1
$\begingroup$

I was reading Gallian's Contemporary Abstract Algebra where I came across the following :

Let $D$ be an integral domain. A polynomial $f(x)\in D[x]$ that is neither the zero polynomial nor a unit is said to be irreducible over $D$ if whenever $f(x)$ is expressed as a product $f(x)=g(x)h(x)$, then $g(x)$ or $h(x)$ is a unit.

In the case that an integral domain is a field $F$, it is equivalent to define a nonconstant $f(x)\in F[x]$ to be irreducible if $f(x)$ cannot be expressed as a product of two polynomials of lower degree.

Two things changed when the integral domain is a field :

  1. He used the term "nonconstant" instead of "neither zero nor unit".

  2. He replaced "$f(x)=g(x)h(x)\implies g(x)$ or $h(x)$ is a unit" by "$f(x)$ cannot be expressed as a product of two polynomials of lower degree".

My questions regarding 1 and 2:

  1. Aren't the nonzero non-invertible elements of $D[x]$ all the nonconstant polynomials anyway? I know that $2x+1$ in $\Bbb Z_4[x]$ is a non constant invertible polynomial, but $\Bbb Z_4$ isn't an integral domain. Is there such an example?

  2. Again, could someone give an example showing that the two statements aren't the same in any $D[x]$?

$\endgroup$
1
$\begingroup$

Consider $4\in\mathbb{Z}[x]$. It cannot be expressed as the product of two polynomials of lower degree. But it is not irreducible, since $4=2\times2$.

$\endgroup$
  • $\begingroup$ Why do we require the polynomial to be non-invertible? $\endgroup$ – Hrit Roy Sep 19 '18 at 14:28
  • $\begingroup$ Why are you asking that? Neither of the definitions that you mentioned requires that. $\endgroup$ – José Carlos Santos Sep 19 '18 at 14:31
  • $\begingroup$ "neither zero nor unit"? $\endgroup$ – Hrit Roy Sep 19 '18 at 14:37
  • 1
    $\begingroup$ You are right. Sorry about that. It's like the integers. We divided into three classes: $0$, those which are invertible ($\pm1$) and the remaing ones. And the remaing ones can alwyas be written as a product of irreducible integers, that is, of prime numbers. We want to have such a classification in $D[x]$. $\endgroup$ – José Carlos Santos Sep 19 '18 at 14:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.