# Thread: Ring homomorphism - isomorphism

1. ## Ring homomorphism - isomorphism

Hello

Could you please guys help me with the following exercise?

Let $\displaystyle x: Z -->Zn$ a function defined by $\displaystyle x(a)=[a]$

Show that$\displaystyle x$ es a homomorphism

Is $\displaystyle x$ a rings isomorphism?

Thanks

2. Originally Posted by osodud
Hello

Could you please guys help me with the following exercise?

Let $\displaystyle x: Z -->Zn$ a function defined by $\displaystyle x(a)=[a]$

Show that $\displaystyle x$ is a homomorphism

Is $\displaystyle x$ a rings isomorphism?

Thanks
What do the brackets represent in this case? The floor function?

To show that its a homomorphism, show that $\displaystyle x(a+b)=\dots=x(a)+x(b)$ and $\displaystyle x(ab)=\dots=x(a)x(b)$.

Now try to see if $\displaystyle \ker x=\{0\}$ and if $\displaystyle x$ is onto. If both are satisfied, its a ring isomorphism.

3. Originally Posted by osodud
Hello

Could you please guys help me with the following exercise?

Let $\displaystyle x: Z -->Zn$ a function defined by $\displaystyle x(a)=[a]$

Show that$\displaystyle x$ es a homomorphism

Is $\displaystyle x$ a rings isomorphism?

Thanks
The homomorphism part follows from the fact that addition of congruence classes modulo $\displaystyle n$ is defined as $\displaystyle [a]+[b]=[a+b]$ (and the operation is well defined).

4. Originally Posted by Chris L T521
What do the brackets represent in this case? The floor function?
The square brackets represent the equivalence class of the thing in the brackets - $\displaystyle x, y \in [a] \iff x \equiv y \equiv a mod n$ (equivalently, $\displaystyle x-y \equiv x-a \equiv y-a \equiv 0 mod n$). Essentially, $\displaystyle [a] = a+\mathbb{Z}n$ (with $\displaystyle \mathbb{Z}n = \{ni : i \in \mathbb{Z} \}$).

Hint for part 2: Notice that $\displaystyle [a+n] = [a]$...So, how many elements are in $\displaystyle im(x)$, the image of the function $\displaystyle x$?