Exercise 5.2.5 (K-linear subspace)

Question

Let $M:K$ be a field extension and $K \subseteq L \subseteq M$. In the proof of Proposition 5.2.4, I said that if $L$ is a
subfield of $M$ then $L$ is a $K$-linear subspace of $M$. Why is that true? And is the converse also true? Give proof or a counterexample.

Richard Ganaye · Accepted Answer

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ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

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ewcommand{\Gal}{\mathrm{Gal}}

\begin{proof} 
\begin{enumerate}
\item[(i)]
Assume that  $K \subseteq L \subseteq M$, and  $L$ is a subfield of $M$.
    \begin{enumerate}
    \item[$\bullet$] $0 \in L$, thus $L 
e \varnothing$.
    \item[$\bullet$] If $a, b \in K$, and $\alpha, \beta \in L$, then $a,b,\alpha, \beta$ are in $L$, because $K \subseteq L$. Since $L$ is a subfield of $M$, $a\alpha + b         \beta \in L$.
    
    (Here the extern law of $M$ is given by
    $$
    \left\{
    \begin{array}{ccl}
    K 	imes M & 	o &M\
    (a,\alpha) & \mapsto & a\cdot \alpha = a \alpha,
    \end{array}
    ight.
    $$
	where $a\alpha$ is the ordinary product of elements of $M$.)
    \end{enumerate}
This shows that $L$ is a $K$-linear subspace of $M$.

\item[(ii)] To give a counterexample, consider the extension $\C : \R$, and $L = \R i = \{a i \mid a \in \R\}$. $L$ is a $\R$-linear subspace of $M$ ($L = \mathrm{span}(i)$), but $L$ is not a subfield of $\C$, because $i \in L$, and $i^2 = -1 
ot \in L$.
\end{enumerate}
\end{proof}

Exercise 5.2.5 (K-linear subspace)

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Exercise 5.2.5 (K-linear subspace)

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