Let $R$ be a domain, $k=Q(R)$ the fraction field, and let $H$ be an $R$-module of rank $2$, i.e. $\dim_k(k \otimes_R H)=2$. Is it right to consider a basis of $k \otimes_R H$ as $(\frac{1}{r}\otimes h_1 ,\frac{1}{s}\otimes h_2 )$? Under what conditions is it reasonable to think that $H$ is a fractional ideal of $R$?


1 Answer 1


If $a$ is any element of $k\otimes_R H$ then by definition one can write $a = \sum_i \frac{r_i}{s_i}\otimes h_i$, which by taking common denominators reduces to something of the form $a=\frac{r}{s}\otimes h$. But one can pass this factor of $r$ on the left through the tensor product to get $a = \frac{1}{s}\otimes h'$ with $s\in R$ and $h' \in H$. So your assertion about the basis is correct.

Edit, re: Eric Wofsey's comments: if $H$ were a fractional ideal, then it would in particular be an $R$-submodule of $k$. But because $k$ is a localisation of $R$, it's flat, and so $k \otimes_R H$ is a $k$-submodule of $k\otimes_R k$, and this latter ring is isomorphic to $k$, again because $R \to k$ is a localisation. So $H$ must be rank 1 (or rank 0).

  • $\begingroup$ $H$ is never a fractional ideal, since a fractional ideal would have rank 1 (or rank 0 if it's trivial). $\endgroup$ Apr 29, 2021 at 7:24
  • $\begingroup$ Fractional ideals need not be principal, right? Just because the base change $k\otimes_R H$ isn't a submodule of $k$, it doesn't mean that $H$ itself is not be a submodule of $k$. $\endgroup$
    – Matt Booth
    Apr 29, 2021 at 13:33
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    $\begingroup$ Tensoring with $k$ is exact, so it preserves submodules. So if $H$ were a submodule of $k$ then $k\otimes H$ would be a submodule of $k\otimes k\cong k$. $\endgroup$ Apr 29, 2021 at 15:18
  • $\begingroup$ Oh yeah, $k$ is a localisation of $R$. Will edit. $\endgroup$
    – Matt Booth
    Apr 29, 2021 at 16:57

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