If an integral domain $R$ is a free $\mathbb{Z}$-module, and $I$ is an ideal of $R$, do $R$ and $I$ have the same rank as $\mathbb{Z}$-modules? Assume $I$ is non-zero.
If domain $R$ is a free $\mathbb{Z}$-module, and $I$ is an integral ideal of $R$, have $R$ and $I$ the same rank as $\mathbb{Z}$-modules?
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abstract-algebra
algebraic-number-theory
3 Answers
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If $R$ is of finite rank, then yes. If $a$ is a nonzero element of $I$, then $Ra$ is a free abelian group inside $I$ of rank equal to that of $R$, so that $I$ has rank at least equal to that of $R$. As it has rank at most equal to that of $R$, we have equality.
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2The same argument works in the infinite case, I guess. – 2017-02-08
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I believe so. Let $\{b_1,\cdots,b_n\}$ be a $\Bbb Z$-basis for $R$, and let $w\in I$, nonzero element. Then $\{wb_i\}$ are still $\Bbb Z$-linearly independent, so the rank of $I$ is at least $n$. And at most $n$ as well, certainly.
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As an aside: through similar arguments offered by the other answers, one also has the result hold for nonzero fractional ideals $I$ of $R$.