4

Let R be a ring, and R[x] be a polynomial ring.

Can we define what it means for a polynomial $p(x) \in R[x]$ to be irreducible over R[x]?

Various sources (such as Wikipedia) only provide such definition for a field or a unique factorization domain.

Sadeq Dousti
  • 3,291
  • 1
    The definition that cleanly generalizes is the notion of prime element, which just means that $(p(x))$ is a prime ideal. Irreducibility is trickier and probably there is no notion which has all the properties you expect, so you need to pick which ones to give up. – Qiaochu Yuan Nov 27 '17 at 18:36

2 Answers2

4

It is not clear what irreducible means in rings that are not domains.

You can define irreducible but it won't have all properties you expect. See this.

In particular, if $R$ is not a domain, it may happen that decomposing a polynomial in $R[x]$ does not simplify it (that is, does not 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: $$ x+1=(2x^2+x+7)(4x^2+6x+7) \bmod 8 $$

lhf
  • 216,483
  • Thanks for the very insightful answer. I just need to ask one more thing: Your answer completely explains why R should be a domain, and the link you provided discussed why R should be an integral domain. I still don't get why R should be a UFD, as per Wikipedia definition? – Sadeq Dousti Dec 04 '17 at 12:43
  • @M.S.Dousti, if $R$ is a domain, $R[x]$ is a subring of $K[x]$, which is a UFD. See also https://en.wikipedia.org/wiki/Gauss%27s_lemma_(polynomial). – lhf Dec 04 '17 at 12:46
  • Thanks a lot for your prompt response. Please correct me if I'm wrong, but I think Wikipedia definition requires R (the underlying ring) to be a UFD, while you are saying that if R is a domain, then R[x] (the polynomial ring itself) is a UFD. So, I think the definition in Wikipedia is much more strict. – Sadeq Dousti Dec 04 '17 at 20:21
  • @M.S.Dousti, I meant, $K[x]$ is a UFD. – lhf Dec 04 '17 at 21:22
  • See here for links to literature on factroization in rings with zero divisors. – Bill Dubuque Oct 13 '18 at 18:15
3

When I taught abstract algebra, I pointed out that any commutative ring with multiplicative unit could be split up into disjoint sets: \begin{align} 1.&\quad\{0\}\\ 2.&\quad\text{nonzero zero-divisors (including nilpotent elements)}\\ 3.&\quad\text{units}\\ 4.&\quad\text{decomposable elements: those writable as}\\ &\quad\text{product two elements not in classes 1, 2, 3}\\ 5.&\quad\text{everything else.} \end{align} The indecomposable elements, also called irreducibles, are those in class 5.

So, for $a$ to be irreducible would mean that $a$ was not writable as product of two non-unit non-zero-divisors, and my “irreducibles” would be the “strong irreducibles” in the categorization mentioned in the link offered by @lhf.

Lubin
  • 62,818