2 "equivalent" definitions of $\mathbf{C}$-algebras.

Question

In my class professor defined a $\mathbf{C}$-algebra (for our purposes) as being a commutative unital ring $R$, that is also a $\mathbf{C}$-vector space in a compatible way (with certain associativity and distributivity identities linking the "scalar multiplication" with the ring operations).

Then he also said we can equivalently define this as being a commutative unital ring equipped with an injective ring homomorphism $f:\mathbf{C} \rightarrow R$. He said we can think of $R$ as containing an isomorphic copy of $\mathbf{C}$.

1) What is the point of the second definition?

2) He didn't give the map explicity, so I tried to "recover" the 2nd definition from the first. I was thinking that we could find such a map by mapping $c \in \mathbf{C}$ to the product $1c \in R$, where $1$ is the unit in $R$.

I'm trying to show this is a ring homomorphism. I can show it is a homomorphism of the additive structures, but Im stuck with showing the same for the multiplicative structure.

Issue is trying to show that $1c_1c_2 = (1c_1)(1c_2)$. Using associativity and distributivity this is the same as showing that $1c_1(c_2 - 1c_2) = 0$. Yet I am not sure that the thing in brackets makes sense, since $c_2 \in \mathbf{C}$ and $1c_2 \in R$.

Thank you.

Answer 1

What is the point of the second definition? There are lots of possible answers. One is that it captures all the equational laws of the first definition in one simple and memorable statement.

As for $(1c_1)(1c_2) = 1(c_1c_2)$, the law that $(ax)(by) = (ab)(xy)$ (where $a, b \in R$ and $x, y \in \Bbb{C}$) should be one of your laws for $R$ to be a $\Bbb{C}$-algebra (possibly phrased as "multiplication in $R$ is bilinear").

(Your transformation of $(1c_1)c_2 - (1c_1)(1c_2)$ into $1c_1(c_2 - 1c_2) = 0$ isn't technically correct for the reason you state - the distributivity law doesn't apply to a mixture of scalar and algebra multiplication. The second definition shows that this doesn't really matter: you can just work as if all the multiplication is going on in $R$, with the field of scalars $\Bbb{C}$ identified with its image in $R$.)