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

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Next exercise: show that there is an epic ring homomorphism that is not surjective. – rschwieb Aug 16 '12 at 19:17
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.

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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.

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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 $g_1\neq g_2$...

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