Exercise 13.4.7

Question

\def\imp{\Rightarrow}
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\def\dcro{\mbox{]\hspace{-.15em}]}}

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ewcommand{\ssi} {si et seulement si }

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

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

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

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

Let $f=x^5+20x+16 \in \mathbb{Q}[x]$ be the polynomial of Example 13.4.6. Show that f is irreducible over $\mathbb{Q}$, and compute its discriminant and irreducible factorization modulo $7$.
\begin{proof}
\be
The given polynomial $f$ is irreducible, since $f(x-1)=(x-1)^5+20(x-1)+16=x^5-5x^4+10x^3-10x^2+25x-5$ is irreducible by Sch"onemann-Eisenstein criterion with $p=5$.

The discriminant may be calculated by the formula (cf. Ex.13.2.15):
$$\Delta(f) = 256a^5+3125b^4,$$
where $a=20,b=16$. Then
\begin{align*}
\Delta(f) &= 2^820^5+5^516^4=2^{18}5^5+5^52^{16}=2^{16}5^5(4+1)=2^{16}5^6.
\end{align*}
Reducing the polynomial modulo 7 gives:
$\bar f(x)=x^5+6x+2$.

We have $\bar f(4)=\bar f(5)=0$ and $\bar f(i)
e 0$ for $i\in \{0,1,2,3,6\}$. Division of $\bar f$ by $(x-4)=(x+3)$ and $(x-5)=(x+2)$ gives:
$$\bar f=(x+2)(x+3)(x^3+2x^2+5x+5) ~~ in~\mathbb{F}_7.$$
\ee\end{proof}

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}}

Exercise 13.4.7

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Exercise 13.4.7

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