I just finished a unit on the uniqueness of factorization in $F[x]$, but my textbook didn't give me an example of non-uniqueness in $R[x]$ where $R$ isn't a field. Is there a simple example of a polynomial in $\Bbb{Z}_{12}[x]$ (or some other non-field) that has multiple factorizations into irreducibles? In other words, I know why the proof depends on $F$ being a field, I just would like to see an example showing why uniqueness doesn't work in $R[x]$. Thanks!

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    Does the notion of irreducible still make sense for rings that are not domains? See for instance https://math.stackexchange.com/questions/1838066/irreducible-elements-for-a-commutative-ring-that-is-not-an-integral-domain/1838593 – lhf Nov 08 '17 at 17:52
  • In the context of polynomial rings it seems like it still makes sense, but I don't really know. My book says that f(x) is reducible if it can be written as a product of two polynomials of lesser degree, so in this context the definition is stronger than just being written as a product of two non-units. – Dan Nov 08 '17 at 18:03

1 Answers1


I would go with $$(x-1)(x+1)=x^2-1=x^2-25=(x-5)(x+5).$$

Jyrki Lahtonen
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    Zero divisors at play here. The product $(x-1)(x+1)$ vanishes at $x=5$ even though neither of the factors does. – Jyrki Lahtonen Nov 08 '17 at 17:48
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    But is it clear that the linear factors are irreducible? They may be the product of two quadratic polynomials for instance. – lhf Nov 08 '17 at 17:51
  • A good question @lhf! I don't have time to think about it now. I'm needed elsewhere in five. – Jyrki Lahtonen Nov 08 '17 at 17:54
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    @JyrkiLahtonen $x+1=(6x+1)(6x^2+7x+1)$ – Jose Brox Nov 08 '17 at 18:02
  • Thanks for the help. – Dan Nov 08 '17 at 18:03
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    Even if the linear factors can be written as products of quadratics, this example works for the definition in my book (Fraleigh). Fraleigh counts all linear polynomials as irreducible. – Dan Nov 08 '17 at 18:05
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    @JoseBrox That is an interesting fact. But because $6$ is nilpotent in $\Bbb{Z}_{12}$ your other factor $6x+1$ is of the form $1+n$ with $n$ nilpotent. Therefore it is a unit in the ring $\Bbb{Z}_{12}[x]$. This means that your factorization does not make $x+1$ reducible. To wit, $(1+6x)^2=1$. – Jyrki Lahtonen Nov 08 '17 at 19:34
  • I don't think I can shed enough light on the "proper" way of defining irreducible elements in a ring with zero divisors. The interested readers can use [Bill Dubuque's old answer](https://math.stackexchange.com/a/732005/11619) and the links in there to learn more. – Jyrki Lahtonen Nov 08 '17 at 19:47
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    @JyrkiLahtonen You are right. That was fun! Next I would play $x+1=(4x+1)(1-3x)$, but now I'm not sure whether the factors are units or not! A straightforward proof does not come to mind, although they are not inverted by linear polynomials. – Jose Brox Nov 08 '17 at 20:20
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    And, @JoseBrox, your second factorization [does not seem to have units](https://math.stackexchange.com/q/19132/11619). I'm at loss here :-) – Jyrki Lahtonen Nov 08 '17 at 20:55
  • @JyrkiLahtonen Thanks for the result on units of $R[X]$! About the "unfortunate sequence of events", I thought that maybe was something public. Never mind. – Jose Brox Nov 08 '17 at 21:55