4

Paolo gives the following proof (reproduced verbatim) of Eisenstein's criterion.

Proposition 5.17. Let $R$ be a (commutative) ring, and let $\mathfrak p$ be a prime ideal of $R$. Let $$ f = a_0 +a_1 x+\cdots+a_n x^n\in R[x] $$ be a polynomial, and assume that

  • $a_n\notin\mathfrak p$;
  • $a_i\in\mathfrak p$ for $i = 0, \ldots, n - 1$;
  • $a_0\notin\mathfrak p^2$.

Then $f$ is not the product of polynomials of degree < $n$ in $R[x]$.

Proof. Argue by contradiction. Assume $f = gh$ in $R[x]$, with both $d = \deg g$ and $e = \deg h$ less than $n = \deg f$; write $$ g = b_0 + b_1 x+\cdots+b_d x^d,\quad h=c_0 +c_1 x+\cdots+c_e x^e, $$ and note that necessarily $d > 0$ and $e > 0$. Consider $f$ modulo $\mathfrak p$: thus $$ \bar f = \bar g \bar h \text{ in } (R/\mathfrak p)[x], $$ where $\bar f$ denotes $f$ modulo $\mathfrak p$, etc. By hypothesis, $f = \bar a_n x^n$ modulo $\mathfrak p$, where $\bar a_n\ne 0$ in $R/\mathfrak p$. Since $R/\mathfrak p$ is an integral domain, factors of $f$ must also be monomials: that is, necessarily $$ \color{blue}{\bar g = \bar b_d x^d,\quad \bar h = \bar c_e x^e}.\tag{How?} $$ Since $d>0$, $e>0$,this implies $b_0\in \mathfrak p$,$c_0 ∈\mathfrak p$. But then $a_0 = b_0 c_0 ∈ \mathfrak p^2$, contradicting the hypothesis.

Question: Why do the highlighted equations hold? I can only conclude that $\bar g = \bar b_i x^i$ for some $0\le i\le d$ for instance. Why does this $i$ have to be $d$?

darij grinberg
  • 17,777
Atom
  • 3,905
  • 3
    This is indeed a non-sequitur when taken literally. You should replace the $d$ and $e$ in (How?) by new numbers $d'\leq d$ and $e'\leq e$, and observe that $d'+e' = n$. Hence, if one of $d'$ and $e'$ is $0$, then the other is $n$, which is impossible because both $d$ and $e$ are $< n$. Thus you get the desired conclusion. – darij grinberg Apr 03 '23 at 15:49
  • 1
    $d'\le d,e'\le e,d'+e'=d+e.$ Hence not only can none of $d',e'$ be $0,$ but $d'=d$ and $e'=e.$ (This does not use that $d,e<n.$) – Anne Bauval Apr 03 '23 at 15:52
  • @darijgrinberg That answers it! Because, for $a_0 = b_0 c_0$ to be in $\mathfrak p^2$, we must also have one of $d'$, $e'$ be $0$, which you show can't happen. So we are done! – Atom Apr 03 '23 at 15:56
  • 1
    You can also see that $b_dc_e \notin \mathfrak{p}$, hence $\overline{b_d} \neq 0, \overline{c_e} \neq 0$, and $\overline{g}$ (resp. $\overline{h}$) has degree $d$ (resp. $e$). – Aphelli Apr 03 '23 at 15:56
  • I'm not convinced that $d+e=n$; we did not require $R$ itself to be an integral domain. – darij grinberg Apr 03 '23 at 16:18
  • Can you give a fuller reference for "Paolo"? Thanks... – paul garrett Apr 03 '23 at 16:53
  • @paulgarrett That's Paolo Aluffi's Algebra: Chapter 0. – Atom Apr 03 '23 at 17:10
  • @paul garrett Atom looks to be very familiar with him : calling him by his christian name... – Jean Marie Apr 03 '23 at 17:29
  • As explained here in the linked dupe, the criterion exploits that $x^n$ is a product of primes so - as in usual proofs of uniquness of such products, we can use Euclid's Lemma to write the factors in terms of these primes. In fact the criterion easily generalizes by replacing $x$ with any prime polynomial. Recall that. here. $x$ is prime in $,D[x]!\iff! D[x]/x\cong D,$ is a domain. – Bill Dubuque Apr 03 '23 at 18:10
  • 1
    @JeanMarie, I guess so! :) – paul garrett Apr 03 '23 at 18:53
  • 1
    @JeanMarie I only wish... :p – Atom Apr 03 '23 at 19:10
  • 3
    Why is this a "duplicate"? This asks for clarification, not motivation, and has a prime ideal instead of a prime element. – darij grinberg Apr 04 '23 at 20:11
  • Wrongly reopened dupe of this – Bill Dubuque Apr 05 '23 at 18:50
  • @BillDubuque: $R$ is not a domain in general. $x$ is not prime in general. – darij grinberg Apr 05 '23 at 19:54
  • @darij It seems you misunderstood my prior comment. $x$ is prime in the domain $D[x],\ D = R/\frak p.,$ It's truly unfortunate that some users repeatedly reverse proper dupe closures before asking for clarification of their misunderstandings. – Bill Dubuque Apr 05 '23 at 20:03
  • @BillDubuque: I know that the ideas in your linked post are roughly the same ideas that answer this question. That doesn't make it a dupe. A nontrivial amount of thinking is required to transfer those ideas. – darij grinberg Apr 05 '23 at 21:00
  • @darij They are certainly not the same since your answer says nothing at all about the key idea at the heart of the matter, i.e. how the irreducibility result arises from unique factorization of prime powers in the modular image. – Bill Dubuque Apr 05 '23 at 21:19
  • 3
    @BillDubuque: That's because OP understands that point already, and it is the degrees that are confusing him. – darij grinberg Apr 05 '23 at 21:21
  • @darij As any experienced teacher will tell you - students often claim to (fully) understand matters when they really do not. Please be more careful when reversing dupe closures (in particular please read the closure comment more carefully to prevent the above mistake from reoccuring). – Bill Dubuque Apr 05 '23 at 21:43
  • 3
    @BillDubuque: The OP made the effort of marking in blue the point he did not understand, and commenting on the part that he did ("I can only conclude that $\bar g = \bar b_i x^i$ for some $0\le i\le d$ for instance" -- do you think he could have concluded that without the unique factorization of $x^n$ or something equivalent?). I am surprised at the effort you are spending to shoehorn your answer to a different question into fitting this one. – darij grinberg Apr 05 '23 at 23:09

1 Answers1

6

(Expanding my comment into an answer:)

Taken literally, the claim (How?) is indeed a non-sequitur, for the reason you observed. But we can modify it slightly to make it work. Namely, we have $\overline g \cdot \overline h = \overline{gh} = \overline f = \overline{a_n}x^n$ in $\left(R/\mathfrak p\right)\left[x\right]$. Since $R/\mathfrak p$ is an integral domain, we thus have $\overline g = \lambda x^i$ and $\overline h = \mu x^j$ for some nonnegative integers $i$ and $j$ with $i+j = n$ and some nonzero elements $\lambda$ and $\mu$ of $R/\mathfrak p$ (if you are unsure about this, embed the integral domain $R / \mathfrak p$ into its field of quotients, and recall that a univariate polynomial ring over a field is a UFD; but there are more elementary ways to prove this too). Consider these $i$ and $j$.

If we had $i = 0$, then we would have $j = i+j = n$, which would entail $\deg \overline h = j = n$, so that $\deg h \geq \deg \overline h = n$, which would contradict $\deg h < n$. Hence, we cannot have $i = 0$. Thus, $i > 0$. Therefore, the polynomial $\overline g = \lambda x^i$ has zero constant term (in $R/\mathfrak p$). In other words, $\overline{b_0} = 0$ (since the constant term of $\overline g$ is $\overline{b_0}$). In other words, $b_0 \in \mathfrak p$. Similarly, $c_0 \in \mathfrak p$. Hence, $b_0 c_0 \in \mathfrak p^2$. But this contradicts $b_0 c_0 = a_0 \notin \mathfrak p^2$, and this contradiction completes the proof.

darij grinberg
  • 17,777