What's your definition of an $R$-algebra for a non-commutative ring $R$?

For me its just a monoid in the category $_RMod$

But if $R$ is non-commutative, what is the tensor product in $_RMod$?

If $M$ and $N$ are $R$-modules, then I consider $N$ as a right module over $R^{op}$ (the opposite ring of R) and I define the tensor product between $N$ and $M$ as $N _{R^{op}}\otimes_R M$ is that way.

But $M\otimes_{R^{op}}N$ only makes sense if $M$ is a right $R^{op}$-module and $N$ is a left $R^{op}$-module.

nono, my notion must not have been too evident.... the symbol $_{R^{op}}$ was not on the right of $N$ but on the left of the tensor product... That is my tensor product $_{R^{op}}\otimes_R$ is from $Mod_{R^{op}} \times _RMod$ to $_RMod$... (formally it takes a pair (N,M) in $_RMod^2$ to the corresponding pair (N,M) in $Mod_{R^{op}} \times _RMod$ and then applies the tensor product $_{R^{op}}\otimes_R$

ok

So .. in this case what do you think?

But then what does $_{R^{op}}\otimes_R$ mean? There isn't a natural monoidal structure on the category of (left, say) modules for a general non-commutative ring. E.g., if $R$ is the ring of $2\times2$ matrices over a field, and $V$ is the module of column vectors, then what is $V\otimes V$?

Sorry, just saw your edit. $M$ and $N$ are not $R^{op}\otimes R$-modules.

why not?

The right action makes sense no (unless I made a mistake in my verification)

But if $R$ isn't commutative then the left $R$-action and the right $R^{op}$-actions don't commute.

Hmm... so if $R$ isn't commutative I guess it would make sense to define an algebra as existing over the center of $R$, that is ..... N \otimes M may be defined as above... except we may first restrict the scalars of $N$ and $M$ to Z(R) and then take the tensor _{Z(R)^{op}} \otimes _{Z(R)}....

That makes sense as a definition of $N\otimes M$ as a $Z(R)$-module, but then there are two different natural $R$-module structures: $r(x\otimes y)$ could be $(rx)\otimes y$ or $x\otimes (ry)$, and neither makes $_RMod$ a monoidal category (there isn't a unit).

Well...

Going back to your original question, you could define an $R$-algebra as a ring $A$ together with a ring homomorphism $R\to A$, but that doesn't agree with the standard definition if $R$ is commutative (a ring $A$ with a ring homomorphism $R\to Z(A)$). And with this definition, $A\otimes_RS$ isn't an $S$-algebra in a natural way if $R$ is a subring of $S$.

Or...

we can define algebras over bimodules exclusivly.. but then that doesn't exaclty agree with the usual definitions either...

If you define an "R-algebra" as an R-bimodule A with an associative multiplication A\otimes_RA\to A, then apart from the fact that this doesn't agree with the usual definition when R is commutative, I don't see any way in general of defining A\otimes_RS as an S-algebra. The obvious was of getting an S-bimodule is B=S\otimes_RA\otimes_RS, but then there's no obvious multiplication for B.

(where R is a subring of S)

Hmm... after giving it some thought, I think this line of thought is not so interesting.... (at least for now) mostly for the reason you pointed out... namely that $_RMod$ is not a monoidal category....

(sorry for the late reply .. i had an appointment to get to)

And thanks again for your help Jeremy :)