# Definition algebra over commutative ring (injectivity needed?)

I'm aware that many differently looking definitions exist for an algebra over a commutative ring $A$. For me, the most natural definition of an algebra over a commutative ring $A$ consists of a tuple $(B, f)$ where $B$ is a ring (not necessarily commutative) and $f: A \to B$ is a ring homomorphism satisfying $f(A) \subseteq Z(B)$, i.e. $bf(a) = f(a)b$ for all $a \in A, b \in B$.

Some authors seem to require that the homomorphism $f$ is also injective, such that $A$ can be embedded into $B$. I can't seem to find whether this assumption is common.

In my undergraduate thesis I want to use that, for an ideal $I$ of a commutative ring $A$ and some $a \in A$ fixed, there exists a unique $A$-algebra homomorphism $$\phi : A[X] \to A/I$$ which maps $X$ to $a + I$. For this I need to be able to consider $A/I$ as an $A$-algebra, which I only can when I drop the injectivity of $f$. Is this clear to a mathematical audience, without having to define what I see as an algebra in my paper?

• Really, some authors require $f$ to be injective? I have never seen that... Commented Mar 28, 2016 at 22:23
• This injectivity assumption is clearly useless and very bad. You certainly want every ring to be a $\mathbb{Z}$-algebra. Commented Mar 28, 2016 at 22:24
• I'm not sure if this counts, but when Shafarevich talks about finite maps of varieties, the integral $k$-algebra extensions they correspond to are always injective, which definitely causes confusion; see, e.g., this MO answer. Commented Mar 29, 2016 at 1:07
• When using the word 'extension' I find it a more natural assumption that the maps need to be injective. Commented Mar 29, 2016 at 8:26

Every ring can be considered as a $\mathbb{Z}$-algebra and, of course, the (unique) homomorphism is not necessarily injective.
It would be indeed quite strange that a quotient of an $A$-algebra is not necessarily an $A$-algebra, which is exactly the example you make.
Of course, if $A$ is a field, then the homomorphism is injective, provided the algebra is not the trivial ring.