I am given the following question:
"Given a random polynomial in $Z\left[x\right]$. We factorise this polynomial in the polynomial rings $\left(R\left[x\right],+,\cdot\right)$, $\left(Z\left[x\right],+,\cdot\right)$ and $\left(Z_3\left[x\right],+,\cdot\right)$ and find respectively $a$ , $b$ and $c$ factors. What is the general relation between $a$, $b$ and $c$?"
The only thing I could think of is $a \geq b \geq c$. Am I missing something?
On
No, the number of factors does not behave like this. Consider some "random" example, like $f(x)=x^2+x+1$. Clearly this has no integer root, so that the factorisation of $f$ over $\mathbb{Z}$ gives $b=1$ factor. The same is true over $\mathbb{R}$. But $x^2+x+1=(x+2)^2$ over $\mathbb{Z}/3$, i.e., $c=2$.
There are the following ring homomorphisms. The inclusion $$ f:\Bbb{Z}[x]\to\Bbb{R}[x], $$ and the homomorphism $$ p:\Bbb{Z}[x]\to\Bbb{Z}_3[x] $$ that reduces all the coefficients modulo three.
Because the homomorphic image of a product is the product of images, we might initially think that the existence of a factorization in $\Bbb{Z}[x]$ implies the existence of a factorization in either of the other rings. And, more generally, think that this implies the inequalities $a\ge b$ and $c\ge b$.
This actually holds in a limited sense with the following special cavets forcing a reformulation. A homomorphism of rings may map a non-unit to a unit. Thus a proper factor may become non-proper. Here this can happen with both these homomorphisms. Both $f$ and $p$ map the product $4(x+1)$ of two non-units of $\Bbb{Z}[x]$ to an irreducible in their target rings, because $f(4)$ and $p(4)$ are both units.
To exclude this possibility we can restrict ourselves to monic polynomials in $\Bbb{Z}[x]$. This implies that there will be no constante factors.
The other caveat only affects $p$. The other homomorphism $f$ is injective, and thus takes distinct factors to distinct factors. For its part $p$ fails in that score. For example $p(x+1)=x+1=p(x+4)$. Therefore we need to be careful in counting the factors of polynomials like $(x+1)(x+4)\in\Bbb{Z}[x]$. It has two non-trivial factors in $\Bbb{Z}[x]$ but only one in $\Bbb{Z}_3[x]$ because that factor has multiplicity two there.
Conclusion. If we restrict ourselves to monic polynomials $f\in\Bbb{Z}[x]$ then $a\ge b$. If we further restrict ourselves to polynomials $f$ such that $p(f)$ has no multiple factors, then also $c\ge b$.
The inequality $c\ge b$ also holds (for monic polynomials) if we for example count only irreducible factors, and those with multiplicities.