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Let $A$ be an integral domain, $\mathfrak{p}$ a prime ideal of $A$. Let $f(X)=a_nX^n+...+a_1X+a_0$ be a primitive and non constant polynomials of $A[X]$. We also suppose that $a_k \in \mathfrak{p}$ for all $k\in \{0,...,n-1\}$ and $a_n \not \in \mathfrak{p}$. We write $f=gh$ with $g,h \in A[X]$ non constants.
Wlog let's write $g(X)=b_mX^m + ...+ b_1X+b_0$ and $h(x)=c_lX^l +...+c_1X +c_0$, with $l,m\ge 1$ and $l+m=n$.

We then look at the previous equality in $(A/\mathfrak{p})[X]$ and we obtain that : $\overline{f}=\overline{gh}=\overline{f}.\overline{g}$ hence $\overline{a_n}X^n=\overline{f}.\overline{g}$.

We want to prove that $\overline{g}=\overline{b_m}X^m$ and $\overline{h}=\overline{c_l}X^l$.

Now in my lecture, it is suggested that as the ring $A/\mathfrak{p}$ is an integral domain we could work in its fraction field $F_{A/\mathfrak{p}}$.
Hence the equality seen in $F_{A/\mathfrak{p}}[X]$ gives automatically (by fields property) that $\overline{g}= \overline{b_m} X^m$ and $\overline{h}=\overline{c_l} X^l$ with $\overline{b_m}, \overline{c_l} \in (F_{A/\mathfrak{p}})^{\times}, \ l,m\ge 1$ and $l+m=n$.

But is it really necessary to work with the fraction field of $A/\mathfrak{p}$ here ?

I mean, if we work on the integral domain $(A/\mathfrak{p})[X]$, we write : $\displaystyle \overline{a_n}X^n=\left(\sum_{r=0}^{m}\overline{b_r}X^r\right).\left(\sum_{j=0}^{l}\overline{c_j}X^j\right)=\overline{b_m}.\overline{c_l}X^{m+l}+\left(\sum_{r=1}^{m-1}\overline{b_r}X^r\right).\left(\sum_{j=1}^{l-1}\overline{c_j}X^j\right) +\overline{b_0}.\overline{c_0}$.

Hence matching coefficients, using the hypothesis on the $a_k$ and using property of integral domains enable us to deduce by induction that for instance, all the $\overline{b_r}=0$ for $r\in\{0,m-1\}$. But on the other hand all the $\overline{c_j}$ for all $j\in\{0,...,l-1\}$are not necessarily $\overline{0}$ and $\overline{h}$ would not be of the form $\overline{h}=\overline{c_l}X^l$.

So do we use the fraction field to ensure that the divisors of $X^n$ are of the form $X^{t}$ with $t\in\{1,...,n-1\}$ here ?

Thanks in advance !

Maman
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  • I would assume that the lemmas the author is referring to work around the fract that $F_{A/\mathfrak p}[X]$ is a UFD, which would still apply if $A/\mathfrak p$ were a UFD, ut not necessarily to a generic domain. It's entirely possible that your procedure works. It should be noted that in your set-up you are missing all the monomials in the form $\overline b_r\overline c_0X^r$, $\overline b_0\overline c_rX^r$, $\overline b_m\overline c_kX^{m+k}$, $\overline b_h\overline c_lX^{h+l}$ for $r\ge 1$, $0\le k<l$, $0\le h<m$. –  Nov 26 '21 at 23:27
  • @SaucyO'Path indeed you are right ! But with that procedure I only ensure that the coefficients of $\overline{g}$ are all $\overline{0}$ except $\overline{b_m}$ but in any case $\overline{h}$ could be written $\displaystyle\sum_{j=0}^{l}\overline{c_j}X^j$ with $\overline{c_j}\neq \overline{0}$. – Maman Nov 27 '21 at 16:26

1 Answers1

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No, in this method of proof of Eisenstein's Criterion we need only that if $D$ is a domain then $X$ is $\rm\color{#c00}{prime}$ in $D[X]$ (by $D[X]/(X)\cong D$ is a domain, or directly by by examining lead coefs), and the further key idea that products (so powers) of $\rm\color{#c00}{primes}$ always have unique prime factorizations. So $\,a X^n = f g\,\Rightarrow f = b X^j,\ g = c X^k,\ a=bc, n = j+k.\,$ The uniqueness claim has an obvious inductive proof using Euclid's Lemma, here using $\,X\mid fg\Rightarrow X\mid f\,$ or $\,X\mid g,\,$ same as in $\Bbb Z$.

Note that the uniqueness proof works even if the domain $D$ is not a UFD (that other irreducibles may not be prime, or that some elements may have no factorizations into irreducbles does not affect this specific case - see the related idea at the heart of Nagata's Lemma).

Bill Dubuque
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  • Thanks a lot for the references ! Indeed in the case of a general domain irreducible elements are not necessarily prime elements. Hence we do not have the factorization into irreducible elements but only into prime elements ! Is that it ? – Maman Nov 27 '21 at 16:46