# An $A$-algebra $B$ carries the same data as a ring map $A \rightarrow B$

I'm trying to show that an $$A$$-algebra $$B$$ has the same data as a ring map $$\phi: A\rightarrow B$$.

An $$A$$-algebra $$X$$ is an $$A$$-module $$B$$ that comes equipped with a bilinear operator $$\times_B: B \times B \rightarrow B$$. So to spell out fully, we have an abelian group $$(B, +_B, 0_B)$$ that is equipped with a bilinear multiplication: $$\times_B$$. We then have that $$B$$ is also a module on $$A$$, so there is an action $$\curvearrowright: A \times B \rightarrow A$$ which obeys the module axioms.

Ring map gives $$A$$-algebra:

Since we have a ring map, this means that $$(A, +_A, \times_A, 0_A, 1_A)$$ and $$(B, +_B, \times_B, 0_B, 1_B)$$ are both rings.

Given a ring map $$\phi: A \rightarrow B$$, we can give $$B$$ an $$A$$-algebra structure by defining the module action to be $$a \curvearrowright b \equiv \phi(a) \times_B b$$. The biliniear operator on $$B$$ is simply the ring multiplication $$\times_B$$.

$$A$$-algebra gives ring map:

Since we have an $$A$$-algebra, $$(A, +_A, \times_A, 0_A, 1_A)$$ is a ring and $$(B, +_B, 0_B)$$ is an abelian group. The $$A$$-module data is given by an action $$\curvearrowright: A \times B \rightarrow B$$, and the algebra / bilinear product data on $$B$$ is given by $$\times_B: B \times B \rightarrow B$$.

I try to define the ring map $$\phi: A \rightarrow B$$. However, the first problem: I don't know that $$B$$ is a ring with unity! So let's assume that the algebra is unital. Then we get a ring $$(B, +_B, \times_B, 0_B, 1_B)$$. Given this, let's define $$\phi(a) \equiv a \curvearrowright 1_B$$.

This lets us prove:

$$\phi(a +_A a') = (a +_A a') \curvearrowright 1_B = (a\curvearrowright 1_B) +_B (a' \curvearrowright 1_B) = \phi(a) +_B \phi(a')$$

Next, we need to show that $$\phi(ab) = \phi(a) \phi(b)$$. I get stuck here:

$$\phi(ab) = (ab) \curvearrowright 1_B = a \curvearrowright (b \curvearrowright 1_B) \\ \phi(a) \times_B \phi(b) = (a \curvearrowright 1_B) \times_B (b \curvearrowright 1_B) \\$$

I have no idea how to proceed. I need some relationship between $$\curvearrowright$$ and $$\times_B$$ which I do not possess. I'd appreciate some help in learning how to continue the proof.

• It would help if you write down explicitly your definition of $A$-algebra. For some people (like me), an $A$ algebra is by definition a ring $B$ endowed with a ring homomorphism $A\rightarrow B$ (everything being commutative). Jul 14, 2020 at 15:56
• Updated. This is approximate, since I'm not sure what the textbook has in mind, this is the first time I've seen the words "A-algebra" in the book. I think it takes it as a pre-requisite. Jul 14, 2020 at 16:00

I think the definition in your textbook means that the operator $$\times_B:B\times B\rightarrow B$$ is $$A$$-bilinear, i.e. it is $$A$$-linear on both operands.
Also, it is probably also assumed that a multiplicative unit $$1_B$$ exists.
This then gives what you want to prove: $$\phi(ab) = (ab)\curvearrowright 1_B = (ab)\curvearrowright (1_B \times_B 1_B) = (a\curvearrowright 1_B)\times_B (b \curvearrowright 1_B) = \phi(a) \times_B \phi(b),$$ where the third equality uses the bilinearity.
• Am I right in understanding that (i) $A$-bilinearity means that $(a \curvearrowright b) \times_B b' = a \curvearrowright (b \times_B b')$, and similarly for the right hand side of $\times_B$? Also, (ii) does $A$-bilinearity also imply $B$-bilinearity? Jul 14, 2020 at 16:11
• Your understanding of $A$-bilinearity is correct. It doesn't always imply $B$-bilinearity. $B$ can be non-commutative or even non-associative. Jul 14, 2020 at 16:17