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Not principal ideal. It's a well known fact, that $\mathbb{Z}[x]$ is not a PID, for example consider the following ideal \begin{align} I = \left(x, x + 2\right) = \{a(x)(x+2) + b(x)x| a(x), b(x) \in \mathbb{Z}[x]\} \end{align} It's not difficult to prove that this ideal isn't principal.

Principal ideal. Also, let's look at the following ideal: \begin{align} J = \left(x^2 - 4, x^2 - x - 2\right) = \{a(x)(x^2 - 4) + b(x)(x^2 - x - 2)| a(x), b(x) \in \mathbb{Z}[x]\} \end{align} I have also proven that this ideal is principal, it is generated by greatest common divisor of polynomials $x^2 - 4$ and $x^2 - x - 2$, namely $J = \left(x-2\right)$

Main confusing ideal. Now let's consider the following ideal \begin{align} V = \left( f_1(x), \ldots f_n(x) \right ), ~~~~~~~ \text{where}~~f_1(x), \ldots f_n(x) \in \mathbb{Z}[x] \end{align} I claim, that this ideal is principal and it's generated by the greatest common divisor of all $f_j(x)$. My proof is the following. Consider the greatest common divisor by $g(x)$. Obviously, that $V \subset \left( g(x) \right)$. Let's proof other inclusion. Consider arbitrary element $ \hat{h}(x) = h(x) g(x) \in \left( g(x) \right)$ Since $g(x)$ is greatest common divisor, $g(x)$ can be written as a linear combination of $f_j(x)$: \begin{align} g(x) = \sum_{j=1}^{j=n} \alpha_j(x) f_j(x) \end{align} Значит: \begin{align} \hat{h}(x) = h (x) g(x) = h(x) \sum_{j=1}^{j=n} \alpha_j(x) f_j(x) = \sum_{j=1}^{j=n} \hat{\alpha_j}(x) f_j(x) ~~~~~~~\text{where}~~~~ \hat{\alpha_j}(x) = h(x) \alpha_j(x) \end{align} So $V$ is principal. Using this proof, if we take $n = 2$ and $f_1(x) = x$ and $f_2(x) = x + 2$ we conclude that $\left(x, x+2\right) = \left( \gcd(x, x+2) \right) = \left(1\right) = \mathbb{Z}[x]$, what is obviusly falce, since $\left(x, x+2\right)$ is firstly proper, and secondly it's not principal. What am i doing wrong? Note that my question isn't answered in linked question, because it's rather obvious fact which i prove by myself that $\left( x, x + 2 \right)$ is not a principal ideal, i am interested in where is the mistake in computations that $\left( f_1, f_2, \ldots, f_n \right)$ is principal, since this computations MUST be wrong, but i don't know where.

NeoFanatic
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  • You already mentioned a counterexample in your first paragraph. See the linked dupe for proofs. – Bill Dubuque May 03 '23 at 18:04
  • @BillDubuque have you read my question? I have asked where is contradiction, and in answer is explanation, that $\gcd$ doesn't work the same way in ring of integer polynomials, as in ring of integers, it's not a duplication if question you mentioned, because i ALREADY proven by myself what is written in question you linked, soit's NOT a duplication of linked question – NeoFanatic May 07 '23 at 09:23
  • if $,g\mid f_i,$ then $I := (f_1,\ldots,f_n) = (g)\iff g = \sum a_i f_i$. In particular $,g = \gcd f_i$ generates $I$ iff it can be written as a linear combination of the generators, i.e. a Bezout identity exists. Your error is assuming that this is always true. There are simple counterexamples, e.g. first linked dupe (same as answer). – Bill Dubuque May 07 '23 at 15:25

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I do not believe you can write the greatest common divisor $g(x)$ as a linear combination of the $f_j(x)$. To see this, just use the example you gave. Suppose $f_1(x) = x$, $f_2(x) = x+2$ and $g(x) = 1$. Suppose also that we can write: \begin{equation} g(x) = \alpha_1(x)\cdot f_1(x) + \alpha_2(x)\cdot f_2(x) \end{equation} With $\alpha_1,\alpha_2\in\mathbb{Z}[x]$. This gives us: \begin{equation} 1 = (\alpha_1(x)+\alpha_2(x))x + 2\alpha_2(x) \end{equation} Just by matching term by term in the polynomial, we see that $\alpha_1(x) = -\alpha_2(x)$ and $\alpha_2(x) = 1/2$. But $\alpha_2\notin\mathbb{Z}[x]$. So this cannot happen and we cannot write $g(x)$ as a linear combination of the $f_j(x)$.

Juri
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  • For integres $a, b \in \mathbb{Z}$ it's always true, that i can write $ax + by = \gcd(a, b)$ and the proof uses Euclidean algorithm, namely you construct the linear combination "from the end". I understand you counterexample, so $\gcd$ in integers doesn't work the same as $\gcd$ in polynomial ring.... – NeoFanatic May 03 '23 at 13:07
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    @NeoFanatic Essentially, yes, the Euclidean algorithm fails in $\Bbb Z[x]$ (it works in rings like $\Bbb Z$ and in $\Bbb Q[x]$), and that's basically what allows ideals to be non-principal. Be a bit careful with this remark, as there are principal ideal domains where the Euclidean algorithm doesn't work (such as $\Bbb R[x, y]/(x^2 + y^2 + 1)$). But if you have an integral domain with Euclidean algorithm, then you necessarily have a principal ideal domain. – Arthur May 03 '23 at 13:10
  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque May 03 '23 at 18:05