regarding the identity element for scalar multiplication on a vector space Is the identity element for scalar multiplication for a vector space the same as the identity element for multiplication operation in the field?
 A: Yes, unless your vector space is the zero vector space:

Suppose that $V$ is a non-zero vector space over a field $F$, and let $a \in F$ such that $av=v$ for any $v \in V$. Then, $(a-1)v$ is the zero vector for any $v \in V$, and if $v$ is not the zero vector, $(a-1)v$ equal the zero vector implies that $a-1=0$, that is, $a=1$.

A: Yes.
Suppose otherwise; let our vector space be $V$ over a field $F$. Let $a \in F$ be the multiplication identity (i.e. $ax=x$ $\forall x \in F$), but suppose there exists $b \in F$ such that instead $b \vec v = \vec v$ $\forall v \in V$ but $b \ne a$.
Scalar multiplication is performed componentwise, so if $v_i$ denote the components of $\vec v$, $bv_i$ denote those of $b \vec v$. Then if $b$ is the identity for scalar multiplication, $bv_i = v_i$ for all $i$. Recall, however, that since $v_i \in F$, then $av_i \in F$ since $a$ is the identity there. Thus, $bv_i = av_i$ for all $i$.
But this is equivalent to
$$bv_i - av_i = 0 \implies (b-a)v_i = 0$$
Since we assume $b \ne a$, then we have that $v_i = 0$ $\forall i$. But this is nonsensical since there should be no restrictions on the vector (if $v_i = 0$ for all $i$, then $\vec v$ is the zero vector).
Indeed, if $b=a$, i.e. $a$ is the identity for multiplication in $F$ as well as scalar multiplication in $V$, we see that
$$a \vec v = \vec v \implies av_i = v_i \; \forall i$$
which absolutely makes sense. Thus, the multiplicative identity element from a field must be the multiplicative identity for scalar multiplication in vector spaces over that field.
