Ring homomorphism

**osodud** · June 23rd 2009, 07:04 AM

Hello

Could you please guys help me with the following exercise?

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

Show that $x$ es a homomorphism

Is $x$ a rings isomorphism?

Thanks

**Chris L T521** · June 23rd 2009, 11:22 AM

Originally Posted by osodud

Hello

Could you please guys help me with the following exercise?

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

Show that $x$ is a homomorphism

Is $x$ a rings isomorphism?

Thanks

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

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

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

**TheAbstractionist** · June 23rd 2009, 12:11 PM

Originally Posted by osodud

Hello

Could you please guys help me with the following exercise?

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

Show that $x$ es a homomorphism

Is $x$ a rings isomorphism?

Thanks

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

**Swlabr** · June 25th 2009, 03:57 AM

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 - $x, y \in [a] \iff x \equiv y \equiv a mod n$ (equivalently, $x-y \equiv x-a \equiv y-a \equiv 0 mod n$ ). Essentially, $[a] = a+\mathbb{Z}n$ (with $\mathbb{Z}n = \{ni : i \in \mathbb{Z} \}$ ).

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

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