Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $R$ and $S$ be rings and $f:R\to S$ a monomorphism. Is $f$ injective?

share|improve this question
    
Next exercise: show that there is an epic ring homomorphism that is not surjective. –  rschwieb Aug 16 '12 at 19:17
add comment

3 Answers 3

up vote 2 down vote accepted

Yes, $f$ is injective:

Let's assume the map of the underlying sets is NOT injective. Then there are distinct $x,y\in R$ such that $f(x)=f(y)$.

Next, consider the monomorphisms $g_1,g_2\colon \mathbf{Z}[X]\to R$ by $g_1(X)=x$ and $g_2(X)=y$. Note that these monomorphisms are different and $f\circ g_1= f\circ g_2$. This implies $f$ is NOT a monomorphism, which leads to a contradiction.

share|improve this answer
add comment

Assume that there is a monomorphism of rings $\varphi:A\longrightarrow B$ which is not injective, i.e. there are $a_1\neq a_2$ s.t. $\varphi(a_1)= \varphi(a_2)$.

Now if you consider the maps $g_1$ and $g_2:\mathbb{Z}[X]\longrightarrow A$ defined by $g_i(X)=a_i$, you can check that $\varphi\circ g_1=\varphi\circ g_2$, but $g_1\neq g_2$...

share|improve this answer
add comment

Yes. Suppose $f$ is a monomorphism which is not injective, ie $f(a)=f(b)$. Now consider $g_1:\mathbb{Z}[x]\to R$ given by sending $x\mapsto a$ and $g_2:\mathbb{Z}[x]\to R$ given by $x\mapsto b$. Then $f\circ g_1=f\circ g_2$ despite the two maps being different. Thus monomorphisms must be injective.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.