sets.
If $x$ is in the first set, then $-x$ is also in the set, and hence $x$ is in the second set. If $x$ is in the second set, then $-x$ is in the first set, and accordingly $x$ is in the first set too. This shows equality between the two sets.
Of course, if you would rather take just the second set to be an additive subgroup of $R$, a similar argument works.
There is no reason to prefer one over the other, AFAIK, and moreover I'm pretty sure I've seen the $1+xy$ definition used in a text or two.
A: I will restrict to commutative $R$ here. The correct definition of the Jacobson radical of $R$ is as the intersection of all maximal ideals (what you have mentioned should be seen as a description). Then the description with $1-xy$ occurs naturally:
$x \notin \mathrm{jac}(R)$ iff $x \in (R/\mathfrak{m})^*$ for some maximal ideal $\mathfrak{m}$ iff $xy=1 \bmod \mathfrak{m}$ for some maximal ideal $\mathfrak{m}$ and some $y \in R$ iff $1-xy \notin R^*$ for some $y \in R$.
I think this is the reason why $1-xy$ is prefered in the literature.
A: The following shows that, not only is the second form equivalent, but it is also more natural for some purposes. The equivalences below shed further light on the radical.
Theorem $\ $ TFAE in ring $\rm\:R\:$ with units $\rm\:U,\:$ ideal $\rm\:J,\:$ and Jacobson radical $\rm\:Jac(R)\:.$
$\rm(1)\quad  J \subseteq Jac(R),\quad $ i.e.  $\rm\:J\:$  lies in every max ideal $\rm\:M\:$ of $\rm\:R\:.$
$\rm(2)\quad  1+J \subseteq U,\quad\ \ $  i.e. $\rm\: 1 + j\:$  is a unit for every $\rm\: j \in J\:.$
$\rm(3)\quad  I\neq 1\ \Rightarrow\  I+J \neq 1,\qquad\ $  i.e.  proper  ideals survive in $\rm\:R/J\:.$
$\rm(4)\quad M\:$ max $\rm\:\Rightarrow\:  M+J \ne 1,\quad\! $  i.e. max ideals survive in $\rm\:R/J\:.$
Proof $\: $ (sketch) $\ $  With  $\rm\:i \in I,\ j \in J,\:$ and max ideal $\rm\:M,$
$\rm(1\Rightarrow 2)\quad  j \in all\ M\ \Rightarrow\ 1+j \in no\ M\ \Rightarrow\ 1+j\:$ unit.
$\rm(2\Rightarrow 3)\quad i+j \,=\, 1\ \Rightarrow\ 1-j = i\:$ unit $\rm\:\Rightarrow\:  I = 1\:.$
$\rm(3\Rightarrow 4)\ \ \ $  Let $\rm\:I = M\:$ max.
$\rm(4\Rightarrow 1)\quad  M+J \ne 1 \Rightarrow\ J \subseteq M\:$  by  $\rm\:M\:$ max.
A: In fact, the usual definition of the Jacobson radical is the intersection of all maximal left ideals. This turns out to be the intersection of all maximal right ideals. You might like to look at P.M.Cohn: Introduction to Ring Theory or R.Ash: Abstract Algebra.  
