Vtyjkyuk

Let us assume E to be a vector space over K. My understanding is that K needs to be a commutative field.

1. Is $mathbbZ$ a commutative field?


2. Assuming that it is, so that I can define E over $mathbbZ$, is there any relationship between scalar multiplication and vector addition?

That is, can I say that $x + x = 2x, forall x in E$?

If so, could I reason on the abelian group $(E,+)$ only by saying that any external operation $mathbbZ times E rightarrow E$ can actually be converted to an internal operation $E times E rightarrow E$ by virtue of this "relationship" between addition and multiplication in $mathbbZ$?

$2x + 3y = x + x + y + y + y, forall x,y in E$

$mathbb Z$ is not a field, since $2$ does not have a multiplicative inverse.

The answer to your second question is YES anyway:

You can still define "vector fields" over arbitrary rings. They are called $R$-modules. In face, a $mathbb Z$-module is precisely an abelian group.

Every $mathbb Z$-module is an abelian group, just forget that there is the $mathbb Z$-multiplication

and

Every abelian group $G$ can be equipped with a $mathbb Z$-multiplication and is then a $mathbb Z$-module: define $zcdot g := underbraceg+g+dots+g_z text times$.

This is the only way to define a $mathbb Z$-module structure: $1cdot g=g$ per definition and $(a+b)cdot g = acdot g + bcdot g$, this shows that the operation is unique.

You specifically asked for $2x=x+x$. We can derive that from the module axioms (which are like the vector space axioms, but the "field" is a ring): $2x = (1+1)x = 1x + 1x = x+x$

$mathbb Z$ is not a field since, for instance, $2$ has not multiplicative inverse in $mathbb Z$. Actually, the only elements of $mathbb Z$ with a multiplicative inverse ar $pm1$. And, by the usual definition of field, all fields are commutative.