Exercise 13.3.1

Proof. (a) Let $f(x)=\frac{a_0}{b_0}{x}^n+\frac{a_1}{b_1}{x}^{n-1}+...+\frac{a_n}{b_n}={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=0}^n\frac{a_i}{b_i}{x}^{n-i}$, where $a_i,b_i\in \mathbb{Z}$, and $\nu =\mathrm{lcm}(b_0,b_1,...,b_n)$, then $\mathit{\nu f}(x)={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=0}^n\frac{a_i}{b_i}\nu x^{n-i}={\sum }_{i=0}^n{c}_i{x}^{n-i}$, where $c_i=\frac{a_i}{b_i}\nu \in \mathbb{Z}$.

After multiplication by $c_0^{n-1}$ we have

Hence $g(x)$ is monic and $g(x)=\mathit{\lambda f}(\mathit{\mu x})\in \mathbb{Z}[x]$, where $\lambda =c_0^{n-1}\nu ,\mu =1∕c_0,c_0=\frac{a_0}{b_0}\nu$, i.e. $\lambda ,\mu \in {\mathbb{Q}}^{\text{∗}}$. (b) If ${\alpha }_1,\dots ,{\alpha }_n$ are the roots of $f$ in a splitting field $L=\mathbb{Q}({\alpha }_1,\dots ,{\alpha }_n)$ of $f$ over $\mathbb{Q}$, then ${\beta }_1={\alpha }_1∕\mu ,\dots ,{\beta }_n={\alpha }_n∕\mu$ are the roots of $g$ in $L$, which splits completely over $L$, so $L$ is also a splitting field of $g$. Then

Thus the Galois groups $G_f,G_g$ of $f,g$ over $\mathbb{Q}$ are isomorphic.

Note: If $\tau \in G_f\subset S_n$ corresponds to $\sigma$ for a given numbering of the roots ${\alpha }_1,\dots ,{\alpha }_n$ of $f$, for the corresponding numbering ${\beta }_1,\dots ,{\beta }_n$, $\tau \in G_g$, so $G_f=G_g$ for these chosen numbering. □