# Example of factorization in a polynomial ring which is not an UFD

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.

• 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 $A$. – Arthur Mar 22 at 14:08
• @Arthur Yes, thank you very much, I didn't know how to write it in a simple way. – Marta Fornasier Mar 22 at 14:19

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.
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: $$x+1=(2x^2+x+7)(4x^2+6x+7) \bmod 8$$
• Uh, I'm sorry, I didn't write it, but I was looking for $A[x]$ with $A$ domain. Now I edit the question. – Marta Fornasier Mar 22 at 16:17