Exercise 2.4.8

Question

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In this exercise, you will prove that although $\Delta$ factors in $F[x_1,\ldots,x_n]$, it is irreducible in $F[\sigma_1,\ldots,\sigma_n]$ when $F$ has characteristic different from 2. To begin the proof, assume that $\Delta = AB$, where $A,B \in F[\sigma_1,\ldots,\sigma_n]$ are nonconstant.
\begin{enumerate}
\item[(a)] Using the definition of $\Delta$ and unique factorization in $F[x_1,\ldots,x_n]$, show that $A$ is divisible in $F[x_1,\ldots,x_n]$ by $x_i-x_j$ for some $1\leq i < j \leq n$.
\item[(b)] Given $1\leq i < j \leq n$ and $1\leq l < m \leq n$, show that there is a permutation $\sigma \in S_n$ such that $\sigma(i)= l$ and $\sigma(j)=m$.
\item[(c)]Use part (a) and (b) to show that $A$ is divisible by $x_l-x_m$ for all $1\leq l < m \leq n$.
\item[(d)] Conclude that $A$ is a multiple of $\sqrt{\Delta}$ and that the same is true for $B$.
\item[(e)] Show that part (d) implies that $A$ and $B$ are constant multiples of $\sqrt{\Delta}$ and explain why this contradicts $A,B \in F[\sigma_1,\ldots,\sigma_n]$.
\item[(f)] Finally, suppose that $F$ has characteristic 2. Prove that $\Delta$ is not irreducible.

\end{enumerate}

Richard Ganaye · Accepted Answer

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

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

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\begin{proof}
\begin{enumerate}
\item[(a)] 
Suppose that $\Delta = A B$, where $A,B \in F[\sigma_1,\cdots,\sigma_n]$ are nonconstant.

As $A$ is not a constant, it is divisible by an irreducible factor $h \in F[x_1,\cdots,x_n]$. This irreducible factor $h$ divides $\Delta$, whose only irreducible factors are associate to $x_i-x_j,1\leq i<j\leq n$. $F[x_1,\cdots,x_n]$ being a factorial domain, there exists a pair of subscripts $(i,j)$ and $\lambda \in F^*$ such that $h =\lambda ( x_i-x_j),1\leq i<j\leq n$.

Conclusion :

$A$ is divisible in $k[x_1,\ldots,x_n]$ by a factor $x_i-x_j$, for some $(i,j), 1 \leq i <j\leq n$.

\item[(b)] 
The set $U = \gcro1,n\dcro \setminus \{i,j\}$ and $V = \gcro1,n\dcro \setminus \{l,m\}$ have same cardinality $n-2$, so there exists a bijection $f : U 	o V$.

Let $\sigma :\gcro1,n\dcro 	o \gcro1,n\dcro$ defined by $\sigma(k) = f(k)$ if $k\in U, \sigma(i) = l, \sigma(j) = m$. Then $\sigma$ is bijective (the application $	au$  defined by $	au(m) = f^{-1}(m)$ if $m \in V, 	au(l)=i,	au(m) = j$ satisfies $	au \circ \sigma = \sigma \circ 	au = e$).

There exists $\sigma \in S_n$ such that $\sigma(i) = l, \sigma(j)=m$.

\item[(c)] 
By (a),
$A = (x_i-x_j) C,\ C \in k[x_1,\cdots x_n]$.

As $A$ is symmetric, using the permutation $\sigma$ of (b), 
\begin{align*}
A &=\sigma\cdot A\
&=\sigma\cdot[(x_i-x_j) C]\
&= \sigma \cdot(x_i-x_j)\ \sigma\cdot C\
&=(x_l-x_m) (\sigma\cdot C).
\end{align*}

So $A$ is divisible by $x_l-x_m,\ 1\leq l<m\leq n$.

\item[(d)] 
As these factors are irreducible and not associate, their product divides $A$, thus 
$$\sqrt{\Delta} = \prod_{1\leq l <m\leq n} (x_l-x_m) \mid A.$$

The same reasoning applies to $B$, which is also divisible by $\sqrt{\Delta}$.

\item[(e)] $A = A_1 \sqrt{\Delta}, B = B_1 \sqrt{\Delta}$, where $A_1,B_1 \in F[x_1,\cdots,x_n]$.

Thus $\Delta = AB = A_1B_1 \Delta$, with $\Delta 
eq 0$, therefore $A_1B_1=1$, which implies that $A_1=a\in F^*,B_1=b\in F^*$ :

$$A = a \sqrt{\Delta}, B = b \sqrt{\Delta},\ a,b\in F^*.$$
But $A\in F[\sigma_1,\ldots,\sigma_n]$, thus for all transposition $	au$ in $S_n$, $$A =	au\cdot A = 	au\cdot(a\sqrt{\Delta})=a\, 	au\cdot \sqrt{\Delta} = -a\sqrt{\Delta}=-A.$$
So $2A = 0$, and as the characteristic of $F$ is not 2, $A=0$, so $\Delta = 0$, which is a contradiction.

Conclusion : $\Delta$ is irreducible in  $F[\sigma_1,\cdots,\sigma_n]$.

\item[(f)] 
If the characteristic of $F$ is 2, then $\sqrt{\Delta}$ is symmetric, since for all transposition $	au$, $	au.\sqrt{\Delta} = - \sqrt{\Delta} = \sqrt{\Delta}$.

Thus $\Delta = (\sqrt{\Delta})^2 = D^2$, where $D = \sqrt{\Delta} \in F[\sigma_1,\cdots,\sigma_n]$ : therefore $\Delta$ is not irreducible in $F[\sigma_1,\cdots,\sigma_n]$ if the characteristic of $F$ is 2.
\end{enumerate}
\end{proof}

Exercise 2.4.8

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Exercise 2.4.8

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