Encountered difficulties in proving the attached image. Greatly appreciate for the help!
A counter example is $\displaystyle \begin{align*} 2\,\textrm{mod}\,4 \times 2\,\textrm{mod}\,4 = 0\,\textrm{mod}\,4 \end{align*}$.
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Prove It
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Alexis87 said:
Encountered difficulties in proving the attached image. Greatly appreciate for the help!
As for the proof: Since $\displaystyle \begin{align*} p \end{align*}$ is prime, it is a positive integer.
$\displaystyle \begin{align*} 0\,\textrm{mod}\,p = k \, p \end{align*}$, where $\displaystyle \begin{align*} k \end{align*}$ is some integer.
As $\displaystyle \begin{align*} p \end{align*}$ is prime, the only way to get a multiple of $\displaystyle \begin{align*} p \end{align*}$ is if that number has $\displaystyle \begin{align*} p \end{align*}$ as a factor. But any multiple of $\displaystyle \begin{align*} p \end{align*}$ is itself $\displaystyle \begin{align*} 0\,\textrm{mod}\,p \end{align*}$, and thus the only way to get $\displaystyle \begin{align*} 0\,\textrm{mod}\,p \end{align*}$ through multiplication in $\displaystyle \begin{align*} \mathbf{Z}_p \end{align*}$ is to multiply by $\displaystyle \begin{align*} 0\,\textrm{mod}\,p \end{align*}$.
It is well known that a vector space always admits an algebraic (Hamel) basis. This is a theorem that follows from Zorn's lemma based on the Axiom of Choice (AC).
Now consider any specific instance of vector space. Since the AC axiom may or may not be included in the underlying set theory, might there be examples of vector spaces in which an Hamel basis actually doesn't exist ?