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I encounter the following confusion regarding coprimality in polynomial rings. Let me first give the definition that I consider. Given $(\mathbb{Z}/n\mathbb{Z})[X]$, where $n$ is composite, I think coprimality is commonly defined as follows:

$f,g\in (\mathbb{Z}/n\mathbb{Z})[X]$ are coprime if $(f)+(g)= (\mathbb{Z}/n\mathbb{Z})[X]$.

That is, the sum of the ideals generated by $f$ and $g$ equals $R$. I think we could give the following equivalent definition:

$f,g\in (\mathbb{Z}/n\mathbb{Z})[X]$ are coprime if the only common divisors are units in $\mathbb{Z}/n\mathbb{Z}$.

This makes sense since the first definition is equivalent to the existence of $h_1,\, h_2\in (\mathbb{Z}/n\mathbb{Z})[X]$ such that $fh_1 +g h_2 =1$. Hence any common divisor of $f$ and $g$ must divide $1$, therefore must be a unit in $\mathbb{Z}/n\mathbb{Z}$.

Now comes the question. I was reading through this paper, where it mentioned in the second paragraph of Section 7 that elements that are coprime in $(\mathbb{Z}/n\mathbb{Z})[X]$ are also coprime in $\mathbb{F}_p[X]$, where $p$ is a prime dividing $n$.

However, suppose we have $n=10$, $p=5$, $f=X+7$ and $g=X+2$. Then $f$ and $g$ are coprime in $(\mathbb{Z}/10\mathbb{Z})[X]$, since their difference is nonzero modulo $10$. Hence they should be coprime in $\mathbb{F}_5[X]$. But $f=g$ in $\mathbb{F}_5[X]$, so this is clearly not the case.

What am I missing? Did I define coprimality incorrectly, or misunderstand the statement in the paper?

Any help is appreciated!

Brendan Darrer
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DisplayName
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    In the first paragraph of section 6 he states that coprime meams comaximal, i.e. $(f)+(g)=(1),,$ which persists in every quotient ring (since the consequent Bezout equality persists). – Bill Dubuque Jan 22 '24 at 10:49
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    But $(x+7,x+2)\neq (1)$ since reduced mod $5$ it is $(x+2)\neq (1),\ \ $ – Bill Dubuque Jan 22 '24 at 10:55
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    Those two notions of "coprime" are not generally equivalent, i.e. comaximal $,\Rightarrow, \gcd !=! 1$ but the converse generally fails, e.g. see here. $\ \ $ – Bill Dubuque Jan 22 '24 at 11:04
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    @BillDubuque Let me try to understand what you are saying (although I'm still quite confused). The paper states that coprime means comaximal. In your first comment, you state that $(f)$ and $(g)$ being comaximal in $(\mathbb{Z}/n\mathbb{Z})[X]$ implies $(f)$ and $(g)$ being comaximal in $\mathbb{F}_p[X]$.

    Then, you state that for the example we have $(f)+(g)$ is not equal to $\mathbb{F}_p[X]$. With this, do you mean to say that $(f)$ and $(g)$ are not coprime/comaximal in $(\mathbb{Z}/n\mathbb{Z})[X]$?

     – DisplayName Jan 22 '24 at 11:36
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    @BillDubuque The reason I stated the second (apparently not equivalent) definition that considers common divisors is that in section 6 of the linked paper, they show which conditions $s$ and $s'$ need to satisfy for $X^2+s$ and $X+s'$ to be coprime. Using the original definition with ideals, I don't really know how to obtain these conditions. Whereas with the other definition, we can reason that these two polynomials are coprime if $-s'$ is not a root of $X^2+s$. – DisplayName Jan 22 '24 at 12:38
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    He is using $I:=(f)+(g) = (1)$ if $I$ contains a unit $u$ in $\Bbb Z_n,,$ i.e. an integer $u$ coprime to $n$. In the earlier examples it is $,u = f-g = b-b'.\ $ – Bill Dubuque Jan 23 '24 at 02:58
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    @BillDubuque I've slept on it and hopefully what I now will be saying makes some sense. I have one final question for you in the end. The definition of $f$ and $g$ being coprime if $(f)+(g)=(1)$ is used because it then also holds in $\mathbb{F}_p[X]$ (with $p\mid n$), which you cannot argue when defining coprime in terms of unit divisors. For $f=X+b$ and $g=X+b'$ it is clear that $(f)+(g)=(1)$ if $f-g=b-b'$ is a unit. But what about $f=X^2+b$ and $g=X+b'$? Clearly $-b'$ must not be a root of $f$ since otherwise $(f)+(g)=(g)\neq (1)$. But is this a sufficient condition? Thanks for all the help! – DisplayName Jan 23 '24 at 08:08
  • The conditions on $b,b'$ needed for coprimality are discussed in section $6$ in the subsections on substituting $1/x$ or $x^2$ for $x.\ \ $ – Bill Dubuque Jan 24 '24 at 02:17
  • @BillDubuque I have looked at those, and understand how the conditions are derived when substituting $1/x$ for $x$, but not when substituting $x^2$ for $x$... – DisplayName Jan 24 '24 at 08:15

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