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Let $R$ be an arbitrary ring (not necessarily commutative) with identity and let $f(x)\in R[x]$ be a polynomial over $R$.

Then we have several analogues for irreducibility of $f(x)$. For example,

  1. $R[x]f(x)$ is maximal left ideal in $R[x]$;
  2. $f(x)R[x]$ is maximal right ideal in $R[x]$;
  3. If $f(x) = g(x)h(x)$ for non-constant polynomials $g,h\in R[x]$ then $g$ or $h$ is invertible polynomial.
  4. If $f(x) = g(x)h(x)$ for non-constant polynomials $g,h\in R[x]$ then $\deg(g)\geq \deg(f)$ or $\deg(h)\geq \deg(f)$.

So, what is the most right or logical way to define iireducible polynomial over arbitrary ring? Maybe there are another good definitions?

Xam
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Mikhail Goltvanitsa
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    Over non-domains standard notions such as irreducible and associate bifurcate into various inequivalent notions, e.g. see the citations here. So there is no unique "most right or logical way" to extend the definitions. – Bill Dubuque Sep 01 '17 at 18:55
  • @BillDubuque, thank you. But I can not find an example, which shows that the forth definition of irreducibility is not correct in other sense. Do you know such example? – Mikhail Goltvanitsa Sep 01 '17 at 19:13

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