Ring Monomorphism from Integers to Rationals
Jump to navigation
Jump to search
Theorem
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.
Proof
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 $a, b \in \Z$.
\(\ds \map \phi {a + b}\) | \(=\) | \(\ds \frac {a + b} 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac a 1 + \frac b 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi a + \map \phi b\) |
\(\ds \map \phi {a \times b}\) | \(=\) | \(\ds \frac {a \times b} 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {a \times b} {1 \times 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac a 1 \times \frac b 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi a \times \map \phi b\) |
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.
$\blacksquare$
Sources
- 1964: Iain T. Adamson: Introduction to Field Theory ... (previous) ... (next): Chapter $\text {I}$: Elementary Definitions: $\S 3$. Homomorphisms: Example $4$