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I need help this question

Prove that every finite ring has the Invariant Dimension Property (IDP). (Assume $1_R \neq 0_R$.)

This is what I know I should do. Let $X$ and $Y$ be two sets such that the free module with basis $X$ is isomorphic to the free module with basis $Y$. Next, I have to show that $|X| = |Y|$. This is where I'm stuck.

Thanks.

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    Using cardinal arithmetic, under certain axioms of set theory, you can show that if $R$ is a finite set, then $|R^X|=|R^Y|$ implies that $|X|=|Y|$. This is elementary if $X$ and $Y$ are finite, but interesting things can happen otherwise, as seen at http://mathoverflow.net/questions/17152/when-2a-2b-implies-ab-a-b-cardinals. Of course, an isomorphism is more than just a bijection, and I believe that the answer to your question does not depend on the result mentioned above. Isomorphism of infinite rank free-modules always implies equality of the rank.2010-12-05
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    Maybe there is a more elementary proof for the case when $R$ is finite, but a reference is Hungerford's *Algebra*, Theorem IV.2.6 for the general case of infinite rank free modules over unital rings.2010-12-05

2 Answers 2

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Just count the number of elements. Since $R$ is finite, you can say exactly how many elements a free modulo of rank $k$ has. For finite $k$, that will do it. In the infinite case, you'll need to do a bit of cardinal work.

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As $R$ is finite, it is Artinian, so the modules $R^m$ and $R^n$ have finite length. You can now use the Jordan-Hölder theorem to deduce that $R^m\cong R^n$ implies $m=n$.

Or you can just count :-)