@RonWeasley Even if you define scalar multiplication as modulo 7, the axioms of a vector space won't hold.For example, one of the axioms is (a#b)*v= a*(b*v). Here # is the multiplication operator of the field F (as a and b are both scalars) and * is scalar multiplication. In this case, # is multiplication mod 5 and * is multiplication mod 7.
[This differentiation of operators is important as you'll realise at the end of this text. This is the standard definition. For clarity you can glance at this -> https://www.math.arizona.edu/~cais/223Page/hout/236w06fields.pdf specifically at page 5 axiom 6 ]Coming to the example, let a = 2, b = 3, andv=6. Here 2,3 belong to Z5 and6belongs to Z7.
Now (2#3)*6 = (6 mod 5)*6 = 1*6 = 6. 2*(3*6) = 2*(18 mod 7) = 2*4 = 8 mod 7 = 1
So LHS =/= RHS.
Here the different values come because you have to take mod 5 when multiplying two scalars, and not mod 7 (as per standard definition of Vector Space)Long story short, Z7 is not a vector space over Z5 no matter what operator you take for scalar multiplication. In fact, according to some comments here (https://math.stackexchange.com/questions/1927048/is-z7-integer-modulo-under-addition-a-vector-space-over-z5), you cannot define a scalar multiplication operator for this.
Also, don't worry about such questions. A quick google search reveals that the proof of why Z7 is not a vector space over Z5 uses concepts and theorems not in our syllabus. So it won't come in the exams.
Thanks a lot. It cleared a lot of my concepts.
