Hello!

I am trying to show that $\displaystyle \alpha$ is an ismorphism,

$\displaystyle \alpha:U(st) \rightarrow U(s) \oplus U(t)$

defined by $\displaystyle \alpha(x) = (x\ mod\ s, x\ mod\ t) $ where $\displaystyle x \in U(st)$

where U(n) is the multiplicative group of units (refresher: numbers relativly prime to n)

and $\displaystyle \oplus$ is the external direct product.

so clearly $\displaystyle \alpha$ is well defined

I have shown it is 1-1

So my problem is...

I am having trouble showing is it onto (to show to that it is a bijection)

so let $\displaystyle b = (y_1,y_2) \in U(s) \oplus U(t)$. So b is an arbitrary element of $\displaystyle U(s) \oplus U(t)$.

so what we need to do is find some $\displaystyle x \in U(st)$ such that

$\displaystyle \alpha(x) = b = (y_1,y_2) $ (we must find a x that maps to b... defn of onto, right?)

Keep in mind that we know that $\displaystyle gcd(s, y_1) = gcd (t, y_2) = 1$ since $\displaystyle gcd(s,t) = 1 $

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So I thought maybe $\displaystyle x = y_1t +y_2s$?

then by the definition of alpha,

$\displaystyle \alpha(x) = ((y_1t +y_2s)\ mod\ s, (y_1t +y_2s)\ mod\ t) = ((y_1t)\ mod \ s, (y_2s)\ mod\ t)$

Now we just need to get rid of the t in the first component and s in the second component to map to that arbitrary element $\displaystyle b = (y_1, y_2) \in U(s) \oplus U(t)$

Yikes sorry for all the description. All help would be much appriciated, this actually stumped our class for a few minutes at the end of class today!

Thanks! in advance!