Exercise 4.1.5

Proof. $\text{∙}$ Let $\gamma \in F[{\alpha }_1,\dots ,{\alpha }_{n-1}][{\alpha }_n]$. Write $R=F[{\alpha }_1,\dots ,{\alpha }_{n-1}]$. By definition, there exists a polynomial $p={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{k=0}^d{a}_k{x}_n^k\in R[x_n]$ such that $\gamma =p({\alpha }_n)$, and for every $a_k\in R,0\le k\le d$, there exists $f_k\in F[x_1,\dots ,x_{n-1}]$ such that $a_k=f_k({\alpha }_1,\dots ,{\alpha }_{n-1})$.

Thus

Let $f=\underset{k=0}{\overset{d}{\sum <!--\; FUNCTION\; APPLICATION\; -->}}{f}_k(x_1,\dots ,x_{n-1})x_n^k$. Then $f\in F[x_1,\dots ,x_n]$, and $\gamma =f({\alpha }_1,\dots ,{\alpha }_n)$, so $\gamma \in F[{\alpha }_1,\dots ,{\alpha }_n]$. We have proved

$\text{∙}$ Conversely, let $\gamma \in F[{\alpha }_1,\dots ,{\alpha }_n]$.

There exists $f\in F[x_1,\dots ,x_n]$ such that $x=f({\alpha }_1,\dots ,{\alpha }_n)$.

As $F[x_1,\dots ,x_n]=F[x_1,\dots ,x_{n-1}][x_n]$, $f=\underset{k=0}{\overset{d}{\sum <!--\; FUNCTION\; APPLICATION\; -->}}{f}_k(x_1,\dots ,x_{n-1})x_n^k$, where $f_k\in F[x_1,\dots ,x_{n-1}]$.

So $\gamma =\underset{k=0}{\overset{d}{\sum <!--\; FUNCTION\; APPLICATION\; -->}}{f}_k({\alpha }_1,\dots ,{\alpha }_{n-1}){\alpha }_n^k=\underset{k=0}{\overset{d}{\sum }}{a}_k{x}_n^k$, with $a_k=f_k({\alpha }_1,\dots ,{\alpha }_{n-1})\in F[{\alpha }_1,\dots ,{\alpha }_{n-1}]=R$.

Let $p={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{k=0}^d{a}_k{x}_n^k$. Then $p\in R[x_n]$ and $x=p({\alpha }_n)$, thus $x\in R[{\alpha }_n]=F[{\alpha }_1,\dots ,{\alpha }_{n-1}][{\alpha }_n]$.

The reciprocal inclusion

is proved, and so

Note: in an alternative way, we could write a lemma analogous to Lemma 4.1.9 and show that $F[{\alpha }_1,\dots ,{\alpha }_n]$ is the smallest subring of $L$ containing ${\alpha }_1,\dots ,{\alpha }_n$ (where $L$ is a ring containing $F$ and ${\alpha }_1,\dots ,{\alpha }_n$), and prove as in Exercise 4 that

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