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I'm looking for a particular example of a polynomial ring $A[x]$, with $A$ integral domain, which is not an UFD. An easy example is $\mathbb{Z}[\sqrt{-5}][x]$, here $6 x^2 = (2x)(3x) = ((1+\sqrt{-5})x)((1-\sqrt{-5})x)$.

I would like to find a ring and a polynomial where the problem is not only in the coefficients. In other words a ring $A[x]$ where there exist a polynomial $f \in A[x]$ with two factorizations that are not of the form $f = (a_1g)(b_1h) = (a_2 g)(b_2h)$, with $g,h \in A[x]$, $a = a_1b_1 = a_2b_2 \in A$ (with $a_1b_1, a_2b_2$ two factorization of $a$). Or not of a similar form with more factors.

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  • $\begingroup$ You're right. That example is a bit tame. We take $6x^2$, and then we take out the $x^2$, factor $6$ in two different ways, and then we redistribute the $x^2$. You can see that if you consider the full factorisation into irreducibles, which is $2\cdot 3 \cdot x\cdot x$ versus $(1+\sqrt{-5})\cdot (1-\sqrt{-5})\cdot x\cdot x$. So the way I see it, what you're asking about is basically an example showing the non-UFD-ness of $A[x]$ for some non-UFD $A$, which is not trivially reucible to an example from $A$. $\endgroup$ – Arthur Mar 22 at 14:08
  • $\begingroup$ @Arthur Yes, thank you very much, I didn't know how to write it in a simple way. $\endgroup$ – Marta Fornasier Mar 22 at 14:19
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Perhaps this is the sort of thing you are looking for: If $a,b,c,d\in A$ are irreducible non-associate elements such that $ab=cd$, then $$ (ax+c)(dx-b) = adx^2 - bc = (ax-c)(dx+b). $$ With $A = \mathbb{Z}[\sqrt{-5}]$, taking $(a,b,c,d) = (2,3,1+\sqrt{-5},1-\sqrt{-5})$, this results in $$ (2x+1+\sqrt{-5})((1-\sqrt{-5})x-3) = (2x-1-\sqrt{-5})((1-\sqrt{-5})x+3). $$ However, you might still see this a kind of cheat, since $ax+c$ and $dx+b$ actually are scalar multiples; it's just that the scalar factor, $d/a = b/c$, isn't in $A$ but is instead in its field of fractions. But this is unavoidable: over the field of fractions, polynomials do have unique factorization; so seemingly distinct factorizations into linear factors over $A$ necessarily become identical aside from units over the field of fractions.

Another possibility is to use a higher-order polynomial that is irreducible over $A$, but not over the field of fractions. For example, if $ab=cd$ then $$ (ax+c)(ax+d) = a(ax^2 + (c+d)x + b). $$ In $\mathbb{Z}[\sqrt{-5}]$, this might be realized as $$ (2x + 1 + \sqrt{-5})(2x + 1 - \sqrt{-5}) = 2(2x^2 +2x + 3). $$ This is, in essence, an example of the failure of Gauss's lemma for a non-UFD. I found the idea for this example in David E Speyer's answer to a MathOverflow question; the other responses may also be of interest.

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When $A$ is not a domain, decomposing a polynomial in $A[x]$ as $f=gh$ does not necessarily simplify it, that is, does not always reduce its degree.

For instance: $$ 5x+1=(2x+1)(3x+1) \bmod 6 $$

It is even possible to decompose a linear polynomial as a product of two quadratic polynomials: $$ x+1=(2x^2+x+7)(4x^2+6x+7) \bmod 8 $$

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  • $\begingroup$ Adapted from math.stackexchange.com/a/2539517/589 $\endgroup$ – lhf Mar 22 at 15:20
  • $\begingroup$ Uh, I'm sorry, I didn't write it, but I was looking for $A[x]$ with $A$ domain. Now I edit the question. $\endgroup$ – Marta Fornasier Mar 22 at 16:17

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