Coprimeness of polynomials over a ring with zero divisors

Asked Apr 21 '17 at 20:05

Active Apr 24 '17 at 19:27

Viewed 40 times

Let $R=\mathbb{Z}_{p^s}[x]$ for a prime number $p$, and $s\in\mathbb{N}$. Let $f(x),g(x)\in \mathbb{Z}_{p^s}[x]$ such that $f(x)R[x]+g(x)R[x]=R[x]$, Show that $gcd(f(x),g(x))=1$.

I've already asked this question, and I try to write a proof but I can't. Someones can give me the proof or some book where I can find it, My attempt's are usefull for that reason I don't wrote it.

edited Apr 24 '17 at 19:27

asked Apr 21 '17 at 20:05

Ragnar1204

1,160

2

What is your definition of coprime? By any reasonable definition, yes, they are coprime. – Ege Erdil Apr 21 '17 at 20:39
The definition is that gcd(f,g)=1 but how can i prove this using the statement? – Ragnar1204 Apr 21 '17 at 20:46
What is your definition of gcd? – Ege Erdil Apr 21 '17 at 20:47
d=gcd(f,g) if d|f, d|g and any other divisor of both polynomials say h, must h|d. – Ragnar1204 Apr 21 '17 at 20:48
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Okay, now, can you think of a way to write $ a(x) f(x) + b(x) g(x) = 1 $ in the ring $ R[X] $, for some polynomials $ a $ and $ b $? Can you now check whether $ 1 $ is the gcd of $ f $ and $ g $? – Ege Erdil Apr 21 '17 at 20:49
I've tried by two days to do that and I don't achieve nothing :( – Ragnar1204 Apr 21 '17 at 20:53
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Well, the constant polynomial $1$ is in the right-hand part of your equation on the second line of your question. Does that help? – Lubin Apr 21 '17 at 21:49
Not much... I write that the degrees of $f$ and $g$ must be 0 but that isn't true always... I feel so dumb... – Ragnar1204 Apr 21 '17 at 21:53

Coprimeness of polynomials over a ring with zero divisors

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