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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.

Son Bi
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  • You're almost there. If $gh \in \mathbb{Z}$, what can you say about $g,h$ individually? – Fredrik Meyer Jul 12 '12 at 06:46
  • @FredrikMeyer That they're both integers? I'm just worried that by adding two polynomials $pf+3r$, maybe all nonconstant terms could cancel. – Son Bi Jul 12 '12 at 06:48
  • What happens when you add two polynomials "$pf+3r$" is irrelevant. The step in your proof where you conclude $3=gh$ is still valid. Multiplying two polynomials always increases the degree (check this), and the degree of constants are 0, so g, h must both be integers, hence $x^3-x^2+2x-1$ can't be inside your ideal. Contradiction. – Fredrik Meyer Jul 12 '12 at 06:55
  • @FredrikMeyer Sorry, I don't follow your point. I know that $\deg(fg)=\deg(f)+\deg(g)$. I see from $3=gh$ that $g$ and $h$ are both integers, so $g=1$ or $3$. I know $g\neq 3$, so $g=1$. Then I'm trying to prove that $(3,x^3-x^2+2x-1)\neq (1)$ by showing that $1\notin(3,x^3-x^2+2x-1)$ say. What ideal are you referring to when you say $x^3-x^2+2x-1$ can't be inside your ideal? – Son Bi Jul 12 '12 at 07:00
  • No, you're not trying to show that 1 is not in your ideal. You want to show that assuming principality leads to a contradiction. See Bill Dubuque's answer for my point about degrees. – Fredrik Meyer Jul 12 '12 at 20:44
  • @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 $\mathbb{Z}[x]$. – Son Bi Jul 13 '12 at 02:53

3 Answers3

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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.

Gerry Myerson
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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 $\mathbf Z[x]$?

Dylan Moreland
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  • Thanks Dylan. I think that the image is $x^3-x^2+2x-1=x+2x^2+2x+2=2x^2+2$, which is not a unit, since it is not a constant in $\mathbb{Z}/(3)[x]$. Could you clarify why this shows $(3,f)\neq\mathbb{Z}[x]$? Does it follow by identifying $(3,f)$ with $(f)$ in $\mathbb{Z}[x]/(3)$, but since $f$ is not a unit in $\mathbb{Z}/(3)[x]$, it generates a proper ideal, and thus a proper ideal of $\mathbb{Z}[x]/(3)$, so $(3,f)$ is proper in $\mathbb{Z}[x]$? – Son Bi Jul 12 '12 at 08:39
  • The proof doesn't need primality of $\rm:3 =: m,:$ only $\rm:m:$ nonunit and $\rm:f:$ monic nonconstant mod $\rm m.\ \ $ – Bill Dubuque Jul 12 '12 at 13:48
  • @BillDubuque Agreed! Perhaps my comment was misleading. I just wanted to point out that the factor ring was particularly nice. – Dylan Moreland Jul 12 '12 at 14:43
  • @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! – Dylan Moreland Jul 12 '12 at 14:48
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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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Bill Dubuque
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