Exercise 4.1.3

Proof. $\text{∙}$ By hypothesis, $F\subset L$ and ${\alpha }_1,\dots ,{\alpha }_n\in L$.

$1\in F[{\alpha }_1,\dots ,{\alpha }_n],$ so $F[{\alpha }_1,\dots ,{\alpha }_n]\ne \varnothing$.

Let $x,y\in F[{\alpha }_1,\dots ,{\alpha }_n]$. By definition, there exist polynomials $p,q\in F[x_1,\dots ,x_n]$ such that

As $p-q,\mathit{pq}\in F[x_1,\dots ,x_n]$, and as $x-y=(p-q)({\alpha }_1,\dots ,{\alpha }_n),\mathit{xy}=\mathit{pq}({\alpha }_1,\dots ,{\alpha }_n)$, so $x-y\in F[{\alpha }_1,\dots ,{\alpha }_n],\mathit{xy}\in F[{\alpha }_1,\dots ,{\alpha }_n]$.

Conclusion: $F[{\alpha }_1,\dots ,{\alpha }_n]$ is a subring of $L$.

$\text{∙}$ The same argument, where we take rational fractions $p,q$ in place of polynomials show that $p,q\in F(x_1,\dots ,x_n)\Rightarrow p-q,\mathit{pq}\in F(x_1,\dots ,x_n)$, so $x-y=(p-q)({\alpha }_1,\dots ,{\alpha }_n),\mathit{xy}=\mathit{pq}({\alpha }_1,\dots ,{\alpha }_n)\in F({\alpha }_1,\dots ,{\alpha }_n)$. Thus $F({\alpha }_1,\dots ,{\alpha }_n)$ is a subring of $L$.

Moreover, if $x\in F({\alpha }_1,\dots ,{\alpha }_n),x\ne 0$, then $x=\frac{p({\alpha }_1,\dots ,{\alpha }_n)}{q({\alpha }_1,\dots ,{\alpha }_n)}$, where $p,q\in F[x_1,\dots ,x_n]$, and $q({\alpha }_1,\dots ,{\alpha }_n)\ne 0$. Since $x\ne 0$, we have also $p({\alpha }_1,\dots ,{\alpha }_n)\ne 0$.

Hence $\frac{1}{x}=\frac{q({\alpha }_1,\dots ,{\alpha }_n)}{p({\alpha }_1,\dots ,{\alpha }_n)}\in F({\alpha }_1,\dots ,{\alpha }_n)$.

Conclusion: $F({\alpha }_1,\dots ,{\alpha }_n)$ is a subfield of $L$. □