# Ring homomorphism is injective if and only if the induced maps on localizations are injective

Suppose $\phi:A\to B$ is a ring homomorphism, where $A,B$ are commutative. Then for every $p\in A$, there is an induced map $\phi_p: A_p\to B_{\phi(p)}$ on the localization defined by $\phi_p(x/q) = \phi(x)/\phi(q)$.

I would like to show that $\phi$ is injective if and only if $\phi_p$ is injective for every $p\in A$, but I have some problems.

Suppose $\phi$ is injective. If $\phi_p(x/q) = 0$, then $\phi(x)/\phi(q) = 0$. Thus there exists an $r\in \{\phi(p)^n | n \in \mathbb N\}$ such that $r\phi(x) =0$. If I could show that this implies that there exists an $s\in \{p^n | n\in \mathbb N\}$ such that $sx =0$, then this would show that $\phi_p$ is injective.

Edit: I got confused between localization at a prime and what I really wanted. Now it is easy to solve the problem.

Any help would be appreciated! Thanks!

• What are your $p$'s? At first, it looks like you are localizing the powers of a single element, but then when you write the condition, it seems like you are localizing at a prime ideal (which need not be principal). – Tobias Kildetoft Jan 23 '13 at 20:25
• The $p's$ are any element of $A$. I am localizing the powers of a single element. – Zoltan Jan 23 '13 at 20:26
• Then how did your condition become $r\not\in (\phi(p))$? – Tobias Kildetoft Jan 23 '13 at 20:28
• I see your point. It should be the set of all powers of $\phi(p)$. – Zoltan Jan 23 '13 at 20:30
• Not only that, it should be $\in$ rather than $\not\in$. You should probably take a closer look at how the equivalence relation on the localized ring is defined again. – Tobias Kildetoft Jan 23 '13 at 20:33

First suppose $\phi$ is injective, and suppose that $$\frac{\phi(x)}{\phi(q)}=\frac{\phi(x)}{\phi(p^n)}=\frac{\phi(x)}{\phi(p)^n}=0\in B_{\phi(p)}$$
This means there is some $m\in\mathbb{N}$ such that $\phi(p)^m\phi(x)=0$ in $B$. By injectivity it follows that $p^mx=0$ in $A$ so that $$\frac{x}{q}=\frac{x}{p^n}=0\in A_p$$
Hence, $\phi_p$ is injective for all $p\in A$.
The reverse direction follows from considering the case $p=1\in A$.