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I know this is already answered here but I am wondering that if the following way to prove is also correct - let

$$f(x) = 4x^2+4x+1$$

$$g(x)=4x^2-1$$ Since this is a UFD so a unique gcd will exist. gcd is $$d(x)=(2x+1)$$

Now to prove this is a principal ideal domain I have to show that a general element of form $rf+sg$ can be generated with some element $d\in \mathbb{Z}[x]$

i.e. there is some ideal $(d) = (f,g)$

So my point is that there isn't any actually d for which this will form the same ideal as $(f,g)$ even if $d$ is gcd of $f$ and $g$

i.e. $$(2x+1)\not= r(4x^2+4x+1) + s(4x^2-1)$$ for any r, x $\in \mathbb{Z}[x]$ although this can happen for $\mathbb{Q}[x]$

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codeomnitrix
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  • How will $;\langle f,g\rangle;$ not contain $;d=a(x)f(x)+b(x)g(x)\in\langle f,g\rangle;$ ?? – Timbuc Oct 07 '14 at 11:52
  • deg(2x+1)<deg(f(x)) as well as deg(g(x)), so I have to choose a(x) and b(x) in such a way that order 2 terms gets removed and for that only choice is to take order of a(x) and b(x) same but it will never make (2x+1) as result – codeomnitrix Oct 07 '14 at 11:55
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    Fair enough, @codeo ...but how do you know $;a(x),b(x)\notin\Bbb Z[x];$ ? Other than actually calculating them (within $;\Bbb Q[x];$, say), I can't see your proof works. – Timbuc Oct 07 '14 at 12:03
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    yeah got it, actually explicitly I can't show that (but I think there would not be...not sure), but I can't prove it in this way. thanks for clarifying. – codeomnitrix Oct 07 '14 at 12:47

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What you have is not yet enough to show that $\mathbb{Z}[x]$ is not a PID. We would have to show that there is no polynomial $d$ such that $(f,g)=(d)$. You only claim that $d=2x+1$ is not in $(f,g)$. But note that we have $d=gcd(f,g)=af+bg\in (f,g)$ for some polynomials $a,b$. The answers of your link have already shown you how to do this kind of proof (with the ideal $(2,x)$, which is not principal).

Dietrich Burde
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  • Yeah there is the confusion, can I write d = af+bg if d is gcd(f, g) even if it is not principal ideal domain? and are there any a, b in z[x] for which d = af+bg? – codeomnitrix Oct 07 '14 at 12:09
  • You are right about problems with the gcd in non-factorial rings - see here. However $\mathbb{Z}[x]$ is a UFD. – Dietrich Burde Oct 07 '14 at 12:14
  • Thanks Dietrich, got it, actually I can never show that there are no two polynomials a and b such that this fails, although I think there would not be but no way to show actually...So the proof is in sufficient. Thanks for clarifying – codeomnitrix Oct 07 '14 at 12:49