I am working through a lecture where two definitions for the extension of a field are given and I have troubles to understand them.
All considered rings are commutative with identity. Let $K$ and $L$ be fields.
Def. 1: $L$ is an extension of $K$, if $K \subset L$.
Def. 1.1: $L$ is an extension of $K$, if $L$ is a $K$-algebra.
Remarks: A $K$-algebra is a ring $A$ which is a $K$-module, s.t. the multiplication $A \times A \to A$ is $K$-bilinear ($\iff (k_1a_1)(k_2a_2)=k_1k_2a_1a_2$ where $k_i \in K, a_i \in A$).
The second definitions seems to be much more involved than the first one (at least for myself).
My thoughts: I first reconsidered the definition of an algebra according to the book "Advanced Linear Algebra, 3rd edition" by Steven Roman. An algebra $\mathcal{A}$ over a field $F$ is a nonempty set $A$, together with three operations, called addition (denoted by $+$), multiplication (denoted by juxtaposition) and scalar multiplication (also denoted by juxtaposition), for which the following properties hold: $(i) \mathcal{A}$ is a vector space over $F$ under addition and scalar multiplication, $(ii) \mathcal{A}$ is a ring under addition and mulitplication, $(iii)$ If $r \in F$ and $a,b \in \mathcal{A}$, then $r(ab)=(ra)b=a(rb)$.
Now, property $(i)$ should be equivalent to saying "$\mathcal{A}$ is an $F$-module", since a vector space is just a special type of module, that is a vector space is a module over a field. It seems to me that it is not necessarily required that $\mathcal{A}$ is a commutative ring with identity, which seems to be assumed in definition $1.1$, but just a ring according to property $(ii)$.
Property $(iii)$ on the other hand, looks different than saying that the multiplication $A \times A \to A$ is $K$-bilinear, which might have something to do with my problem as well.
In summary, however, it is completely unclear to me in essence why all this should be the same as saying "$K \subset L$".
Question: Why are the above definitions considered to be equivalent (my algebra skills are very rusty, so please keep that in mind)?
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 $K\subseteq L$.