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If $R$ is a finite boolean ring then $R \cong \mathbb{Z}_2 \times \mathbb{Z}_2 \times \dots \times \mathbb{Z}_2$.

So this makes sense for a lot for some obvious reasons, and I feel like I could exhibit an isomorphism by defining a map by having the $ith$ element of $R$ make the $ith$ tuple a one instead of a zero, etc. It seems strange though that this isomorphism suggests a certain linear independence of the elements of $R$, can somebody give me some insight into this phenomonon? Thanks!

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    $\begingroup$ What are the operations in a boolean ring? $\endgroup$ – Wuestenfux Sep 13 '18 at 15:40
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    $\begingroup$ @Wuestenfux A boolean ring is a ring (with unit) in which every element is idempotent ($x^2 = x$). $\endgroup$ – Alex Kruckman Sep 15 '18 at 22:17
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Hint If $R$ is a boleean ring, then $\{ 0,1 \}$ is a subring which is isomorphic to $\mathbb Z_2$. Show that $R$ becomes a vector space over $\mathbb Z_2$.

Simpler approach If $R \neq \{ 0,1 \}$ then for each $x \in R \backslash \{ 0, 1 \}$ you have $$R= Rx \oplus R(1-x)$$

Now each of these are boolean rings (the units are $x$ and $1-x$ respectively) which are smaller than $R$, use strong induction.

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  • $\begingroup$ So to do this would I show that if $a_1x_1 + \dots + a_nx_n=0$ where $a_i \in \mathbb{Z}$ and $x_i \in R$ then $a_i=0 \forall i$? I'm having trouble doing this, care to help? $\endgroup$ – Math is hard Sep 15 '18 at 17:22
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    $\begingroup$ @MichaelVaughan if $R$ is any ring and $k$ is a subfield of $R$, then you can check with the axioms that $R$ is a vector space over $k$. Then, you simply USE the fact that every vector space has a basis, you don't need to prove linear independence. $\endgroup$ – N. S. Sep 15 '18 at 18:11
  • $\begingroup$ how do we know that the basis is finite dimensional without knowing linear independence though? Like.... idk given $\{x_i\}$ an enumeration of the elements of the Boolean ring $R$, how come $x_1 + x_2$ seems like it should equal some $x_i$, yet when I look at the isomorphism i'm trying to prove it seems like all these elements are linearly independent and you can't get to one of them from a linear combination of the others. $\endgroup$ – Math is hard Sep 15 '18 at 18:32
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    $\begingroup$ @MichaelVaughan First, since $R$ is finite, any basis $B$ is finite. Second, you seem to believe that $R$ is the basis, but this is obviously not true. Instead, you should attempt to do the following: Since $R$ is a vector space over $\mathbb Z_2$ then it has some basis $B$. Next, since $B \subset R$ and $R$ is finite then $B$ is finite. Let $B=\{x_1,..,x_n \}$, then $$a_1x_1+....+a_nx_n \leftrightarrow (a_1,...,a_n)$$ is the isomorphism. Note that since $B$ is a basis, it is by definition linearly independent. $\endgroup$ – N. S. Sep 15 '18 at 19:03
  • $\begingroup$ Ah yes, for some reason I was thinking $R$ was the basis. Thanks for being patient with me. For some reason this result still surprises me, like I understand each theorem that leads to the result but for some reason it's still strange. Thanks! $\endgroup$ – Math is hard Sep 15 '18 at 20:41

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