# Difference between rings and algebras

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?

• You'll have to remind me what $k1$ is defined to be. As far as I can see if one takes $\mathbb{F}_4$ as an $\mathbb{F}_4$ algebra, throws away the algebra structure, then the resulting unital ring can be made into an $\mathbb{F}_4$ algebra in two different ways (swapping the actions of the two $3$-roots of unity.) Sep 3 '19 at 16:48
• For a ring $R$ and a field $F$ that $R$ is a $F$-algebra means $R$ is a left $F$-module and the $F$-module action commutes with the multiplication of $R$. This is what you need to obtain a ring homomorphism $P(x) \in F[x] \mapsto P(r) \in R$ for all $r \in R$. Sep 3 '19 at 17:10

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.
• It's rather misleading to call what OP is describing "canonical", though, since it requires you to know what $k1$ is. Sep 3 '19 at 17:27
• @EricWofsey That is true. I think the only time you have a canonical reconstruction is when $\Bbb F$ is contained in the ring. Or at least a unique or distinguished homomorphism from $\Bbb F$ to the ring. Sep 3 '19 at 18:42