On the proof of Eisenstein's criterion given in Abstract Algebra by Dummit & Foote

Question

My question specifically refers to the proof of Proposition 3 given in chapter 9, section 9.3 of 'Abstract Algebra' by Dummit & Foote. I give the proposition here for convenience:

Let P be a prime ideal of the integral domain $R$, and let $f(x)$ be a monic polynomial of degree greater than or equal to $1$ in $R[x]$. Suppose, bar the leading coefficient, that the coefficients of $f$ are elements of $P$, and that the constant term of $f$ is not in $P^2$. Then $f$ is irreducible in $R[x]$.

It's assumed, for contradiction, that $f(x)$ is reducible in $R[x]$ so there exists non-zero polynomials $g,h \in R[x]$ such that $f$ is the product of these polynomials. We then reduce this equation $\bmod P$, where $P$ is a prime ideal in $R$. By the assumptions on the coefficients of $f(x)$, one is left with $x^n$ equal to the product of $g$ and $h$ where these polynomials now have coefficients in $(R/P)[x]$. Since $P$ is a prime ideal, $R/P$ is a domain and so it follows that $g$ and $h$ must have $0$ constant term i.e. the constant terms of $g$ and $h$ must both be in $P$. But we also have, in $R[x]$, that the constant term of $f$ is equal to the product of the constant terms of g and h implying that the constant term of $f$ is in $P^2$, a contradiction.

I don't understand why we can assume that both the constant terms of $g$ and $h$ must both be in $P$ by the above argument. Couldn't it be the case that say, for example, the constant term of $g$ was in $P$ but the constant term of $h$ was not. In this case, the constant term of the product of $g$ and $h$ would still be in $P$, but we would not be able to conclude with the contradiction given above.

Answer 1

I'll use an $\overline{\alpha}$ to denote the image of $\alpha\in R$ in the quotient $R/P$.

Because $(R/P)$ is a domain, and you have $$\overline{a}x^n = \overline{g}\overline{h},$$ then passing the the field of quotients of $R/P$ (and polynomials over fields have unique factorization) we conclude that $\overline{g}$ and $\overline{h}$ must be constant multiples of a power of $x$ in $(R/P)[x]$. That means that if, for example, $$g(x) = a_mx^m+\cdots + a_1x+a_0,$$ then reducing modulo $P$ you just get $\overline{g}(x)=\overline{a_m}x^m$. So $a_0,\ldots,a_{m-1}$ must lie in $P$. The same argument holds for $h$. So both the constant terms of $g$ and $h$ must both lie in $P$.

Answer 2

It seems to me that the proof, as you have conveyed it, is indeed deficient. The proof I give below uses the hypothesis that $f(0)\notin P^2$ in the beginning, instead of D&F’s, which seems to want to use it at the end.

I use the bar notation, as @ArturoMagidin has in his response. Let $n=\deg f$.

We have our two monic polynomials $\bar g$ and $\bar h$ in $(R/P)[x]$, where $R/P$ is a domain. The hypothesis is that one of these (say $\bar g$) has constant term zero, the other, $\bar h$, has nonzero constant. Let us then write $\bar g=x^s\bar g_0$ with $1\le s\le\deg g<n$ and $\bar g_0$ having nonzero constant term. Thus $\bar f=x^s\cdot\bar g_0\bar h$, which has a nonzero $x^s$-coefficient, equal to the product of the nonzero constant terms of $\bar h$ and $\bar g_0$. But this contradicts the hypothesis is that $\bar f=x^n$.

Answer 3

Here's an argument with author approval:

Let $f(x)=x^n+c_{n-1}x^{n-1}+\cdots+c_1 x+c_0 \in R[x]$, $R$ an integral domain, $P$ a prime ideal of $R$, where $n\geq 1$ and $c_{n-1},...,c_1,c_0 \in P$ but $c_0 \not\in P^2$.

Since $n\geq 1$, $f(x)\not=0$.

Suppose $f(x)$ were reducible, then there exist non-units, $a(x),b(x) \in R[x]$, such that $f(x)=a(x)b(x)$ and since $a(x),b(x)$ are not units, if one is constant, then $f(x)\not=a(x)b(x)$ because the leading coefficient on the left is $1$$(f$ monic$)$ whereas the leading coefficient on the right cannot be $1$. Thus $a(x),b(x)$ must be nonconstant polynomials in $R[x]$.

Reducing the coefficients modulo $P$, with the assumptions on the coefficients of $f(x)$

$$x^n = \overline{a(x)b(x)} \in (R/P)[x]$$

Since $P$ is a prime ideal, $R/P$ is an integral domain, and has no zero-divisors, the constant term $a_0b_0$ of $a(x)b(x)$ is $0$ in $R/P$, at least one of $a_0,b_0$ must be $0$ in $R/P$, i.e. at least one of $a_0,b_0$ is in $P$.

Suppose $a_0 \in P$ while $b_0 \not\in P$, then if

$a(x) = a_i x^i + a_{i-1}x^{i-1}+\cdots +a_1 x+a_0, b(x)=b_j x^j+b_{j-1}x^{j-1}+\cdots+b_0$, where $i+j=n$, and $i,j \geq 1$,

there must be some least indexed nonzero constant $a_k \bmod P$, $1\leq k \leq i<n$, otherwise $\overline{a(x)}$ would be identically $0 \bmod P$, and $\overline{a(x)b(x)} = 0 \not= x^n $. But then $a_k\not=0 \bmod P$ and $b_0\not= \bmod P$, so $a_kb_0\not = 0 \bmod P$ since $R/P$ is a domain, and $\overline{a(x)b(x)}$ would have lowest order term $a_kb_0x^k \not= 0\bmod P$ for $k<n$, and cannot be equal to $x^n$. Contradiction. $b_0 \in P$.