# A puzzling point in proof of Eisenstein Criterion for irreducible polynomials on Integral Domain

Theorem (Eisenstein Criterion). Let $$P$$ be a prime ideal of the integral domain $$R$$ and let $$f(x) = x^n +a_{n-1}x^n+\cdots+a_1 x +a_0$$ be a polynomial in $$R [x]$$ (here $$n \geq 1$$). Suppose $$a_{n-1},\dots, a_1, a_0$$ are all elements of $$P$$ and suppose $$a_0$$ is not an element of $$P^2$$. Then $$f(x)$$ is irreducible in $$R[x]$$.

Proof. Suppose $$f(x)$$ were reducible, say $$f(x) = a(x)b(x)$$ in $$R[x]$$, where $$a(x)$$ and $$b(x)$$ are nonconstant polynomials. Reducing this equation modulo $$P$$ and using the assumptions on the coefficients of $$f(x)$$ we obtain the equation $$x^n = \overline{a(x)b(x)}$$ in $$(R/ P)[x]$$, where the bar denotes the polynomials with coefficients reduced mod $$P$$. Since $$P$$ is a prime ideal, $$R/P$$ is an integral domain, and it follows that both $$\overline{a(x)}$$ and $$\overline{b(x)}$$ have $$0$$ constant term, i.e., the constant terms of both $$a (x)$$ and $$b(x)$$ are elements of $$P$$ . But then the constant term $$a_0$$ of $$f(x)$$ as the product of these two would be an element of $$P^2$$, a contradiction. $$\Box$$

This proof is from Abstract Algebra by Dummit & Foote. The puzzling point is bold-italic in the proof. Why it is not possible that only one constant of $$\overline{a(x)}, \overline{b(x)}$$ is $$0$$ and the other is not? I don't think it violates the rule of integral domain in this case.

Could any one give me some idea ? Thank you in advance!

• The proof as given is either incomplete or incorrect. This is often a stumbling point for students first learning the proof, so it is pedagogically inexcusable to omit justification of this crucial point. – Bill Dubuque Nov 3 '20 at 12:26

Match coefficients. You obtain a system of equations where if $$\deg a = i$$ and $$\deg b = j$$ then $$i+j = n$$, and, denoting their coefficients by $$\alpha_s$$ and $$\beta_t$$, you have $$\alpha_i \beta_j = 1$$ and then intermediary sums of cross terms, all of which must be $$0$$. Then the proof as in Dummit and Foote resumes - if these are zero, they are elements of $$P$$. Ideal arithmetic says that such products are elements of $$P^2$$, obtaining the contradiction.

Edit: I did not see your more refined question at the end. Suppose that $$a(x)$$ has $$0$$ constant term, and it's $$b(x)$$ that doesn't. Then $$a \mod P$$ would have to be identically $$0$$. If not, you could look at its lowest nonzero coefficient and this would eventually be multiplied by the constant term in $$b \mod P$$. The result is a nonzero monomial with $$\deg < n$$ in the product $$ab \mod P$$, since $$P$$ is a prime ideal. But that means that the whole polynomial $$a(x)$$ is in $$P$$, and therefore the whole polynomial is in $$P$$ since ideals are closed under multiplication. But $$f$$ is monic, so this is impossible.

• By matching coefficients I get $$\begin{cases} \alpha _0\beta _0=0,\\ \alpha _1\beta _0+\alpha _0\beta _1=0,\\ \alpha _2\beta _0+\alpha _1\beta _1+\alpha _0\beta _2=0,\\ \cdots\\ \alpha_i \beta_j=0\\ \end{cases}$$ The coefficients of cross sums are zero indeed, but how to make sure those $\alpha_s$ and $\beta_t$ when $\alpha<i$ and $\beta<j$ are all zero? – atlantic0cean Nov 3 '20 at 3:10
• I've answered what your question actually is now, I think. Sorry about that.. – Alfred Yerger Nov 3 '20 at 3:39
• @Algred Yerger Thank you very much for your contribution! I think the proof by Dummit & Foote is actually deficient while your proof works. – atlantic0cean Nov 3 '20 at 4:47
• Certainly they could have elaborated some more on this point. – Alfred Yerger Nov 3 '20 at 5:19
• @atlantic0cean I added a proof which greatly clarifies the inferences you ask about above. – Bill Dubuque Nov 3 '20 at 11:14

That proof of Eisenstein's criterion is clearer using unique factorization of prime products.

$$\rm\color{#0a0}{Assume}$$ that $$f$$ is reducible $$\,f = g\,h,\,$$ $$\,i = \deg g,\ j = \deg h,\ \color{#c00}{i,j\ge 1}$$, wlog $$\,g,h\,$$ monic.

This factorization maps to $$\,x^n = \bar g\,\bar h\,$$ in $$\,\bar R := R/P,\,$$ a domain, so $$\,x\,$$ is prime in $$\bar R[x]$$.

Since it is a prime power: $$\,x^n = x^k\, x^{n-k}\,$$ are the only possible monic factorizations.

Thus $$\,\bar g = x^i,\ \bar h = x^j,\,$$ so $$\,\color{#c00}{i,j\ge 1}\Rightarrow\,\bar g(0)\!=\!0\!=\!\bar h(0),\,$$ i.e. $$\,g(0),h(0)\in P$$.

Therefore $$\,f(0) = g(0)h(0)\in P^2,\,$$ contra $$\rm\color{#0a0}{hypothesis},\,$$ hence $$f$$ is irreducible.

Alternative direct inductive proof that $$\,\bar g(0)\!=\!0\!=\!\bar h(0),\,$$ i.e. $$\,x\mid \bar g,\bar h.\,$$ If not, wlog $$\, x\nmid \bar g,\,$$ then $$\,x^n = \bar g\,\bar h\,$$ $$\Rightarrow\,x\mid \bar g\,\bar h,\,$$ so $$\,x\nmid \bar g\Rightarrow x\mid \bar h,\,$$ by $$\,x\,$$ prime. Repeating this shows all $$\,n\,$$ factors of $$\,x\,$$ must divide into $$\,\bar h,\,$$ so $$\,n = \deg \bar h = \deg h,\,$$ so $$\,\color{#c00}i = \deg g \color{#c00}{= 0},\,$$ contra hypothesis.

• Thanks for your explicit answer! BTW how do you change the color of text? – atlantic0cean Nov 3 '20 at 12:49