# Showing that a set is a basis of a field as a vector space over a subset of that field

Let $K \subseteq L \subseteq F$ be fields and assume that $\{\alpha_1,\ldots,\alpha_m\}$ is a basis of $F$ as a vector space over $L$ and $\{\beta_1,\ldots,\beta_n\}$ is a basis of $L$ as a vector space over $K$.

Prove that $\{\alpha_i\beta_j : 1\leq i \leq m; 1 \leq j \leq n\}$ is a basis of $F$ as a vector space over $K$.

Would someone please explain to me what exactly it is that I need to show? My lecturer hasn't given us any resources on how to do this and I am quite lost as on how to start this proof. I will be grateful for any tips and hints. Just a brief guideline on how to start this proof would be great. Thanks in advance.

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Do you know the definition of a basis of a vector space? – Michael Hardy Sep 1 '14 at 17:51
Take an arbitrary element of $F$, express it as an $L$-linear combination of the basis $\{\alpha_1, \ldots, \alpha_m\}$, then expand the coefficients as $K$-linear combinations of $\{\beta_1, \ldots, \beta_n\}$. – Dustan Levenstein Sep 1 '14 at 17:53
It means that the defintion of basis of vector space applies. Btw, it is not needed that $F$ is a field; vector space over $L$ is enough. – Hagen von Eitzen Sep 1 '14 at 17:53
because $K,L,F$ are fields, you can use those as the scalars. $F$ as a vector space over $K$ means that the scalars are the elements of $K$. Think $K,L,F$ as $\mathbb{Q},\mathbb{R},\mathbb{C}$ – yess Sep 1 '14 at 17:54
@Henry : It means all of the coefficients in the linear combinations involved are in the field $K$. ${}\qquad{}$ – Michael Hardy Sep 1 '14 at 17:59

You need to show the following two things.

a. Linear combinations of the elements of $\{a_ib_j\}_{i,j}$ span $F$.

b. $\{a_ib_j\}_{i,j}$ are linearly independent.

For a., Let $f\in F$. Then as $\{b_1,\ldots,b_n\}$ is a basis of $F_L$, then $$f=\ell_1 b_1+\cdots+\ell_n b_n, \quad \ell_i\in L.$$ As $\{a_1,\ldots,a_m\}$ is a basis of $L_K$, then $$\ell_i=k_{1i}a_1+\cdots+k_{mi}a_m,\quad k_{ji}\in K$$ and hence $$f=\sum_{i,j}k_{ij}a_ib_j.$$

For b., assume that $$\sum_{i,j}c_{ij}a_ib_j=0.$$ Then $$\sum_{j}b_j\left(\sum_{i}c_{ij}a_i\right)=0,$$ and as the $b_j$'s are linearly independent over $L$, then $$\sum_{i}c_{ij}a_i=0, \quad j=1,\ldots,n,$$ and as the $a_i$'s are linearly independent over $K$, then $$c_{ij}=0,$$ for all $i,j$.

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"$\{\alpha_1,\ldots,\alpha_m\}$ is a basis of $F$ as a vector space over $L$."

This means every member of $F$ can be written in exactly one way as a linear combination of members of $\{\alpha_1,\ldots,\alpha_m\}$, the coefficients in the linear combination being members of $L$.

"$\{\beta_1,\ldots,\beta_n\}$ is a basis of $L$ as a vector space over $K$."

This means every member of $L$ can be written in exactly one way as a linear combination of members of $\{\beta_1,\ldots,\beta_m\}$, the coefficients in the linear combination being members of $K$.

"$\{\alpha_i\beta_j : 1\leq i \leq m; 1 \leq j \leq n\}$ is a basis of $F$ as a vector space over $K$."

This means every member of $F$ can be written in exactly one way as a linear combination of members of $\{\alpha_i\beta_j : 1\leq i \leq m; 1 \leq j \leq n\}$, the coefficients in the linear combination being members of $K$.

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Let $K = \mathbb{Q}$, $L$ the extension $\mathbb{Q}(\sqrt{2})$, and $F=\mathbb{Q}(\sqrt{2},\sqrt{3})$. Then the set $\{1,\sqrt{3}\}$ forms a basis over $L$, meaning that the coefficients used are taken from $L$, but $L$ is also a vector space over $\mathbb{Q}$ with basis $\{1,\sqrt{2}\}$. Your statement now says that $F$ has basis $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$ using coefficients in $\mathbb{Q}$.

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an element of $F$ can be written as $\alpha+\beta\sqrt(2)+\gamma \sqrt(3)+\delta\sqrt(6)$ with rational coefficients. But this can also be written as $\alpha+\beta\sqrt(2)+(\gamma+\delta\sqrt(2))\sqrt(3)$. – Bogaerts Marc Sep 1 '14 at 18:10