An isomorphism

**Capillarian** · February 22nd 2011, 06:09 PM

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

Deduce that $\mathbb{Z}^r / \langle (x_1,\ldots,x_r)\rangle \cong \mathbb{Z}^{r-1} \oplus (\mathbb{Z}/h\mathbb{Z})$ .
__________________________________________________ __

The first part isn't a problem. But the deduction is something I'm having trouble with... help?

**Swlabr** · February 23rd 2011, 12:28 AM

Originally Posted by Capillarian

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

Deduce that $\mathbb{Z}^r / \langle (x_1,\ldots,x_r)\rangle \cong \mathbb{Z}^{r-1} \oplus (\mathbb{Z}/h\mathbb{Z})$ .
__________________________________________________ __

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 $\phi:\mathbb{Z}^r \rightarrow \mathbb{Z}^r$ with $(x_1, \ldots, x_r)\mapsto (h, 0, 0, \ldots, 0)$ where $h=gcd(x_1, x_2, \ldots, x_r)$ .

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

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

**Capillarian** · February 23rd 2011, 11:07 AM

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 $\phi:\mathbb{Z}^r \rightarrow \mathbb{Z}^r$ with $(x_1, \ldots, x_r)\mapsto (h, 0, 0, \ldots, 0)$ where $h=gcd(x_1, x_2, \ldots, x_r)$ .

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 ' $\implies$ ' direction still works, but if h = 1, how do you go about choosing an explicit isomorphism that maps $(x_1, \ldots, x_r)\mapsto (1, 0, 0, \ldots, 0)$ ?

**Swlabr** · February 24th 2011, 12:33 AM

The isomorphism is basically just $(x_1, x_2, \ldots, x_n) \mapsto (x_1i_1+x_2i_r \dlots x_ni_n, 0, \ldots, 0)$ where $x_1i_1+x_2i_2+\ldots+x_ni_n=gcd(x_j)=1$ . This is an isomorphism as one can easily prove that the $i_j$ 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.

Thread: An isomorphism

LinkBack

Thread Tools

Display

An isomorphism

Similar Math Help Forum Discussions

Isomorphism

isomorphism

isomorphism

Isomorphism proves: f and g are isomorphisms, prove f + g is an isomorphism

Isomorphism

Search Tags