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

  1. Is $\mathbb{Z}$ a commutative field?
  2. Assuming that it is, so that I can define E over $\mathbb{Z}$, 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 $\mathbb{Z} \times E \rightarrow E$ can actually be converted to an internal operation $E \times E \rightarrow E$ by virtue of this "relationship" between addition and multiplication in $\mathbb{Z}$?

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

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$\mathbb Z$ 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 $z\cdot g := \underbrace{g+g+\dots+g}_{z \text{ times}}$.

This is the only way to define a $\mathbb Z$-module structure: $1\cdot g=g$ per definition and $(a+b)\cdot g = a\cdot g + b\cdot g$, 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): $2x = (1+1)x = 1x + 1x = x+x$

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$\mathbb Z$ 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.

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