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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?

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  • $\begingroup$ 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.) $\endgroup$ – ancientmathematician Sep 3 '19 at 16:48
  • $\begingroup$ 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 $r \in R$. $\endgroup$ – reuns Sep 3 '19 at 17:10
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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.

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The main difference is this: Algebras have a scalar multiplication defined, rings don't. The fact that the multiplication can be recovered to make some rings into algebras in a canonical way is more or less irrelevant.

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  • $\begingroup$ It's rather misleading to call what OP is describing "canonical", though, since it requires you to know what $k1$ is. $\endgroup$ – Eric Wofsey Sep 3 '19 at 17:27
  • $\begingroup$ @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. $\endgroup$ – Arthur Sep 3 '19 at 18:42

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