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Question: If $R$ is a finite boolean ring, then show $R \cong \mathbb Z_2 \times \mathbb Z_2 \times \cdots\times \mathbb Z_2$.

I know that

  1. $\mathrm{char}(R) =2$

  2. $R$ has $2^k$ elements for some $k$ (this is because, if $p\neq 2$ is a prime with $p\mid|R|$, then there exists a non-zero $x \in R$ s.t. $px=0$ and $p $ is odd $\implies x=0$, contradiction).

Does this suffice to answer my question?

Zev Chonoles
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jim
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2 Answers2

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Once you know that $R$ has characteristic $2$, it will be a vector space (an algebra, really) over the field $F$ with $2$ elements, of dimension $d$, say. So $R$ has $2^d$ elements.

For each $a \in R$, consider the linear map $T_a : R \to R$ given by $u \to au$. Note that $a \mapsto T_a$ gives a morphism of rings from $R$ to $\operatorname{End}(R)$ (these are the endomorphisms of $R$ as a vector space over $F$). This morphism is injective because $T_a(a) = a$, so its kernel is $0$. If $S$ is its image, then $R$ is isomorphic to $S$.

Since $S$ is commutative, and all of its elements are roots of $x^2-x$, the elements of $S$ can be simultaneously diagonalized, with $0$ and $1$ on the diagonal. There are $2^d$ matrices like this, and $R \cong S$ has $2^d$ elements, so $S$ is the whole ring of diagonal matrices. Choose the (canonical) basis $T_{a_{1}}, \dots, T_{a_{d}}$ such that all $T_{a_{i}}$ have only one $1$ on the diagonal (which corresponds to $T_{a_i}(a_i) = a_i^2=a_i$).

Then with respect to the basis ${a_{1}}, \dots, {a_{d}}$ the ring $R$ has the form you require, because $a_i a_j = 0$ for $i \ne j$.

I guess this is just a version for this case of Stone's representation theorem.

Andreas Caranti
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  • sir thanks for the proof . is their any other way to show this result because sir some of the results you have used are not familiar to me? – jim Feb 16 '13 at 15:48
  • @jim, you're welcome. An alternate proof depends on what you have covered so far. However, I am really using some linear algebra only. – Andreas Caranti Feb 16 '13 at 15:55
  • sir more of a abstract algebra oriented proof rather than linear algebra – jim Feb 16 '13 at 15:57
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No, it does not suffice, because (for example) the finite field with $2^k$ elements, $\mathbb{F}_{2^k}$, satisfies both of the properties listed, and this is not isomorphic to $(\mathbb{Z}/2\mathbb{Z})^k$ when $k>1$ because in $\mathbb{F}_{2^k}$, the zero ideal is prime.

Another counterexample would be $\mathbb{F}_2[x]/(x^k)$. We can see that this is not isomorphic to $(\mathbb{Z}/2\mathbb{Z})^k$ when $k>1$ because any element of $(\mathbb{Z}/2\mathbb{Z})^k$ is idempotent, while this is not true of $\mathbb{F}_2[x]/(x^k)$.

Zev Chonoles
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    $(R,+)$ is an abelian group and for every non zero element a , $2a=0$ holds so it cannot have elements of additive order more than $2$ .does this help to show the result? – jim Feb 16 '13 at 12:05
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    @jim, it helps a bit, but that has only to do with the additive structure, to prove that there is an isomorphism you have to tackle the multiplicative structure as well. – Andreas Caranti Feb 16 '13 at 15:30