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in the book "etale cohomolgy" milne is using two notions: coprime ideals and strictly coprime ideals. It seems to me that both the notions are same. because (f(t))+(g(t))=(f(t),g(t)).

What am i doing wrong?

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    $\begingroup$ I believe we have coprime = have no common factors and strictly coprime = together generate the whole ring. The difference can be seen in the case of $\mathbb Z[x]$, where $2,x$ have no common factors, but $1\not\in(2,x)$. $\endgroup$ – Wojowu Feb 9 '18 at 17:01
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    $\begingroup$ @Wojowu i see...but how would you define "this common factor " condition for general commutative ring?? $\endgroup$ – user111251 Feb 9 '18 at 17:04
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    $\begingroup$ The notion of divisibility makes sense in every commutative ring. Then we just say there is no non-unit which divides both elements. $\endgroup$ – Wojowu Feb 9 '18 at 17:05
  • $\begingroup$ Wojowu i see...but every reference i saw says that I and J are said to be coprime if I+J=the whole ring ...this condition is same as saying that (I,J)=whole ring...anyway thanks $\endgroup$ – user111251 Feb 9 '18 at 17:13
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    $\begingroup$ Actually, Milne doesn't use "two notions: coprime ideals and strictly coprime ideals". He only talks about strictly coprime polynomials. See Billy's post. $\endgroup$ – anon Feb 9 '18 at 20:29
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I believe Milne's definitions are as follows. Let $R$ be a ring, and $f, g\in R[x]$ two polynomials. $f$ and $g$ are coprime if they share no factors in $R$; they are strictly coprime if $(f,g) = R[x]$ as ideals. Strictly coprime implies coprime, and if $R$ is a field then they're equivalent, but in general coprime doesn't imply strictly coprime. (The polynomials $x+2$ and $x+4$ are coprime, but not strictly coprime, in $\mathbb{Z}[x]$.)

The point is: if $R$ is, say, a complete DVR with maximal ideal $\mathfrak{m}$ and residue field $\overline{R} = R/\mathfrak{m} = k$, and $h\in R[x]$ is monic, and we can find a factorisation of $\overline{h}\in k[x]$ as a product of two monic coprime polynomials $\overline{f}, \overline{g}\in k[x]$ (i.e. no common roots), then we can lift this to a factorisation $h = fg$ where $f,g$ are strictly coprime (i.e. their ideal generates $R[x]$). This is Hensel's lemma.

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