The statement is the following:
Let $R$ be a ring with identity, $I(\neq R)$ an ideal of $R$,$F$ a free $R$-module with basis $X$ and $\pi:F\rightarrow F/IF$ the canonical epimorphism. Then $F/IF$ is a free $R/I$-module with basis $\pi(X)$ and $|\pi(X)|=|X|$.(The action of $F/IF$ over $R/I$ is given by $(r+I)(a+IF)=ra+IF$)
The sketch of proof is the following:
- Prove that $\pi(x_1),\cdots,\pi(x_n)$ can span the image.
- Prove that $\pi(x_1),\cdots,\pi(x_n)$ is linear independent.
- Prove that the canonical epimorphism is one to one to conclude $|\pi(X)|=|X|$.
There is two things that I do not understand:
- Since both the basis $x_1,\cdots,x_n$ and the basis $\pi(x_1),\cdots,\pi(x_n)$ is linear independent. Why can't conclude $|\pi(X)|=|X|$ directly?
- I can not understand the one to one intuitionally(although I understand the proof) because I think the number of element of $F/IF$ and the $R/I$ is obviously smaller than that of $F$ and $R$.
Thanks for the answer! Since the theorem is found in Algebra by Hungerford, maybe I midunderstand the proof in the book when writing the above sketch of proof:
If there is something wrong with my understanding, could you point out?
Indeed, showing that $\pi(x_1),\ldots, \pi(x_n)$ are linearly independent suffices to show that $\lvert X\rvert=\lvert \pi(X)\rvert$. On the other hand, note that this quotient map is in general not injective unless you are quotienting by the zero ideal. For instance, $\Bbb{Z}$ is a free rank $1$ module over itself with basis $1$. Quotient by $(2)$ to get $\Bbb{Z}\to \Bbb{Z}/2\to 0$. This is clearly not injective (indeed, all even numbers are sent to $0\in \Bbb{Z}/2$).