Exercise 5.1.12

Proof. $\bar{\phi }:L_1\to L_2$ is a field isomorphism, whose restriction to $F_1$ (and co-restriction to $F_2$) is the field isomorphism $\phi :F_1\mapsto F_2$.

$[L_1:F_1]<\infty$. Let $(f_1,\dots ,f_d)$ a basis of $L_1$ over $F_1$. We show that $(\bar{\phi }(f_1),\dots ,\bar{\phi }(f_d))$ is a basis of $L_2$ over $F_2$.

$\text{∙}$ If   ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^d{b}_i\bar{\phi }(f_i)=0$, where $b_i\in F_2$, then, since $\phi :F_1\to F_2$ is surjective, $b_i=\phi (a_i),a_i\in F_1,i=1,\dots d$.

As the restriction of $\bar{\phi }$ to $F_1$ is $\phi$, $b_i=\phi (a_i)=\bar{\phi }(a_i)$. $\bar{\phi }$ being a ring homomorphism,

As the kernel of $\bar{\phi }$ is $0$, ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^d{a}_i{f}_i=0$, where the family ${(f_i)}_{1\le i\le d}$ is free, thus $a_1=\cdots =a_d=0$, and since $b_i=\phi (a_i)$, $b_1=\cdots =b_d=0$. So the family ${(\bar{\phi }(f_i))}_{1\le i\le d}$ is free.

$\text{∙}$ Let $y$ be any element in $L_2$. As $\bar{\phi }$ is surjective, there exists $x\in L_1$ such that $y=\bar{\phi }(x)$. $(f_1,\cdots ,f_d)$ being a basis, there exists $(a_1,\cdots ,a_d)\in F_1^d$ such that $x={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^d{a}_i{f}_i$.

Then

where $b_i=\phi (a_i)\in F_2$. Consequently ${(\bar{\phi }(f_i))}_{1\le i\le d}$ is a basis of $L_2∕F_2$, and so

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