Two linear algebra questions (finitely generated, isomorphism)

**Ester** · September 17th 2010, 08:24 AM

I have two exercises which are driving me crazy.

1) Is Q finitely generated Z-module?

2) Let $d,e \in N/\{0\}$ . Show, that $Hom_Z(Z_d,Z_e) \cong Z_f$ , where $f = gcd (d,e)$ .
I have previously proved that $Hom_Z(Z_d, G) \cong \{x \in G | dx = 0\}$ , where G is an Abelian group and $d \in N/\{0\}$ , so it's enough to show that $H:=\{\overline{n} \in Z_e | d\overline{n} = 0\} \cong Z_f$ . And I have been given the rule $\overline{k} \mapsto \overline{k(e/f)}$ . So I have to show that $f: Z_f \rightarrow H$ , $f(\overline{k}) = \overline{k(e/f)}$ , is isomorphism.

I have been able to show that this function is homomorphism and injective, but the problem here is how I can show that this function is surjective?

**Bruno J.** · September 17th 2010, 09:11 AM

Originally Posted by Ester

I have two exercises which are driving me crazy.

1) Is Q finitely generated Z-module?

2) Let $d,e \in N/\{0\}$ . Show, that $Hom_Z(Z_d,Z_e) \cong Z_f$ , where $f = gcd (d,e)$ .
I have previously proved that $Hom_Z(Z_d, G) \cong \{x \in G | dx = 0\}$ , where G is an Abelian group and $d \in N/\{0\}$ , so it's enough to show that $H:=\{\overline{n} \in Z_e | d\overline{n} = 0\} \cong Z_f$ . And I have been given the rule $\overline{k} \mapsto \overline{k(e/f)}$ . So I have to show that $f: Z_f \rightarrow H$ , $f(\overline{k}) = \overline{k(e/f)}$ , is isomorphism.

I have been able to show that this function is homomorphism and injective, but the problem here is how I can show that this function is surjective?

For the first one, take a finite set of rational numbers and look at the $\mathbb{Z}$ -submodule which they generate. What are the possible denominators of its elements?

**Ester** · September 18th 2010, 01:14 AM

Originally Posted by Bruno J.

For the first one, take a finite set of rational numbers and look at the $\mathbb{Z}$ -submodule which they generate. What are the possible denominators of its elements?

Okey... I have no idea, but I came up something like this:

$S = \{x_1, ... , x_n\}$ where $x_1, ... , x_n \in Q$ . So $S \subset Q$ . Z -submodule, which they generate, is <S>, right? And if $a_1, ... , a_n \in Z$ and $x_1, ... , x_n \in S$ , then because $S \subset <S>$ and $<S>$ is submodule, $a_1x_1 + ... + a_nx_n \in <S>$ .

Am I even close?

**tonio** · September 18th 2010, 03:23 AM

Originally Posted by Ester

Okey... I have no idea, but I came up something like this:

$S = \{x_1, ... , x_n\}$ where $x_1, ... , x_n \in Q$ . So $S \subset Q$ . Z -submodule, which they generate, is <S>, right? And if $a_1, ... , a_n \in Z$ and $x_1, ... , x_n \in S$ , then because $S \subset <S>$ and $<S>$ is submodule, $a_1x_1 + ... + a_nx_n \in <S>$ .

Am I even close?

You´re close, in fact on it, to the definition of Z-module but not to show that the finitely generated Z-submodule S of Q cannot be the hole Q...
Read carefully what Bruno wrote you: in any Z-combination $a_1z_1+\ldots+a_nz_n$ of the elements of S, how many

posible denominators can there be? How many primes are there?...

**Ester** · September 18th 2010, 06:01 AM

Originally Posted by tonio

You´re close, in fact on it, to the definition of Z-module but not to show that the finitely generated Z-submodule S of Q cannot be the hole Q...
Read carefully what Bruno wrote you: in any Z-combination $a_1z_1+\ldots+a_nz_n$ of the elements of S, how many

posible denominators can there be? How many primes are there?...

Well... How many primes? Infinite amount.
Does this mean that if I choose an arbitrary $k \in N$ and an arbitrary $a_1x_1+ ... + a_nx_n \in <S>$ , then $(a_1x_1+ ... + a_nx_n)/k \in Q$ but it doesn't belong in <S>?

**tonio** · September 18th 2010, 10:33 AM

Originally Posted by Ester

Well... How many primes? Infinite amount.
Does this mean that if I choose an arbitrary $k \in N$ and an arbitrary $a_1x_1+ ... + a_nx_n \in <S>$ , then $(a_1x_1+ ... + a_nx_n)/k \in Q$ but it doesn't belong in <S>?

I don't understand what you meant

1) how many possible primes can divide ANY denominator in a linear combination

$a_1x_2+\ldots+a_nx_n\in S$ , with $a_i\in\mathbb{Z}$ ? After you answer this question pass on to the infinitude of primes...

Tonio

**Ester** · September 19th 2010, 01:18 AM

I came up something like this:

Let's choose arbitrary $s \in <S>$ , so $s = a_1x_1 + ... + a_nx_n$ , where $a_i \in Z$ and let's choose $w \in N/\{0,1\}$ so that $gcd(w,x_i) = 1$ . Now $s/w \in Q$ .

Antithesis: $s/w \in <S>.$
$s/w = (a_1x_1 + ... + a_nx_n)/w = (a_1x_1)/w + ... + (a_nx_n)/w$
Now $a_1/w, ... , a_n/w \in Z$ iff w=1. But this is contradiction. So s/w does not belong in <S>.

**tonio** · September 19th 2010, 04:46 AM

Originally Posted by Ester

I came up something like this:

Let's choose arbitrary $s \in <S>$ , so $s = a_1x_1 + ... + a_nx_n$ , where $a_i \in Z$ and let's choose $w \in N/\{0,1\}$ so that $gcd(w,x_i) = 1$ . Now $s/w \in Q$ .

Antithesis: $s/w \in <S>.$
$s/w = (a_1x_1 + ... + a_nx_n)/w = (a_1x_1)/w + ... + (a_nx_n)/w$
Now $a_1/w, ... , a_n/w \in Z$ iff w=1. But this is contradiction.

No contradiction at all: $w$ was chosen to be pairwise coprime with the $x_i's$ , not with the $a_i\s$ ...

You insist in not paying attention to the hint you've been given already three times...

Tonio

So s/w does not belong in <S>.

.

**Ester** · September 19th 2010, 05:26 AM

Okey.

I'm not paying attention to the hint, because I don't get it.

**tonio** · September 19th 2010, 03:30 PM

Originally Posted by Ester

Okey.

I'm not paying attention to the hint, because I don't get it.

1) How many different primes can be part of any of the denominators of $\{x_1,\ldots,x_n\}$ ?

2) If $a_i\in\mathbb{Z}$ , can the number of primes in the denominator of $a_1x_1+\ldots+a_nx_n$ be different from the number in (1)?

3) Since there are INFINITE primes then...

Tonio

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