Notice the following tower of fields.
Base field is $K=Q$, intermediate field is $L=Q(\sqrt{3})$ and top field is $M=Q(\sqrt{7})$.
Field $L=Q(\sqrt{3})=\{a+b\sqrt{3}|a,b \in Q\}$.
Field $M=Q(\sqrt{3})=\{a+b\sqrt{3}+c\sqrt{7}+d\sqrt{7x3=21}|a,b,c,d \in Q\}$.
Note that the field $M$ is a vector space over $L$ and $K$. Indeed, from the expression $M=Q(\sqrt{3})=\{a+b\sqrt{3}+c\sqrt{7}+d\sqrt{7x3=21}|a,b,c,d \in Q\}$ one can see that $M$ as vector space over $K$ has dimension $4$.
That is $[M:K]=4$.
There is another way of expression $M$ as a vector space over $L$.
Letting $u_1,u_2$ be elements of this form $u_1=a+b\sqrt{3}$ and $u_2=c+d\sqrt{3}$, one can express elements of $M$ as $u_1+\sqrt{7}(u_2)$.
That is $[M:L]=2$. Similarly $[L:K]=2$. Thus,
$[M:K]=[M:L][L:K]=2*2=4$.
First theorem of Field extensions formalizes these observations.
Theorem 1:Let $K,L,M$ be fields with $K\subset L \subset M$. Then $[M:K]$ is finite if and only if both $[M:L]$ and $[L:K]$ are finite and in that case $[M:K]=[M:L][L:K]$.
Proof is a generalization of above observations.
Assume that $[M:K]$ is finite, then consider $L$ as subspace of $M$ over $K$. As a subspace of finite vector space $[L:K]$ is finite. Any basis that spans $M$ over $K$ (for example, $\{a,b,c,d\}$ in above fields), also spans $M$ over $L$; hence, $[M:L]$ is also finite.
Assume the vector space $M$ over $L$ has $[M:L]=m$ and $L$ over $K$ has $[L:K]=n$, then we need to show that $[M:K]=mn$.
Let $z \in M$. $z$ can be expressed as $z=\sum_i u_i h_i$ where $h_i \in L$. And $h_i=\sum_j c_{ij}v_j$. Putting all this together, $z = \sum_i \sum_j c_{ij}u_i v_j$. $i,j$ range over $m$ and $n$. This is the spanning set for $M$ over $K$.
To show this is independent, we need to show that whenever $\sum_{ij} c_{ij} u_i v_j = 0$ we need to have $c_{ij}=0$.
$\sum_{ij} c_{ij} u_i v_j$ can be rewritten as $\sum_i h_i u_i$ where $h_i \in L$. Since $u_i$ form a basis for $M$ over $L$, we must have each $h_i=0$ and since this means $\sum c_{ij} v_j=0$ and independence of $v_j$ leads to each $c_{ij}=0$.
One consequence of this thoerem is that if the order of $M/L$ is a prime, then this precludes any fields between $M$ and $L$.
Field $L=Q(\sqrt{3})=\{a+b\sqrt{3}|a,b \in Q\}$.
Field $M=Q(\sqrt{3})=\{a+b\sqrt{3}+c\sqrt{7}+d\sqrt{7x3=21}|a,b,c,d \in Q\}$.
Note that the field $M$ is a vector space over $L$ and $K$. Indeed, from the expression $M=Q(\sqrt{3})=\{a+b\sqrt{3}+c\sqrt{7}+d\sqrt{7x3=21}|a,b,c,d \in Q\}$ one can see that $M$ as vector space over $K$ has dimension $4$.
That is $[M:K]=4$.
There is another way of expression $M$ as a vector space over $L$.
Letting $u_1,u_2$ be elements of this form $u_1=a+b\sqrt{3}$ and $u_2=c+d\sqrt{3}$, one can express elements of $M$ as $u_1+\sqrt{7}(u_2)$.
That is $[M:L]=2$. Similarly $[L:K]=2$. Thus,
$[M:K]=[M:L][L:K]=2*2=4$.
First theorem of Field extensions formalizes these observations.
Theorem 1:Let $K,L,M$ be fields with $K\subset L \subset M$. Then $[M:K]$ is finite if and only if both $[M:L]$ and $[L:K]$ are finite and in that case $[M:K]=[M:L][L:K]$.
Proof is a generalization of above observations.
Assume that $[M:K]$ is finite, then consider $L$ as subspace of $M$ over $K$. As a subspace of finite vector space $[L:K]$ is finite. Any basis that spans $M$ over $K$ (for example, $\{a,b,c,d\}$ in above fields), also spans $M$ over $L$; hence, $[M:L]$ is also finite.
Assume the vector space $M$ over $L$ has $[M:L]=m$ and $L$ over $K$ has $[L:K]=n$, then we need to show that $[M:K]=mn$.
Let $z \in M$. $z$ can be expressed as $z=\sum_i u_i h_i$ where $h_i \in L$. And $h_i=\sum_j c_{ij}v_j$. Putting all this together, $z = \sum_i \sum_j c_{ij}u_i v_j$. $i,j$ range over $m$ and $n$. This is the spanning set for $M$ over $K$.
To show this is independent, we need to show that whenever $\sum_{ij} c_{ij} u_i v_j = 0$ we need to have $c_{ij}=0$.
$\sum_{ij} c_{ij} u_i v_j$ can be rewritten as $\sum_i h_i u_i$ where $h_i \in L$. Since $u_i$ form a basis for $M$ over $L$, we must have each $h_i=0$ and since this means $\sum c_{ij} v_j=0$ and independence of $v_j$ leads to each $c_{ij}=0$.
One consequence of this thoerem is that if the order of $M/L$ is a prime, then this precludes any fields between $M$ and $L$.
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