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How can we show this is true? Here $f(x)$ is any degree $1$ and higher polynomial in $\mathbb Z[x].$ I've tried saying suppose I is not equal to $\langle n\rangle$ or $\langle f(x)\rangle$, and then through a series of arguments show that it must be $\langle f(x),n\rangle$ for some f and n.I have chosen f to be the smallest positive degree polynomial in my ideal. Then choose an element in the ideal that is not a multiple of $f(x).$ Then $f$ doesn't divide $g,$ and hence $g= fq + r$, where $r$ must be of degree 0 by minimality of the degree of $f.$ So then we have a constant $r$ (not equal to $0$ or $1$) that's in the ideal. Now any other poly $h$ in I that isn't a multiple of $f$ must have remainder a constant by the same reasoning. Now if this constant remainder is coprime with $r,$ then $1$ is in the ideal and we get a contradiction. So only elements whose constant remainder with $f$ has a factor in common with $r$ are in $I$. If $r$ is prime, we would be done as then all remainders would be a multiple of $r$, hence we would have $I= \langle f(x), r\rangle$. But if $r$ is not prime, I don't know how to proceed. Could someone help or give an easier proof? Thanks! Edit: Sorry there was formatting issues, I've fixed it.

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Hmm, you have a problem with your use of division. For the division-with-remainder to work in $\Bbb{Z}[x]$ you need the denominator $f$ to be monic (leading coefficient 1, or at least leading coefficient a unit), but it might not be. Can you write $(4,2x,x^2)$ in your desired format? So I think you are trying to prove something that isn't true.

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To complete your proof, think about the collection of all the remainders you get from all polynomials which are not multiples of $f$; they form an ideal in $\mathbb{Z}$, and all ideals in $\mathbb{Z}$ are generated by one integer (the greatest common divisor). So you can still get an integer $n$.

You should be a bit more careful with your statement $g = fq+r$ though; consider the case $f = 2x$ and $g = 3x^2 + 1$. How can you express $g = fq+r$?

If you know about Gauss's Lemma, you can modify your proof to consider polynomials in $\mathbb{Q}[x]$, where you can divide polynomials like this, and then argue from them to polynomials in $\mathbb{Z}[x]$.

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  • $\begingroup$ Thank you, I do know about Guass Lemma..however the other comment says the statement isn't true? I am confused. $\endgroup$ – PaulDavis Apr 30 '18 at 1:09
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    $\begingroup$ @PaulDavis C Monsour is correct. Indeed more is true. She/he points out that there are ideals in $\Bbb Z[x]$ that require three generators. In fact for any $n\in\Bbb N$, there are ideals which require $n$ generators. $\endgroup$ – Lord Shark the Unknown Apr 30 '18 at 1:28
  • $\begingroup$ @PaulDavis Try to use Gauss's Lemma and see if the proof goes through! And/or investigate the example C Monsour gave. Looks to me like C Monsour gave a good counterexample, but you should check it out for yourself! $\endgroup$ – Y. Forman Apr 30 '18 at 1:29

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