Can we say anything about the unit of a $k$-algebra $A$ in terms of the unit $1\in k$? Context: Being confused about new concepts and trying to make new distinction to better understand it.
Let's say we have have associative $k$-algebra $A$. Where $k$ denotes a field. An algebra is a $k$-vector space, let's assume that we have a finite basis $V=\{v_1,...,v_n\}$. So now my question is, the unit of $A$, can we say anything about in terms of the unit of $1\in k$ ? I mean must we have that the unit is of the form $(1,0,0,\ldots,0)$ ? Or something like that ?
I now realize that the reason I want to understand this, is probably because I get a bit confused about things like $\alpha \cdot1$ where $\alpha \in k$ and $1\in A$. I first confused $1$ for being in $k$, and thought we have $\alpha \cdot 1 = \alpha$. And then I was like, but wait $\alpha$ is not a vector, this can't be true. But how can I picture $1$ a vector then?
 A: I assume throughout that $A\ne\{0\}$.
You don't know anything about the elements of $A$ in general - even if you pick a basis and identify the elements with traditional vectors, you can't necessarily pick out which one is an identity - for example, it can only be $(1,0,0,\dotsc,0)$ if the identity was the first element of your basis! If you have written the multiplication in terms of the basis, you could solve some equations to find the identity, but this would probably not be pleasant.
For convenience, it might help at first to write $1_k$ for the identity of $k$ and $1_A$ for the identity of $A$. Then you have identites:
$$1_k\cdot a=a\:\forall\:a\in A$$
$$a\cdot 1_A=1_A\cdot a=a\:\forall\:a\in A$$
There is nothing particularly special about $\alpha\cdot 1_A$ from this point of view.
However, an alternative point of view is that you can get a $k$-linear map $k\to A$ by extending the map $1_k\to 1_A$. This embeds $k$ inside $A$, so you can treat scalar multiplication exactly like multiplication in $A$! Under this identification, the element $\alpha$ of $k$ is identified with $\alpha\cdot 1_A$ in $A$, and you have
$$\alpha\cdot a=\alpha\cdot 1_A\cdot a\:\forall\:a\in A$$
so the two points of view are compatible.
Important: Not everybody requires that algebras have multiplicative identities, so make sure that yours really do!
