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I have a small hitch in showing $(3,x^3-x^2+2x-1)$ is not principal in $\mathbb{Z}[x]$. Towards the contrary, I suppose $(3,x^3-x^2+2x-1):=(3,f)=(g)$ is principal. Then $3\in (g)$, so $3=gh$ for some $g,h\in\mathbb{Z}[x]$. Thus $g,h$ must be constant, and $g\mid 3$, so $g=1,3$. But $g$ cannot be $3$, since $f\neq 3p$ for any $p\in\mathbb{Z}[x]$, since the coefficients are not all divisible by $3$.

If $g=1$, then $(3,f)=\mathbb{Z}[x]$. I don't think this is true, but I don't know how to make it rigorous. I tried supposing $1=pf+3r$ where $p,r\in\mathbb{Z}[x]$ to reach a contradiction and show that $1\notin(3,f)$, but I don't how to more formally prove it. What can I do? Thanks.

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    @FredrikMeyer Right, I see that assuming $(3,f)$ is principal implies either $(3,f)=(3)$, or $(3,f)=(1)$ just by looking at $3=gh$. The first case is of course not true, since $f\notin (3)$. However, $f\in (1)$ obviously, so I don't see how the contradiction is reached without first proving that $(3,f)$ is proper in $\mathbb{Z}[x]$.2012-07-13

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Say $pf+3r=1$. Divide $r$ by $f$; $r=qf+s$, where $s$ is of degree at most 2. Now $1=pf+3r=pf+3(qf+s)=(p+3q)f+3s{\rm\qquad so\qquad}(p+3q)f=1-3s$ The right side has degree at most 2, so the left side has degree at most 2, but the left side is a multiple of $f$, which has degree 3, so the left side is identically zero, so the right side is identically zero, but the constant term on the right side is 1 modulo 3, contradiction.

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    @yoyo, I don't know. Why not try a few examples and see what you can work out?2016-11-05
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Another way of viewing this: look at the quotient ring \[ \mathbf Z[x]/(3) \simeq \mathbf (\mathbf Z/3\mathbf Z)[x], \] which is a polynomial ring over a field. Is the image of $f$ in this ring a unit? Is it clear why this settles the question of whether $(3, f)$ is all of $\mathbf Z[x]$?

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    @SonBi Yes, this is what$I$had in mind. Just to be clear: if $A$ is a commutative ring and $I \subset J$ are ideals in that ring, then taking images and preimages give a bijection between ideals of $A$ containing $I$ and ideals of $A/I$. So $J$ is proper if and only if its image in $A/I$ is proper. And here the image of $(3, f)$ in the quotient is principal, generated by the image of $f$, as you noted!2012-07-12
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Hint $\rm\,\ fg = 1 + 3h\ \Rightarrow\ mod\ 3\!:\ fg\equiv 1\:\Rightarrow\:deg(fg) = 0\:\Rightarrow\: deg(f) = 0\:$ contra $\rm\: f \equiv x^3 +\:\cdots$

Remark $\ $ The same proof works for $\rm\:3\to m > 1\:$ and any $\rm\:f\:$ both monic and nonconstant mod $\rm\:m.$ Generally over any ring R, a polynomial $\rm\:f\in R[x]\:$ is a unit iff $\rm\,f_0 = f(0)\,$ is a unit in R and all higher coefficients are nilpotent in R, i.e. for all $\rm\:i>0,\,\ f_i^n = 0\:$ for some $\rm\:n\in \Bbb N.$ In particular, if $\rm\:f\in \mathbb Z[x]\:$ has degree $> 1$ and leading coefficient $\rm\:c\:$ coprime to $\rm\:m>1\:$ then in $\rm\:R = \Bbb Z/m\:$ the leading coefficient of $\rm\:f\:$ becomes a unit, so $\rm\:f\:$ remains a nonunit over R (as the hint shows more simply).

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    @Gerry: "mod $3$: ..." means over $\,\Bbb Z/3\ $ (for all following objects on the line)2012-07-13