Exercise 2.4.4

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Let $f \in F[x]$ be monic, and suppose that $f = (x-\alpha_1)\cdots(x-\alpha_n)$ in some field $L$ containing $F$. Prove that $\Delta(f) 
e 0$ if and only if $\alpha_1,\ldots,\alpha_n$ are distinct. This shows that $f$ has distinct roots if and only if its discriminant is nonvanishing.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
Let $f \in F[x]$ such that $f = (x-\alpha_1)\cdots(x-\alpha_n)$ in an  extension $L$ of $F$.

By Proposition 2.4.3, 
\begin{align}
\Delta(f) = \prod_{1\leq i < j \leq n} (\alpha_i - \alpha_j)^2. \label{eq2.4.4:10}
\end{align}

$\bullet$ If $\Delta(f) 
eq 0$, by \eqref{eq2.4.4:10}, for all pairs $(i,j), 1\leq i < j \leq n, \alpha_i - \alpha_j 
eq 0$. The roots $\alpha_i$ are so distinct.

$\bullet$ If the roots $\alpha_i,1 \leq i \leq n,$ are distinct roots, then $\alpha_i - \alpha_j 
eq 0$ for all $(i,j)$ such that $1\leq i < j \leq n$, thus $\Delta(f) 
eq 0$.

\end{proof}

Exercise 2.4.4

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Exercise 2.4.4

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