Let \(K\) be a subfield of a field \(L\), and let \(L\) be a subfield of a field \(E
.\) (Thus, \(K \subseteq L \subseteq E\), and \(K\) is a subfield of \(E .\) )
Suppose \(E\) is of dimension \(n\) over \(L\), and \(L\) is of dimension \(m\) over \(K
.\) Show that \(E\) is of dimension \(m n\) over \(K .\)
Suppose \(\left\\{v_{1}, \ldots, v_{n}\right\\}\) is a basis of \(E\) over \(L\) and
\(\left\\{a_{1}, \ldots, a_{m}\right\\}\) is a basis of \(L\) over \(K .\) We claim
that \(\left\\{a_{i} v_{j}: i=1, \ldots, m, j=1, \ldots, n\right\\}\) is a basis
of \(E\) over \(K\). Note that \(\left\\{a_{i} v_{j}\right\\}\) contains \(m n\)
elements.
Let \(w\) be any arbitrary element in \(E .\) Because \(\left\\{v_{1}, \ldots,
v_{n}\right\\}\) spans \(E\) over \(L, w\) is a linear combination of the \(v_{i}\)
with coefficients in \(L\) :
$$
w=b_{1} v_{1}+b_{2} v_{2}+\cdots+b_{n} v_{n}, \quad b_{i} \in L
$$
Because \(\left\\{a_{1}, \ldots, a_{m}\right\\}\) spans \(L\) over \(K\), each
\(b_{i} \in L\) is a linear combination of the \(a_{j}\) with coefficients in \(K\)
:
$$
\begin{aligned}
&b_{1}=k_{11} a_{1}+k_{12} a_{2}+\cdots+k_{1 m} a_{m} \\
&b_{2}=k_{21} a_{1}+k_{22} a_{2}+\cdots+k_{2 m} a_{m} \\
&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\
&b_{n}=k_{n 1} a_{1}+k_{n 2} a_{2}+\cdots+k_{m n} a_{m}
\end{aligned}
$$
where \(k_{i j} \in K\). Substituting in (1), we obtain
$$
\begin{aligned}
w &=\left(k_{11} a_{1}+\cdots+k_{1 m} a_{m}\right) v_{1}+\left(k_{21}
a_{1}+\cdots+k_{2 m} a_{m}\right) v_{2}+\cdots+\left(k_{n 1} a_{1}+\cdots+k_{n
m} a_{m}\right) v_{n} \\
&=k_{11} a_{1} v_{1}+\cdots+k_{1 m} a_{m} v_{1}+k_{21} a_{1} v_{2}+\cdots+k_{2
m} a_{m} v_{2}+\cdots+k_{n 1} a_{1} v_{n}+\cdots+k_{n m} a_{m} v_{n} \\
&=\sum_{i, j} k_{j i}\left(a_{i} v_{j}\right)
\end{aligned}
$$
where \(k_{j i} \in K\). Thus, \(w\) is a linear combination of the \(a_{i} v_{j}\)
with coefficients in \(K\); hence, \(\left\\{a_{i} v_{j}\right\\}\) spans \(E\) over
\(K\). The proof is complete if we show that \(\left\\{a_{i} v_{j}\right\\}\) is
linearly independent over \(K .\) Suppose, for scalars \(x_{j i} \in K\), we have
\(\sum_{i, j} x_{j i}\left(a_{i} v_{j}\right)=0 ;\) that is,
$$
\left(x_{11} a_{1} v_{1}+x_{12} a_{2} v_{1}+\cdots+x_{1 m} a_{m}
v_{1}\right)+\cdots+\left(x_{n 1} a_{1} v_{n}+x_{n 2} a_{2} v_{n}+\cdots+x_{n
m} a_{m} v_{m}\right)=0
$$
or
$$
\left(x_{11} a_{1}+x_{12} a_{2}+\cdots+x_{1 m} a_{m}\right)
v_{1}+\cdots+\left(x_{n 1} a_{1}+x_{n 2} a_{2}+\cdots+x_{n m} a_{m}\right)
v_{n}=0
$$
Because \(\left\\{v_{1}, \ldots, v_{n}\right\\}\) is linearly independent over
\(L\) and the above coefficients of the \(v_{i}\) belong to \(L\), each coefficient
must be 0 :
$$
x_{11} a_{1}+x_{12} a_{2}+\cdots+x_{1 m} a_{m}=0, \quad \ldots, \quad x_{n 1}
a_{1}+x_{n 2} a_{2}+\cdots+x_{m m} a_{m}=0
$$
But \(\left\\{a_{1}, \ldots, a_{m}\right\\}\) is linearly independent over \(K\);
hence, because the \(x_{j i} \in K\),
$$
x_{11}=0, x_{12}=0, \ldots, x_{1 m}=0, \ldots, x_{n 1}=0, x_{n 2}=0, \ldots,
x_{n m}=0
$$
Accordingly, \(\left\\{a_{i} v_{j}\right\\}\) is linearly independent over \(K\),
and the theorem is proved.
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