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I computed the gcd of the 2 polynomials in the ideal and their gcd is equal to 1 ,using euclid algorithm.

I want to know if the ideal in the title is a principal ideal and why ?

I know a principal ideal is the ideal generated by only one element.So $\langle x^2,x+1 \rangle = \langle 1 \rangle$ but if that was a principal ideal then any ideal generated by 2 or more ideals would be principal because he could be generated by $1$ or some gcd of those polynomials. I think the same for non polynomial rings .. since the gcd can be 1 or any number , any ideal can be generated by one element so any ideal would be principal.What am I missing ?

Eduard Valentin
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  • How would you compute their gcd using Euclid's algorithm, exactly? Note that $\Bbb Z[X]$ is not a Euclidean ring. – Jesko Hüttenhain Jun 18 '17 at 08:44
  • @JeskoHüttenhain $x^2 = (x+1)(x-1)+1$ since the last nonzero remainder is 1 , their gcd is 1 – Eduard Valentin Jun 18 '17 at 08:46
  • Aside: $\gcd(f,g)$ is not necessarily an element of $\langle f, g \rangle$. For example, $\gcd(x, 2) = 1$, but it's not too hard to see that $1 \notin \langle 2, x \rangle$. In particular, $1$ cannot be written as a linear combination (with coefficients from $\mathbb{Z}[x]$) of $x$ and $2$. –  Jun 18 '17 at 08:49
  • So , it's not true that the ideal generated by the gcd of 2 polynomials contains the ideal generated by those two polynomials ? @Hurkyl – Eduard Valentin Jun 18 '17 at 08:50
  • @Eduard: However, your argument can easily be adapted to show $1 \in \langle x^2, x+1 \rangle$, –  Jun 18 '17 at 08:50
  • @Eduard: No, that containment is true. It's the reverse: it is possible that $\langle \gcd(f,g) \rangle \not\subseteq \langle f, g \rangle$. The point to take away from this aside is that greatest common divisors are a red herring; they have nothing to do with the argument you're making. –  Jun 18 '17 at 08:51
  • I understand, but what's the approach on this type of problems ? I compute the gcd then i verify if the gcd is in the ideal generated by the 2 (or more) polynomials ? @Hurkyl – Eduard Valentin Jun 18 '17 at 08:55
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    The application of the Euclidean algorithm is problematic if you have an underlying ring like $\mathbb Z$ and not a field, since if you divide $f$ by $g$ the leading coefficient of $g$ needs to be invertible. – Wuestenfux Jun 18 '17 at 11:21

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Your ideal is indeed principal. An example of a not principal ideal in your ring is the ideal $\langle 2,x\rangle$.

Patrick
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