# Why injection on element $0$ in a ring homomorphism implies injection on the others?

If I am not mistaken I have (implicitly) seen for several times that in a ring homomorphism if $\phi(0)=0$ and $\phi(a)\ne 0$ for any other element $a\ne 0$ so all the elements of $A$ in $\phi : A \to B$ are injectively mapped to $B$.

But I don't know its proof! Simple detailed explanation would be much appreciated.

Added: For example the following is written in Ch. 7 Abstract Algebra by Dummit and Foote :

Corollary 10. If R is a field then any nonzero ring homomorphism from R into another ring is an injection.

Proof: The kernel of a ring homomorphism is an ideal. The kernel of a nonzero homomorphism is a proper ideal hence is $0$ by the proposition.

• Suppose not..... see where that takes you Aug 28, 2016 at 9:24
• @DanRust, yes I tried argument by contradiction but no much success! :(
– user200918
Aug 28, 2016 at 9:26
• I really think you should spend some more time on this yourself. The contradiction is fairly easy to spot by using the fact that homomorphisms respect addition. Aug 28, 2016 at 9:28
• @DanRust, the problem is that I can't prove $\phi(-a) =-\phi(a)$ by knowing that $\phi(0) =0$
– user200918
Aug 28, 2016 at 9:31
• $\phi$ is a homomorphism, so $\phi(-a) = -\phi(a)$ always. Aug 28, 2016 at 9:33

If there are two elements $a_1,a_2\in A$ such that $\phi(a_1)=\phi(a_2)$, then the element $a_1-a_2\in R$ must have $$\phi(a_1-a_2)=\phi(a_1)-\phi(a_2)=0_B\in B$$ (where we have used the fact that $\phi$ is a ring homomorphism). So if $0_A$ is the only element of $A$ mapped to $0_B$ by $\phi$, then we must have $a_1-a_2=0_A$, i.e., $a_1=a_2$.

Therefore $\phi$ is injective.

A stronger fact that comes from this is the first isomorphism theorem, which says that for any ring homomorphism $\phi:A\to B$, the set $$\ker(\phi)=\{a\in A:\phi(a)=0_B\}$$ is in fact an ideal of the ring $A$, and that $\phi$ can be written as a composition of three maps, $$A \xrightarrow[\;\;\;\text{(surjective)}\;\;\;]{\text{quotient map}} A/\ker(\phi) \xrightarrow[\;\;\;\text{(bijective)}\;\;\;]{\text{induced map}} \mathrm{im}(A) \xrightarrow[\;\;\;\text{(injective)}\;\;\;]{\text{inclusion map}} B$$

• Since $\phi$ is a homomorphism, $\phi(a_1+a_2)=\phi(a_1)+\phi(a_2)$ for any $a_1,a_2\in A$. Therefore $$\phi(a)+\phi(-a)=\phi(a+(-a))=\phi(0_A)=0_B$$ and therefore $\phi(-a)$ is the additive inverse of the element $\phi(a)$ in the ring $B$. Aug 28, 2016 at 9:35
• Its enough if $\phi$ is group homomorphism Aug 28, 2016 at 9:38
• @L.G. No, that can never happen. In fact, the definition of group homomorphism is only that $\psi:G\to H$ satisfies $$\psi(g_1\cdot g_2)= \psi(g_1)\star\psi(g_2)$$ where $\cdot$ is the operation of $G$ and $\star$ is the operation of $H$, and the definition of ring homomorphism is only that $\phi:A\to B$ satisfies $$\phi(a_1 \mathbin{+_A}a_2)=\phi(a_1) \mathbin{+_B}\phi(a_2)\qquad \phi(a_1 \mathbin{\cdot_A} a_2)=\phi(a_1) \mathbin{\cdot_B}\phi(a_2)$$ where $\mathbin{+_A},\mathbin{\cdot_A}$ are the operations of $A$ and $\mathbin{+_B},\mathbin{\cdot_B}$ are the operations of $B$. Aug 28, 2016 at 10:01
• If $\psi:G\to H$ is a group homomorphism, and $e_G\in G$ and $e_H\in H$ are the groups' respective identity elements, then because $e_G=e_G\cdot e_G$, we have $$\psi(e_G)=\psi(e_G\cdot e_G)= \psi(e_G)\star\phi(e_G)$$ and since $\psi(e_G)$ is some element of $H$, it has an inverse, and multiplying both sides of the above by the inverse of $\psi(e_G)$ produces \begin{align*} \psi(e_G)\star (\psi(e_G))^{-1}&= \psi(e_G)\star\psi(e_G)\star (\psi(e_G))^{-1}\\ e_H&=\psi(e_G) \end{align*} Aug 28, 2016 at 10:04
• The same argument shows that a ring homomorphism $\phi:A\to B$ necessarily must have $\phi(0_A)=0_B$, it is a theorem that you can prove about homomorphisms, not an assumption. Aug 28, 2016 at 10:06

This fact should be familiar from linear algebra. In a sense, kernel determines behavior of the map on whole domain.

The proof is simple:

If $f$ is injective and $x\in\ker f$, then $$f(x) = 0 = f(0) \implies x = 0 \implies \ker f = 0.$$

If $\ker f = 0$ and $f(x) = f(y)$, then $$f(x)-f(y) = 0 \implies f(x-y) = 0 \implies x-y\in\ker f\implies x - y = 0,$$ and thus $f$ is injective.

Thus $f$ is injective if and only if $\ker f$ is trivial.

Note that this is closely related to the First Isomorphism Theorem which states that for $f\colon A\to B$ we have $A/\ker f \cong f(A)$, thus any homomorphism discriminates elements of $A$ only up to its kernel.

Let $G$ and $K$ be two groups and let $\varphi:G\rightarrow K$ be a group homomorphism.

Suppose that the kernel of $\varphi$ consists of just the identity element $e_G$ of $G$. We will prove that this implies that $\varphi$ is injective. Let $g$ and $h$ be elements of $G$ such that $\varphi(g)=\varphi(h)$. Multiplying the inverse of $\varphi(h)$ by both sides of the equation, we get $\varphi(g)\varphi(h)^{-1}=e_H$, but since $\varphi$ is a homomorphism, then $\varphi(g)\varphi(h)^{-1}=\varphi(gh^{-1})=e_H$. This means that the element $gh^{-1}$ is an element of the kernel of $\varphi$. But remember that the only element in the kernel is $e_G$. This means that $gh^{-1}=e_G$. Therefore $g=h$ and so $\varphi$ is injective.

On the other hand, suppose that $\varphi$ is injective. Let $g$ and $h$ be elements of the kernel of $\varphi$. This means that $\varphi(g)=\varphi(h)=e_H$. Since $\varphi$ is injective, then $g=h$. This means that there is only one element in the kernel of $\varphi$ and since $\varphi(e_G)=e_H$, then $e_G$ is the only element in the kernel.

Thus we have proven that a group homomorphism being injective is the same as it having a trivial kernel.

Since the kernel of a ring homomorphism is the same the as the kernel of the corresponding group homomorphism, then this also applies to ring homomorphisms.