# Thread: Two linear algebra questions (finitely generated, isomorphism)

1. ## Two linear algebra questions (finitely generated, isomorphism)

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?

2. 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?

3. 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 $ and $$ is submodule, $a_1x_1 + ... + a_nx_n \in $.

Am I even close?

4. 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 $ and $$ is submodule, $a_1x_1 + ... + a_nx_n \in $.

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?...

5. 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 $, then $(a_1x_1+ ... + a_nx_n)/k \in Q$ but it doesn't belong in <S>?

6. 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 $, then $(a_1x_1+ ... + a_nx_n)/k \in Q$ but it doesn't belong in <S>?

I don't understand what you meant1) 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

7. I came up something like this:

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

8. Originally Posted by Ester
I came up something like this:

Let's choose arbitrary $s \in $, 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: $s/w \in .$
$s/w = (a_1x_1 + ... + a_nx_n)/w = (a_1x_1)/w + ... + (a_nx_n)/w$
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>.
.

9. Okey.

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

10. 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