$(P, +, ·, 0, 1)$ is a subfield of $(F, +, ·, 0, 1)$.
Define $+': F × F → F$
by $x +' y = x + y $ and
$*: P × F → F$
by $r * x = r · x $.
Prove that $(F, +' , * , 0,1)$ is a P-vector space.
I am really un-happy with this notation and surprisingly i already cleaned it up from the original question
Given $a,b,c \in P $ and $u,v,z \in F $ where $a,b,c,u,v,z$ are arbitrary elements consider the following:
1) $u+'v= u+v = v+u =v +'u$ [Commutative]
2) $u+'(v+' z)= u+(v+z) = (v+u) +z =(u +'v)+' z$
$(a*b)*v=(ab)v=a(bv)=a*(b*v) \space$ Hence [Associative]
3) $0 \in F $ consider $v+'0 = v+0 =v\space$ [Additive identity].
4) Given $v\in F, \space$ $\exists(-v) \in F $ s.t $ v+(-v)=0=v+' (-v) \space $ [Additive inverse]
5) $1\in P $ consider $1*v=1v=v$ [Multiplicative identity]
6) $a*(v+' z)= a(v+z) = av +az =(a*v)+' (a*z)$
$(a+'b)*v = (a+b)v = av+ bv = (a*v)+' (b*v)$ [Distribuitive properties]
So if i have done that right that means that $(F, +' , * , 0,1)$ is a vector space with those two binary operations.
The thing i am less sure about is that it is a vector space over P.
i also don't like how i showed property's 4 and 5 they feel really wishy washy the way i showed them is there some nice way to clean this up?