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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?

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    Is the first sentence supposed to say "such that the operations on $K$ are a restriction of the operations on $L$"? – 2017-02-19
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    @TannerSwett yes. – 2017-02-19

2 Answers 2

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@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.

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    in the my example, I think, $K$ should be a vector space over $L$ but not. only, I did not understand this statement. – 2017-02-19
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    @Kahler If $K=\Bbb Q$ and $L=\Bbb R$. Then $K$ is not a vector space over $L$. – 2017-02-19
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    @VikrantDesai Why? – 2017-02-19
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    Because we have $2 \in \Bbb Q$. And if we take $\pi \in \Bbb R$, then $2\pi \notin \Bbb Q$. – 2017-02-19
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    @VikrantDesai Thanks. – 2017-02-19
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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: $$\forall x\in L,1_K\cdot x=x.$$