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I have a problem about a line inside the following proof which is actually from the book Abstract Algebra by Dummit & Foote

Proposition (Eisenstein's Criterion). Let $P$ be a prime ideal in the integral domain $R$ and let $~f(x)=x^n+a_{n-1}x^{n-1}+\dots +a_0$ be a polynomial in $R[x]$ $(n\ge1)$. Suppose $a_{n-1},\dots ,a_0$ are all elements of $P$ and $a_0$ is not an element of $P^2$. Then $f(x)$ is irreducible.

Proof. Suppose $f(x)$ is reducible, say $f(x)=a(x)b(x)$ in $R[x]$, where $a(x),b(x)$ are non-constant 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 bar denotes the polynomials with coefficients reduced mod $P$. Since $P$ is prime ideal, $R/P$ is an integral domain, and it follows that both of $\overline{a(x)}$ and $\overline{b(x)}$ have $0$ constant term, i.e, the constant term of both $a(x)$ and $b(x)$ are elements of $P$. But then constant term of $a_0$ of $~f(x)$ as the product of these two would be an element of $P^2$, a contradiction. This completes the proof.

But I cannot understand the line which I have bold in the proof. What I understand is that since $x^n=\overline{a(x)b(x)}$, so comparing the coefficients on both sides we get the product of the constant terms of $~\overline{a(x)}$ and $~\overline{b(x)}$ is zero in $(R/P)[x]$, which is a Integral domain.....then either the constant term of $~\overline{a(x)}=0$ or constant term of $~\overline{b(x)}=0$...But how does both of them is zero?

Please help.

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    $\begingroup$ Ahh don't you just "love" Dummit and Foote? They just suffocate you with mind-numbing details and write many trivial things, and then they skip steps like this in proofs which give you a headache and make you wonder if you are missing something as obvious as the other details they provide. $\endgroup$ – Ovi Jul 25 '18 at 22:49
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Suppose $b$ mod $P$ has nonzero constant term. Then $a$ mod $P$ must be identically zero, since otherwise its lowest nonzero coefficient would multiply by the constant term in $b$ mod $P$ to produce a nonzero term of degree less than $n$ in the product $ab$ mod $P$, since $P$ is prime. (It wouldn't be cancelled by anything else since all lower degrees of $a$ mod $P$ are $0$.) But that means $a$ is in $P$, and yet the original polynomial is not in $P$, so this can't happen.

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    $\begingroup$ +1 Thanks, was really struggling with this. $\endgroup$ – Ovi Jul 25 '18 at 23:01
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Since $R/P$ is a domain, $R/P[x]$ is a domain. In this ring you have $$x^n = \overline{a(x)} \cdot \overline{b(x)}$$ and $\overline{a(x)},\overline{b(x)}$ have degree $<n$.

Call $$\overline{a(x)} = \sum_{k \le i \le K} \overline{a_i} x^i \\ \overline{b(x)}=\sum_{h \le j \le H} \overline{b_j} x^j $$ Then $$x^n = \left( \sum_{k \le i \le K} \overline{a_i} x^i\right) \cdot \left( \sum_{h \le j \le H} \overline{b_j} x^j\right) = \sum_{m=0}^n \left(\sum_{i+j=m} \overline{a_i b_j} \right) x^m$$

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This can be done by recursion: We now have $a_0b_0\in P$. By the definition of prime ideal, one of $a_0$ and $b_0$ must be in $P$. Suppose $a_0\in P$ but $b_0\notin P.$ Then since $\overline{a(x)}\cdot \overline{b(x)}=\overline{x^n}$. Then $a_1b_0+a_0b_1\in P$. By our assumption, $a_1b_0\in P$ since ideals are closed under subtraction. But $b_0\notin P$. This forces $a_1\in P$. Similarly, $a_0b_2+a_1b_1+a_2b_0\in P$ forces $a_2$ to be in $P$. Proceeding by the same method, then all coefficients of $a(x)$ are in $P$. Finally, we have $b_0+b_1a_{m-1}+\cdots +b_ma_0\in P$ where $m$ is the degree of $a(x)$.Notice that we let $b_i=0$ if $i>\deg b(x)$. Anyway, we know that $a_{m-1},a_{m-2},\cdots,a_0\in P$. Then this forces $b_0\in P$ which contradicts to our assumption.

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