show/hide this revision's text 3 edited body

Thanks for your answers!!

If you take $A=\mathcal{O}_K$, the invertible $A$-modules are exactly the non-zero fractional ideals of $A$, so I guess we can reformulate the exercise to be

There is a bijection of sets $\{(n+1)$-tuples of elements of A such that $\exists i: a_i\neq 0 \}$ modulo equivalence, where equivalence is multiplication by a non-zero element of $A$K$

and

$\{A$-valued points of $\mathbb{P}^n_A\}$

Thus, for $\mathcal{O}_K$, the "classical" definition of points of projective $n$-space coincides with the definition of $\mathcal{O}_K$-valued points of $\mathbb{P}^n_{\mathcal{O}_K}$.
show/hide this revision's text 2 I also forgot "fractional" in the above...

Thanks for your answers!!

If you take $A=\mathcal{O}_K$, the invertible $A$-modules are exactly the non-zero fractional ideals of $A$, so I guess we can reformulate the exercise to be

There is a bijection of sets $\{(n+1)$-tuples of elements of A such that $\exists i: a_i\neq 0 \}$ modulo equivalence, where equivalence is multiplication by a unitnon-zero element of $A$

and

$\{A$-valued points of $\mathbb{P}^n_A\}$

Thus, for $\mathcal{O}_K$, the "classical" definition of points of projective $n$-space coincides with the definition of $\mathcal{O}_K$-valued points of $\mathbb{P}^n_{\mathcal{O}_K}$.
show/hide this revision's text 1

Thanks for your answers!!

If you take $A=\mathcal{O}_K$, the invertible $A$-modules are exactly the non-zero ideals of $A$, so I guess we can reformulate the exercise to be

There is a bijection of sets $\{(n+1)$-tuples of elements of A such that $\exists i: a_i\neq 0 \}$ modulo equivalence, where equivalence is multiplication by a unit

and

$\{A$-valued points of $\mathbb{P}^n_A\}$

Thus, for $\mathcal{O}_K$, the "classical" definition of points of projective $n$-space coincides with the definition of $\mathcal{O}_K$-valued points of $\mathbb{P}^n_{\mathcal{O}_K}$.