$L$ is an extension of $K$, if $L$ is a $K$-algebra

Solution 1:

If $K\subseteq L$ then $L$ is indeed an algebra over $K$, where the scalar multiplication is obtained by restricting the usual field multiplication to $K\times L$. All the three axioms of algebra clearly hold.

Conversely, suppose $L$ is an algebra over $K$. Then we can define $\varphi:K\to L$ by $\varphi(k)=k\cdot 1_L$, where here $\cdot$ means the scalar multiplication. You can check this is a homomorphism of fields. Indeed, the only non trivial part is why is it closed under multiplication. This holds because:

$\varphi(k_1k_2)=(k_1k_2)\cdot 1_L=k_1\cdot (k_2\cdot 1_L)=1_L(k_1\cdot(k_2\cdot 1_L))=(k_1\cdot 1_L)(k_2\cdot 1_L)=\varphi(k_1)\varphi(k_2)$

Where we used the third axiom of an algebra in your list.

Alright, so $\varphi$ is a field homomorphism, and in particular it is injective. So up to injection we indeed have $K\subseteq L$.