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.

  • 1
    $\begingroup$ 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). $\endgroup$
    – WhatsUp
    Jul 14, 2020 at 15:56
  • $\begingroup$ 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. $\endgroup$ Jul 14, 2020 at 16:00

1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$ Jul 14, 2020 at 16:11
  • 1
    $\begingroup$ Your understanding of $A$-bilinearity is correct. It doesn't always imply $B$-bilinearity. $B$ can be non-commutative or even non-associative. $\endgroup$
    – WhatsUp
    Jul 14, 2020 at 16:17

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