# Question about a module of rank 2 over a domain

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 $$R$$?

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).
• $H$ is never a fractional ideal, since a fractional ideal would have rank 1 (or rank 0 if it's trivial). Apr 29, 2021 at 7:24
• 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 $k$. Apr 29, 2021 at 13:33
• 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 $k\otimes k\cong k$. Apr 29, 2021 at 15:18
• Oh yeah, $k$ is a localisation of $R$. Will edit. Apr 29, 2021 at 16:57