Two questions on irreducibilities of polynomials over an integral domain. These are two problems 5.10, 5.11 from Falko, Algebra I, Fields and Galois Theory. 5.10 I do not know how to progress. 5.11 is solved but I do not think I understood the problem in an intuitive way.
5.10. $R$ integral, $f=ax^n+\dots +a_1X+a_0$ primitive in $R[x]$. Suppose there is $\pi$ prime of $R$ and polynomials $\phi,\psi\in R[x]$ s.t $f=a\phi(x)^m+\pi\psi(x),m\in N$. Let $x\to \bar{x}\in R/(\pi)=\bar{R}$. Prove if $\bar{\phi}\in\bar{R}[x]$ is prime and $\bar{\psi}\neq 0($mod $\bar{\phi})$ then $f$ is irr in $R[x]$.
Under quotient map $f\to \bar{f}=\bar{a}\bar{\phi(x)}^n\neq 0$ as $f$ is primitive. If $\bar{f}=0$, then $\pi\vert a_i,a$ for all $i$ contradicts with primitiveness of $f$. Pass to $K[x]=Frac(R)[x]$ which is UFD. Suppose $f=gh$. I see that $\bar{g}=\epsilon \phi(x)^l$ and similarly for $\bar{h}=\epsilon'\phi(x)^{n-l}$ in $K[x]$. 


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*How do I proceed next? Normally I would evaluate at $x=0$ to reach contradiction for Eisenstein criterion proof.

*What is the purpose for $\bar{\psi}\neq 0($mod $\bar{\phi})$? I would expect considering $\bar{R}[x]/(\bar{\phi})$ at some time point
5.11 Let $f=x^n+a_{n-1}x^{n-1}+\dots+a_0\in Z[x]$ and $a_0\neq 0$. $f$ factorizes linearly over complex number $C$. If there are $n-1$ roots of $f$ with absolute value less than 1, then $f$ is irreducible over $Z$.
Suppose $f$ is reducible over $Z[x]$ say $f=gh$. Since $C[x]$ is UFD, I know one of the polynomials, say $g$ will have all roots absolute value less than 1. Since $g\in Z[x]$, the constant coefficient of $g$ must be in $Z$ and its absolute value must be greater than or equal to 1. It looks like this problem is solved by norm map. 


*I normally expect that norm map is not sufficient to determine the behavior of irreducibility and norm accompanied by other invariants of the field extension may solve irreducibility. What is the reason I expect this is the case?

 A: For 5.10:
Suppose that $f=gh$ in $R[X]$, then in $\bar{R}[X]$ this becomes $\bar{a}\bar{\phi}^m=\bar{g}\bar{h}$. Then, since $\bar{\phi}$ is prime, we see that $\bar{g}=\bar{c}\bar\phi^l$ and $\bar{h}=\bar{c'}\bar{\phi}^k$ with $k+l=m$ and $\bar{c}\bar{c'}=\bar{a}$. We can suppose that $cc'=a$ and write $g=c\phi^l+\pi q$ and $h=c'\phi^k+\pi q'$ for some $q,q'\in R[X]$.
Now we have $$f\equiv a\phi^m+\pi(c\phi^lq'+c'\phi^kq)\bmod \pi^2,$$ so $$ \bar{\psi}=\overline{\left(\frac{f-a\phi^m}{\pi}\right)}=\bar{c}\bar{\phi}^l\bar{q'}+\bar{c'}\bar{\phi}^k\bar{q}.$$ Passing to $\bar{R}[X]/(\bar{\phi})$, the hypothesis for $\bar{\psi}$ shows that $l$ or $k$ is $0$, meaning that, say, $\bar{g}$ is constant. But since $a$ is prime with $\pi$, the dominant coefficient of $g$ is prime with $\pi$, so $\deg g=\deg\bar{g}$. Therefore, $g$ is constant and must be a unit in $R$ since otherwise $f$ couldn't be primitive.  
5.11: You have almost solved it yourself.
If $g\in Z[X]$ has its all complex roots of absolute value strictly less than 1, then their product has absolute value strictly less than $1$ also. But the constant coefficient of $g$ is of absolute value at least 1 and is equal to $(-1)^{\deg g}\times$ product of the roots of $g$. Therefore a contradiction.
You can use the norm if $g$ is irreducible, which is completely legit to assume since any polynomial in $Z[X]$ is divisible by an irreducible polynomial (here we should remember that $Z[X]$ is a UFD). 
