arithmetic modulo field with real numbers vector space

Question

if i choose modulo3 to be the field, and the real numbers to be the vector space. how do i Multiplier vector in scalar? for example i take "4" from the vector space of real numbers and want to Multiplier him with "2" from the field of modolo3 how do i write\calculate it thanks

Answer 1

$\mathbb{R}$ (the real numbers) does not have a vector space structure over $\mathbb{F}_3$ (the field of integers modulo 3) that is compatible with its addition structure.

The reason is that since $1+1+1=0$ in $\mathbb{F}_3$, every vector space over $\mathbb{F}_3$ is "3-torsion", in other words, it is always true that $x+x+x=0$. But in $\mathbb{R}$, this is not true. So there will be a contradiction between the addition structure of $\mathbb{R}$ and one of the axioms of a vector space if you try to define a multiplication of elements of $\mathbb{R}$ by elements of $\mathbb{F}_3$.

To see why: since $1$ is the identity of the field $\mathbb{F}_3$, the vector space axioms say we should have $1\cdot x = x$ for all real numbers $x$. But then we will have

$$0\cdot x = (1+1+1)\cdot x = (1\cdot x) + (1\cdot x) + (1\cdot x) = x + x + x$$

The second equality is because the scalar multiplication should distribute over addition in the field.

You can also use the vector space axioms to prove that $0\cdot x$ must be defined to be $0\in \mathbb{R}$. But $x+x+x$ is not zero in $\mathbb{R}$ unless $x$ is zero. So this is a contradiction.

Answer 2

The abelian group $\mathbb{R}$ carries no vector space structure over $\mathbb{F}_p$ since $p \cdot 1 = 0$ does not hold.

Answer 3

If you want the addition on you vector space to be the same as the usual addition of real numbers, then you cannot define such a vector space structure, because for any real number $a$ you would get $$ a+a+a = 1_3\cdot a + 1_3\cdot a + 1_3\cdot a = (1_3+1_3+1_3)\cdot a = 0_3\cdot a = 0. $$ (Here $1_3$ and $0_3$ are elements of the modular field.)