Why is $9x^{2}-3$ reducible over integers? I am not able to understand what are the necessary and sufficient condition for the irreducibility of a polynomial.
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1Well $9x^2 - 3 = 3(3x^2 -1)$ : is $3$ a unit in $\Bbb{Z}[X]$ ? – Maxime Ramzi Jun 20 '17 at 07:41
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In fact, there are different definitions of irreducibility. The confusion comes mainly from the context.
Some literature has defined irreducible polynomials (over some ring, in this case, $\mathbb{Z}$) as the polynomials that cannot be factored into the product of two NON-CONSTANT polynomials. In this way $9x^2-3$ is irreducible.
However if we consider the irreducibility in the domain $\mathbb{Z}[x]$, the definition of irreducibility is that when $f$ is factored into $g\dot\ h$, either $g$ or $h$ is a unit (has an inverse). In this way, $3$ is definitely not a unit, so $9x^2-3$ is not irreducible.
I have encountered both definitions, and they have also brought confusion. Hope my explanation helps.
Junkai Dong
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There are no different definitions of irreducibility. There are only wrong interpretations of the definition of an irreducible element in an integral domain when use it for polynomials. – user26857 Jun 23 '17 at 09:17
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@user26857 Thanks, but I really encountered both definitions in Chinese translations of math books, as well as standard Chinese math textbooks. So maybe it's a translation error. – Junkai Dong Jun 23 '17 at 15:18
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I didn't say that you invented such things. I'm sure these different definitions there exist. The point is that those authors could be wrong. (Please read egreg's answer to see how natural the things are in fact.) – user26857 Jun 23 '17 at 16:00
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The polynomial is irreducible in $\mathbb{Q}[x]$, but reducible in $\mathbb{Z}[x]$.
Definition. Let $R$ be an integral domain. An element $a\in R$, $a\ne0$ and not a unit, is reducible if (and only if) there exist non units $b,c\in R$ such that $a=bc$. An element $a\in R$, $a\ne0$ and not a unit, is irreducible if and only if it is not reducible.
Equivalently, $a$ ($a\ne0$ and not a unit) is irreducible if and only if, for every $x,y\in R$, if $a=bc$ then either $b$ is a unit or $c$ is a unit.
In your case $$ 9x^2-3=3(3x^2-1) $$ is a factorization where neither factor is a unit in $\mathbb{Z}[x]$.
On the other hand, the polynomial is irreducible in $\mathbb{Q}[x]$ because it has no roots (which is a sufficient condition for polynomials of degree $1$ or $2$ over a field).
What’s a necessary and sufficient condition for a nonzero polynomial $f(x)$ in $\mathbb{Z}[x]$ to be irreducible?
First, define the content $c_f$ of $f(x)$ to be the positive greatest common divisor of its coefficients. Any nonzero polynomial can be written as $$ f(x)=c_f \hat{f}(x) $$ where the content of $\hat{f}(x)$ is $1$ (such a polynomial is called primitive).
If the polynomial is constant (and not $\pm1$), then it is irreducible if and only if $c_f$ is prime.
Assume $f$ is not constant. Then, in order to be irreducible in $\mathbb{Z}[x]$, it is necessary that $c_f=1$ and that $\hat{f}(x)$ is irreducible.
Now (a consequence of) Gauss’ lemma tells you that a primitive polynomial is irreducible in $\mathbb{Z}[x]$ if and only if it is irreducible in $\mathbb{Q}[x]$.
egreg
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Readers may find of interest this related result over a domain $,\rm D,$ with quotient field $,\rm K$ $$\rm f,\ is\ prime\ in\ D[x]\iff f,\ is\ prime (= irreducible)\ in\ K[x]\ and,\ f,\ is\ superprimitive $$
$$\rm where,\ f,\ is\ {\bf superprimitive}\ in\ D[x],\ :=,\ d,|,cf, \Rightarrow, d,|,c,\ \ for\ all,\ c,d\in D^*$$
– Bill Dubuque Jun 23 '17 at 18:37