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I know the standard definition of an irreducible polynomial. But, in the book "a course in abstract algebra" by authors "Khanna and Bhambri", irreducible polynomial is defined as following:

Let $R$ be an integral domain with unity. A polynomial $f(x)\in R[x]$ of positive degree(i.e. deg≥1) is said to be an irreducible polynomial over $R$ if it can not be expressed as product of two polynomials of positive degree. In other words if whenever $f(x)=g(x)h(x)$, then $\deg g(x)=0$ or $\deg\text h(x)=0$

According to usual definition of an irreducible polynomial, $f(x)=2x^2+4=2(x^2+2)$ is irreducible over $\mathbb{Q}$ but reducible over $\mathbb{Z}$(reducible over $\mathbb{Z}$ because neither $2$ is unit nor $x^2+2$ is unit in $\mathbb{Z}[x]$) . Hence according to our usual definition, above polynomial is reducible over smaller set $\mathbb{Z}$ but irreducible over bigger set $\mathbb{Q}$

Whereas, if I use above definition in the book then, $f(x)=2(x^2+2)$ then $f(x)$ is irreducible over $\mathbb{Q}$ and over $\mathbb{Z}$ too. I think this looks more appropriate! Irreducible in bigger set and hence it must be irreducible in smaller set!

So which definition is more correct? and is there is disadvantage of using above definition in book?

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    $\begingroup$ Both definitions are equally valid, just make sure you stick to one. I have seen mathematicians use both definitions. $\endgroup$ May 6 '20 at 16:33
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    $\begingroup$ We had several discussions on the definitions here. You can search for such posts. I found this post, but it only considers one definition. See also this one. $\endgroup$ May 6 '20 at 16:33
  • $\begingroup$ @DietrichBurde Sir thanks for the link. $\endgroup$ May 6 '20 at 16:40
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I think that the definition that $a$ is irreducible in $R$ if $a=xy$ in $r$ means one of $x$ and $y$ is a unit is more correct. $x^4 +1$ is irreducible in $\mathbb{Z}$ but is reducible in $\mathbb{Z}/{n\mathbb{Z}}$ for all $n$ >$1$

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