In Gallian there is a theorem which states that
"Let $f\in\mathbb Z[x]$. If $f$ is reducible over $\mathbb Q$, then it is reducible over $\mathbb Z$."
This is an implication statement, i.e., its converse is not true. But my algebra teacher said that if f(x) is a PRIMITIVE polynomial then it will become a bi-implication statement. Also, "All primitive polynomials are irreducible". Since there is no primitive polynomial in Z[x] which is reducible over it, how can the converse be true?
If there is any example which favors the converse part, please mention it.
I am a new user of Maths Stack Exchange, I am using it for the first time. Any help will be greatly appreciated.
Thanks!!
Best Answer
The statement
is indeed correct, but the converse is not generally true, because the polynomial $2x\in\mathbb{Z}[x]$ is reducible in $\mathbb{Z}[x]$, but it becomes irreducible in $\mathbb{Q}[x]$.
However, any nonzero polynomial over $\mathbb{Z}[x]$ can be written in the form $$ cf(x) $$ where $c\in\mathbb{Z}$, $f(x)=a_0+a_1x+\dots+a_nx^n\in\mathbb{Z}[x]$ and $\gcd(a_0,a_1,\dots,a_n)=1$: just compute the gcd of the coefficients and collect it. A polynomial such that the gcd of its coefficients is $1$ is called primitive.
For primitive polynomials, the converse implication also holds:
Let's see why. Suppose $f(x)=g(x)h(x)$, with both factors noninvertible in $\mathbb{Z}[x]$, but $f(x)$ is irreducible as an element of $\mathbb{Q}[x]$. Then one of the factors must be invertible in $\mathbb{Q}[x]$, so a constant polynomial and different from $\pm1$, because otherwise it would be invertible in $\mathbb{Z}[x]$. This would make $f(x)$ into a non primitive polynomial.
Usually the statement is written in terms of irreducibility:
The statement all primitive polynomials are irreducible is false. Consider $f(x)=x^2$: it is primitive, but reducible.