# An isomorphism

• Feb 22nd 2011, 06:09 PM
Capillarian
An isomorphism
Let $\displaystyle (x_1,\ldots,x_r)$ be a non-zero element of $\displaystyle \mathbb{Z}^r$, and let h be the highest common factor of $\displaystyle x_1,\ldots,x_r$. Show that there is an isomorphism $\displaystyle \mathbb{Z}^r \to \mathbb{Z}^r$ taking $\displaystyle (x_1,\ldots,x_r)$ to (1, 0, 0,..., 0) if and only if h = 1.

Deduce that $\displaystyle \mathbb{Z}^r / \langle (x_1,\ldots,x_r)\rangle \cong \mathbb{Z}^{r-1} \oplus (\mathbb{Z}/h\mathbb{Z})$.
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The first part isn't a problem. But the deduction is something I'm having trouble with... help?
• Feb 23rd 2011, 12:28 AM
Swlabr
Quote:

Originally Posted by Capillarian
Let $\displaystyle (x_1,\ldots,x_r)$ be a non-zero element of $\displaystyle \mathbb{Z}^r$, and let h be the highest common factor of $\displaystyle x_1,\ldots,x_r$. Show that there is an isomorphism $\displaystyle \mathbb{Z}^r \to \mathbb{Z}^r$ taking $\displaystyle (x_1,\ldots,x_r)$ to (1, 0, 0,..., 0) if and only if h = 1.

Deduce that $\displaystyle \mathbb{Z}^r / \langle (x_1,\ldots,x_r)\rangle \cong \mathbb{Z}^{r-1} \oplus (\mathbb{Z}/h\mathbb{Z})$.
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The first part isn't a problem. But the deduction is something I'm having trouble with... help?

I like this question. It is subtle. So, you have proven the first part. Now, can you use this to generalise the result? That is, use it to prove that there is an isomorphism $\displaystyle \phi:\mathbb{Z}^r \rightarrow \mathbb{Z}^r$ with $\displaystyle (x_1, \ldots, x_r)\mapsto (h, 0, 0, \ldots, 0)$ where $\displaystyle h=gcd(x_1, x_2, \ldots, x_r)$.

HINT: This isomorphism is the same as the one you've found already...

Can you see how what you want follows from this result?
• Feb 23rd 2011, 11:07 AM
Capillarian
Quote:

Originally Posted by Swlabr
I like this question. It is subtle. So, you have proven the first part. Now, can you use this to generalise the result? That is, use it to prove that there is an isomorphism $\displaystyle \phi:\mathbb{Z}^r \rightarrow \mathbb{Z}^r$ with $\displaystyle (x_1, \ldots, x_r)\mapsto (h, 0, 0, \ldots, 0)$ where $\displaystyle h=gcd(x_1, x_2, \ldots, x_r)$.

HINT: This isomorphism is the same as the one you've found already...

Can you see how what you want follows from this result?

Ah, then I think I've made a mistake in the proof of the first part. The '$\displaystyle \implies$' direction still works, but if h = 1, how do you go about choosing an explicit isomorphism that maps $\displaystyle (x_1, \ldots, x_r)\mapsto (1, 0, 0, \ldots, 0)$?
• Feb 24th 2011, 12:33 AM
Swlabr
The isomorphism is basically just $\displaystyle (x_1, x_2, \ldots, x_n) \mapsto (x_1i_1+x_2i_r \dlots x_ni_n, 0, \ldots, 0)$ where $\displaystyle x_1i_1+x_2i_2+\ldots+x_ni_n=gcd(x_j)=1$. This is an isomorphism as one can easily prove that the $\displaystyle i_j$ are coprime (I'll leave you to do this) and so they `act like' primes. Again, I will leave you to prove that this is an isomorphism.