Why $A \to A \otimes \mathbb{Q}$ is injective Let $A$ be a characteristic zero ring. 
Why $A \to A \otimes \mathbb{Q}$ is injective and what is tensoring with $\mathbb{Q}$ meaning here ?
Can someone briefly explain the importance of tensor product on which I have weakness.
I want to understand the basic concept. 
 A: No one has yet written a counterexample, so I am now writing one below.
The map $- \otimes 1:A \to A \otimes_{\mathbb{Z}} {\mathbb{Q}}$ is not necessarily injective even if $A$ has characteristic zero. For example, the quotient $\mathbb{Z}[x]/(2x)$ is a ring of characteristic zero with a non-torsion-free additive group, so the map "tensor with 1" is not injective in this case.
A: Let $A$ be an Abelian group under addition. Then $A\otimes \Bbb Q$
means $A\otimes_{\Bbb Z}\Bbb Q$, the tensor product of $A$ and $\Bbb Q$
of these groups considered as $\Bbb Z$-modules. Then $a\mapsto a\otimes 1$
is a group homomorphism from $A$ to $A\otimes \Bbb Q$ whose kernel
is the torsion of $A$. So $A\mapsto A\otimes\Bbb Q$ is injective iff $A$
is torsion free.
If in addition $A$ is a commutative ring, then $A\otimes \Bbb Q$ is
a commutative $\Bbb Q$-algebra.
If you are asking why consider such tensor products, suppose $A$ is a ring.
Then each ring homomorphism $\phi:A\to B$ where $B$ is a $\Bbb Q$-algebra
factors uniquely through $i:A\mapsto A\otimes\Bbb Q$, that is $\phi
=\psi\circ i$ where $\psi$ is a uniquely determined homomorphism
from $A\otimes\Bbb Q$ to $B$.
