Understanding of an example of vector space 
Let $L\supseteq K$ be a field such that the operations on $K$ are a restriction of the operations on $L$. Then $L$ is a vector space over $K$.

I didn't understand this example. How was $L$ a vector space over $K$? Can you explain clearly? 
 A: @CFalcon 's answer tells the abstract story. Here's an example you are familiar with that might help you understand. 
The real numbers $\mathbb{R}$ are a subset of the complex numbers $\mathbb{C}$. In fact they are a subfield. Now suppose you temporarily forget that you can multiply complex numbers but remember that you can multiply a complex number by a real number. Then you can think of $\mathbb{C}$ as just $\mathbb{R}^2$ - a two dimensional real vector space. The vectors $(1,0) = 1$ and $(0,1)= i$ are a basis.
For a second somewhat more complicated example prompted by the comments, consider the field of rational numbers $\mathbb{Q}$ as a subfield of the real numbers $\mathbb{R}$. You can add real numbers and multiply them by rationals, so the real numbers are a vector space with scalars the field of rationals. What makes this example harder to understand is that the dimension of $\mathbb{R}$ as a vector space over $\mathbb{Q}$ is infinite - even uncountable. You don't want to try to think about a basis.
A: The setting means that $K$ is a subfield of $(L,+,\times)$, therefore one can define an external multiplication by elements of $K$ on $L$ in the following manner:
$$\cdot\colon\begin{cases}K\times L&\rightarrow & L\\(\lambda,x) & \mapsto &\lambda\times K\end{cases}.$$
It is easy to check that $(L,+,\cdot)$ is a vector space over $K$, namely:


*

*The couple $(L,+)$ is an abelian group.

*The map $\cdot$ is compatible with the multiplication in $K$ i.e. one has:
$$\forall (\lambda,\mu)\in K^2,\forall x\in L,\lambda\cdot(\mu x)=(\lambda\mu)\cdot x.$$

*The map $\cdot$ is distributive with respect to addition in $L$ i.e. one has: 
$$\forall \lambda\in K,\forall (x,y)\in L^2,\lambda\cdot(x+y)=\lambda x+\lambda y.$$

*The map $\cdot$ is distributive with respect to addition in $K$ i.e. one has:
$$\forall (\lambda,\mu)\in K^2,\forall x\in L,(\lambda+\mu)\cdot x=\lambda x+\mu x.$$

*The identity element of $K$ is an identity element for $\cdot$ i.e. one has:
$$\forall x\in L,1_K\cdot x=x.$$
