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Kevin Ventullo
Kevin Ventullo
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Here's an elementary proof that doesn't require ultrafilters, but uses axiom of choice.

The group

$\prod_{n=1}^\infty \mathbb{Z} / \bigoplus_{n=1}^\infty \mathbb{Z}$ \begin{equation*} \prod_{n=1}^\infty \mathbb{Z} / \bigoplus_{n=1}^\infty \mathbb{Z} \end{equation*}

clearly surjects onto

$\prod_{n=1}^\infty (\mathbb{Z}/p\mathbb{Z}) / \bigoplus_{n=1}^\infty (\mathbb{Z}/p\mathbb{Z})$. \begin{equation*} \prod_{n=1}^\infty (\mathbb{Z}/p\mathbb{Z}) / \bigoplus_{n=1}^\infty (\mathbb{Z}/p\mathbb{Z}). \end{equation*}

The latter is a nontrivial $\mathbb{F}_p$-vector space, being a $p$-torsion abelian group. Therefore, we can choose a basis $\{e_i\}_{i\in I}$. Finally, we get a map into $\mathbb{Z}/p\mathbb{Z}$ by killing all but one of the $e_i$.

Here's an elementary proof that doesn't require ultrafilters, but uses axiom of choice.

The group

$\prod_{n=1}^\infty \mathbb{Z} / \bigoplus_{n=1}^\infty \mathbb{Z}$

clearly surjects onto

$\prod_{n=1}^\infty (\mathbb{Z}/p\mathbb{Z}) / \bigoplus_{n=1}^\infty (\mathbb{Z}/p\mathbb{Z})$.

The latter is a nontrivial $\mathbb{F}_p$-vector space, being a $p$-torsion abelian group. Therefore, we can choose a basis $\{e_i\}_{i\in I}$. Finally, we get a map into $\mathbb{Z}/p\mathbb{Z}$ by killing all but one of the $e_i$.

Here's an elementary proof that doesn't require ultrafilters, but uses axiom of choice.

The group \begin{equation*} \prod_{n=1}^\infty \mathbb{Z} / \bigoplus_{n=1}^\infty \mathbb{Z} \end{equation*}

clearly surjects onto \begin{equation*} \prod_{n=1}^\infty (\mathbb{Z}/p\mathbb{Z}) / \bigoplus_{n=1}^\infty (\mathbb{Z}/p\mathbb{Z}). \end{equation*}

The latter is a nontrivial $\mathbb{F}_p$-vector space, being a $p$-torsion abelian group. Therefore, we can choose a basis $\{e_i\}_{i\in I}$. Finally, we get a map into $\mathbb{Z}/p\mathbb{Z}$ by killing all but one of the $e_i$.

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Kevin Ventullo
Kevin Ventullo
  • 4.8k
  • 1
  • 30
  • 41

Here's aan elementary proof that doesn't explicitly require ultrafilters, but uses axiom of choice.

The group

$\prod_{n=1}^\infty \mathbb{Z} / \bigoplus_{n=1}^\infty \mathbb{Z}$

clearly surjects onto

$\prod_{n=1}^\infty (\mathbb{Z}/p\mathbb{Z}) / \bigoplus_{n=1}^\infty (\mathbb{Z}/p\mathbb{Z})$.

The latter is a nontrivial $\mathbb{F}_p$-vector space, being a $p$-torsion abelian group. Therefore, we can choose a basis $\{e_i\}_{i\in I}$. Finally, we get a map into $\mathbb{Z}/p\mathbb{Z}$ by killing all but one of the $e_i$.

Here's a proof that doesn't explicitly require ultrafilters, but uses axiom of choice.

The group

$\prod_{n=1}^\infty \mathbb{Z} / \bigoplus_{n=1}^\infty \mathbb{Z}$

clearly surjects onto

$\prod_{n=1}^\infty (\mathbb{Z}/p\mathbb{Z}) / \bigoplus_{n=1}^\infty (\mathbb{Z}/p\mathbb{Z})$.

The latter is a nontrivial $\mathbb{F}_p$-vector space, being a $p$-torsion abelian group. Therefore, we can choose a basis $\{e_i\}_{i\in I}$. Finally, we get a map into $\mathbb{Z}/p\mathbb{Z}$ by killing all but one of the $e_i$.

Here's an elementary proof that doesn't require ultrafilters, but uses axiom of choice.

The group

$\prod_{n=1}^\infty \mathbb{Z} / \bigoplus_{n=1}^\infty \mathbb{Z}$

clearly surjects onto

$\prod_{n=1}^\infty (\mathbb{Z}/p\mathbb{Z}) / \bigoplus_{n=1}^\infty (\mathbb{Z}/p\mathbb{Z})$.

The latter is a nontrivial $\mathbb{F}_p$-vector space, being a $p$-torsion abelian group. Therefore, we can choose a basis $\{e_i\}_{i\in I}$. Finally, we get a map into $\mathbb{Z}/p\mathbb{Z}$ by killing all but one of the $e_i$.

Source Link
Kevin Ventullo
Kevin Ventullo
  • 4.8k
  • 1
  • 30
  • 41

Here's a proof that doesn't explicitly require ultrafilters, but uses axiom of choice.

The group

$\prod_{n=1}^\infty \mathbb{Z} / \bigoplus_{n=1}^\infty \mathbb{Z}$

clearly surjects onto

$\prod_{n=1}^\infty (\mathbb{Z}/p\mathbb{Z}) / \bigoplus_{n=1}^\infty (\mathbb{Z}/p\mathbb{Z})$.

The latter is a nontrivial $\mathbb{F}_p$-vector space, being a $p$-torsion abelian group. Therefore, we can choose a basis $\{e_i\}_{i\in I}$. Finally, we get a map into $\mathbb{Z}/p\mathbb{Z}$ by killing all but one of the $e_i$.