Krdytkyu

I am using commutative ring with unity in the sense that there exists at least one non-zero element in the ring which doesn't have a multiplicative inverse. Can't we define scalar multiplication on a vector space with elements of commutative ring with unity, instead of field?

Yes, we can do that. We just don't call them “vector spaces”. We call them “modules” instead.

You could, but there a lot of things you couldn't do with them, for instance normalization.

If $R$ is a commutative unital ring, then an $R$-module is an abelian group $M$ with a "scalar multiplication" map $R times M to M$, written $r cdot m = rm$, which satisfies

1. $(r + s)m = rm + sm$


2. $r(m + n) = rm + rn$


3. $r(sm) = (rs)m$

If $R$ is a field, then an $R$-module is precisely a vector space over that field, and we get the usual notions of dimension, bases, and so on. However, if $R$ is not a field, then an $R$-module is usually a slightly more complicated thing.

For example if $R = mathbb{Z}$, then $mathbb{Z}, mathbb{Z}^2, ldots$ are all $mathbb{Z}$-modules in the usual way, and any basis of $mathbb{Z}^n$ will have size $n$. However, $mathbb{Z}/3mathbb{Z}$ is also a $mathbb{Z}-module$, and it has no basis. (No element $m$ of $mathbb{Z}/3mathbb{Z}$ is linearly independent, since $3m = 0$). So the theory of modules is more complicated than the theory of vector spaces.