Last edited by hbghlyj at 2023-3-15 15:49:00Lecture notes
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Notes on transcendence theory |
Corollary 6.6. The polynomial $x^{5}-6 x+3 \in \mathbb{Q}[x]$ is not solvable by radicals.
Proof. The polynomial $P(x):=x^{5}-6 x+3$ is irreducible by Eisenstein's criterion (for $p=3$ ). Furthermore, we compute $P(-1)=8>0$ and $P(1)=-2<0$. Also $\lim _{x \rightarrow \infty} P(x)=\infty$ and $\lim _{x \rightarrow-\infty}=-\infty$. Hence $P(x)$ has roots in $(-\infty,-1),(-1,1)$ and $(1, \infty)$ (by the intermediate value theorem). In particular, $P(x)$ has at least three roots in $\mathbb{R}$. Furthermore, we compute
$$
\frac{\mathrm{d}}{\mathrm{d} x} P(x)=5 x^{4}-6$$and the real roots of $\frac{\mathrm{d}}{\mathrm{d} x} P(x)$ are $\pm \sqrt[4]{\frac{6}{5}}$. If $P(x)$ had more than three roots in $\mathbb{R}$, the polynomial $\frac{\mathrm{d}}{\mathrm{d} x} P(x)$ would have at least three roots in $\mathbb{R}$ by the mean value theorem, which is not possible. We conclude that $P(x)$ has precisely $3=5-2$ roots in $\mathbb{R}$. We can thus conclude from Proposition 6.5 that $\operatorname{Gal}(P) \simeq S_{5}$. Since $S_{5}$ is not solvable (see the end of subsubsection 5.3.1), we conclude from Theorem 5.18 that $P(x)$ is not solvable by radicals. $□$
Czhang271828
Posted at 2022-7-22 17:26:17
