It is clear that, if we forget scalar multiplication, associative algebras reduce to rings. It the resulting ring is unital however, one can recover scalar multiplication $ka:=(k1)a$, for all $k\in\mathbb{F}$, the field where the algebra is defined, and $a$ in the algebra. So what is the difference between unital rings and associative unital algebras?
Difference between rings and algebras
abstract-algebraalgebrasring-theory
Related Solutions
What kind of ring is $U(L)$?
Since representations of Lie algebras behave like representations of groups (the category has tensor products and duals, for example), you should expect that the universal enveloping algebra $U(\mathfrak{g})$ has some extra structure which causes this, and it does: namely, it is a Hopf algebra (a structure shared by group algebras). The comultiplication is defined on basis elements $x \in \mathfrak{g}$ by $$x \mapsto 1 \otimes x + x \otimes 1$$
(this is necessary for it to exponentiate to the usual comultiplication $g \mapsto g \otimes g$ on group algebras) and the antipode is defined by $$x \mapsto -x$$
(again necessary to exponentiate to the usual antipode $g \mapsto g^{-1}$ on group algebras).
This is an important observation in the theory of quantum groups, among other things.
Thus, via the envelopping algebra Lie algebras and their represnetations cn be studied from a ring theoretic point of view. Is the cconverse true in some sense?
Not in the naive sense, the basic problem being that if $A$ is an algebra and $L(A)$ that same algebra regarded as a Lie algebra under the bracket $[a, b] = ab - ba$, then a representation of $L(A)$ does not in general extend to a representation of $A$, but to a representation of $U(L(A))$, which may be a very different algebra (take for example $A = \text{End}(\mathbb{C}^2)$).
Of course there are other relationships between ring theory and Lie theory. For example, if $A$ is a $k$-algebra then $\text{Der}_k(A)$, the space of $k$-linear derivations of $A$, naturally forms a Lie algebra under the commutator bracket. Roughly speaking this is the "Lie algebra of $\text{Aut}(A)$" in a way that is made precise for example in this blog post.
Algebra objects or monoid objects can be defined in any monoidal category. When $R$ is a commutative ring, then the category of left $R$-modules has a monoidal structure given by $\otimes_R$, and algebras in that category coincide with $R$-algebras. But the category of left $R$-modules has no "natural" monoidal structure when $R$ is not commutative - this is the conceptional reason why one usually assumes $R$ to be commutative (of course it has lots of monoidal structures, for example the one given by the categorical product, but this has no interesting algebras).
However, if $R$ is any ring, then the category of $R$-bimodules is a monoidal category (although not symmetric monoidal ...). According to the general definition, an algebra in that monoidal category is an $R$-bimodule $_R A _R$ equipped with a homomorphism $\mu : {}_R A \otimes_R A _R \to {}_R A _R$ of $R$-bimodules and a homomorphism $\eta : {}_R R_R \to {}_R A _R$ of $R$-bimodules such that certain diagrams commute. Equivalently, we have a ring structure $(U(A),*,1)$ on the underlying abelian group $U(A)$ of $A$ such that for all $r \in R$ and $a,b \in A$:
- $r \cdot (a * b) = (r \cdot a) * b$
- $(a * b) \cdot r = a * (b \cdot r)$
- $a * (r \cdot b) = (a \cdot r) * b$
- $r \cdot 1 = 1 \cdot r$
The last two axioms are missing in your definition. We obtain a variety of algebras in the sense of universal algebra and hence limits & colimits exist.
Notice that we really need the axiom $r \cdot 1 = 1 \cdot r$ in order to ensure that $R$ is the initial $R$-algebra.
Caution. If $S$ is a commutative ring with underlying noncommutative ring $R$ (usually, one writes $S=R$, which is confusing here), then any $S$-algebra yields an $R$-algebra, but not every $R$-algebra is of this form! In fact, we only get those $R$-algebras for which $r \cdot a = a \cdot r$ holds.
See MO/21899 for a discussion about definitions of algebras over non-commutative rings.
Best Answer
Notice that to define $ka=(k1)\cdot a$, you have to already know what $k1$ is. If all you have is a ring, then you don't know what $k1$ is yet. The same (unital) ring can have different $\mathbb{F}$-algebra structures, since $k1$ could be defined differently in them.