primitivity of a polynomial over a field Suppose that we have a polynomial 
$f(x)=ax^3-bx^2+cx-d\in\mathbb{Z}[x]$ then $f(x)$ will be called primitive if $(a,b,c,d)=1$ 
I have been told that over a field $F$ there is no notion of primitivity? Could someone explain to me why is that? 
Thank you
 A: Perhaps we should say every nonzero polynomial over a field is primitive, since any nonzero coefficient generates the ideal $F$ (blows up).
All nonzero elements being invertible means we can take the polynomial to be monic (unless it's zero) by factoring out the unit (invertible) leading coefficient.
A: For any integral domain $R$, one can define a primitive polynomial over $R$ as one whose coefficients have no common $R$-divisors besides units. That is, $f=\sum_{i=0}^n a_i x^i \in R[x]$ is primitive if $b | a_i$ for all $i$ implies $b$ is a unit.  (See here. It's the second definition). 
Note that, if If $R = \mathbb Z$, this coincides with your definition.
This definition makes perfect sense for a field, which is in particular an integral domain. But that doesn't mean the definition is interesting: the coefficients of any nonzero polynomial over a field have no common divisors besides units, since any nonzero element of the field is a unit. Hence, every nonzero polynomial over a field is primitive. 
If someone has told you that there is "no notion of a primitive polynomial over a field," my guess would be that they really meant "there is no meaningful notion of a primitive polynomial over a field." It doesn't mean that you can't define primitivity over a field, just that it's not going to give you anything new so it's superfluous. 
A: Over a field, the gcd of any collection of elements is $1$, unless the elements are all zero, and then the gcd is $0$. So over a field every non-zero polynomial would be primitive.
