A field $K$ is an algebra

Question

I learned this definition of an algebra recently. The definition is:

A vectorspace $V$ over a field $K$ is is an algebra if there exists $K$-bilineair map $\varphi\colon V\times V\rightarrow V$ which is also associative.

Now I wanted to check why we can take $V=K$ as a vectorspace and that this becomes an algebra over $K$. I thought the obvious map $\varphi$ must be just multiplying. So if $(x_1,x_2)\in K\times K$ then $\varphi(x_1,x_2)=x_1x_2$. But this is not $K$-bilineair because if $\alpha\in K$ then $\varphi(\alpha(x_1,x_2)=\varphi(\alpha x_1,\alpha x_2)=\alpha^2x_1x_2\neq \alpha x_1x_2=\alpha\varphi(x_1,x_2)$.

My question is what the multiplication must be if I am wrong here. Thanks.

Answer 1

You have shown that the map is not linear. It is, however, bilinear, because $\varphi(\alpha x_1, \beta x_2) = \alpha \beta \varphi(x_1,x_2)$ and so on.

Answer 2

You've proven that the multiplication in $K$ isn't linear, not that it isn't bilinear.

A map $\phi : K \times K \to K$ is bilinear if $\phi(ax, y) = \phi(x, ay) = a\phi(x, y)$, which multiplication does satisfy, by commutativity and associativity.