Ring Monomorphism from Integers to Rationals

From ProofWiki
Jump to navigation Jump to search


Let $\phi: \Z \to \Q$ be the mapping from the integers $\Z$ to the rational numbers $\Q$ defined as:

$\forall x \in \Z: \map \phi x = \dfrac x 1$

Then $\phi$ is a (ring) monomorphism, but specifically not an epimorphism.


First note that:

$\forall a, b \in \Z: a \ne b \implies \dfrac a 1 = \map \phi a \ne \map \phi b = \dfrac b 1$

and so clearly $\phi$ is an injection.

However, take for example $\dfrac 1 2$:

$\not \exists a \in \Z: \map \phi a = \dfrac 1 2$

as $\dfrac 1 2 \notin \Img \phi$.

So $\phi$ is not a surjection.

Next let $a, b \in \Z$.

\(\displaystyle \map \phi {a + b}\) \(=\) \(\displaystyle \frac {a + b} 1\)
\(\displaystyle \) \(=\) \(\displaystyle \frac a 1 + \frac b 1\)
\(\displaystyle \) \(=\) \(\displaystyle \map \phi a + \map \phi b\)

\(\displaystyle \map \phi {a \times b}\) \(=\) \(\displaystyle \frac {a \times b} 1\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {a \times b} {1 \times 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac a 1 \times \frac b 1\)
\(\displaystyle \) \(=\) \(\displaystyle \map \phi a \times \map \phi b\)

So $\phi$ can be seen to be an injective, but not surjective, ring homomorphism.

Hence the result by definition of ring monomorphism and ring epimorphism.