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Definition from ProofWiki:

An algebra over a ring $(G_R,\oplus)$ is an $R$-module $G_R$ over a commutative ring $R$ with a bilinear mapping $\oplus:G^2→G$.

Context: I want to show that an algebra over a ring with many units has many units, but that seems to require that an algebra over a ring with unity has unity (with respect to $\oplus$). I don't why this should be true, but for some obvious examples it is (e.g. products of rings with unity have unity).

Does an algebra over a ring with unity have unity? why?

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    $\begingroup$ No, for instance a Lie algebra does not have a unit, even over a field. Although often people consider unital, associative algebras and these have a unit of course. $\endgroup$ – Myself Dec 12 '15 at 12:46
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    $\begingroup$ The simplest bilinear mapping is the zero mapping. If you use that, do you get a unity? $\endgroup$ – GEdgar Dec 12 '15 at 13:17
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No. Consider the algebra $\mathcal{A}$ of functions $f: \mathbb{R} \to \mathbb{R}$ which have compact support over the ring $\mathbb{R}$, and let $\oplus$ be the usual multiplication of functions. This ring has no identity (since for any given nonzero function $f$, we can find a nonzero function $g$ so that $\operatorname{supp} f \cap \operatorname{supp} g = \emptyset$, so $f\oplus g = 0$).

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Another example with nonzero associative multiplication is $\oplus_{j=1}^\infty F$ which is the same thing as the subset of $\prod_{j=1}^\infty F$ consisting of elements which are zero outside of a finite set.

Strictly speaking it is the same idea as Strants solution except I'm using the functions $\Bbb N\to F$, but I feel like it is a bit trimmer and easier to see what is going on.

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