# Projective coordinates over a non UFD ring

Is it true that when the integers of a number field are not a UFD then not every point in projective $n$-space over that field can be given by relatively prime algebraic integer coordinates?

When a point can be given by integer coordinates generating a non-principle ideal as GCD then it seems that point cannot be given by relatively prime integers.

But I wonder if I am missing something? Marcello Robbiani "Rational points of bounded height on Del Pezzo surfaces of degree six" at

http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=GDZPPN002069458&IDDOC=217550

page 411 seems to say that by taking suitable account of the (finitely many) ideal classes of the ring of integers, every point can be represented with relatively prime integer coordinates. Apparently I am misunderstanding either Robbiani or something about algebraic integers.

Does Robbiani just mean that each point of projective $n$ space can be specified as a relatively-prime-integer $n+1$-tuple times an ideal class?

## 1 Answer

The author explains what the "relatively prime" terminology means: "we fix once for all a family of ideals $\mathfrak{a}_1,\ldots,\mathfrak{a}_h$ representing the $h$ classes of ideals $\mathfrak{R}_i$ in $O_K$ and additionally require from our coordinates to satisfy $(a_0,a_1,a_2) = \mathfrak{a}_i$" for some $i$.

That is, given $a_0,a_1,a_2$, we know that the ideal they generate is a multiple of some $\mathfrak{a}_i$, and the author simply requires you to normalize $a_0,a_1,a_2$ so that the ideal they generate is exactly $\mathfrak{a}_i$. Thus, when the author says the coordinates are relatively prime, it just means that they generate one of the chosen ideals $\mathfrak{a}_i$. It does not mean they are relatively prime in the sense of generating the entire ring. However, if the $\mathfrak{a}_i$ are chosen to be maximal in their class, then I guess it does imply that their gcd in the ring is 1. (That is, there is no $b \in O_K$, such that every $a_i$ is divisible by $b$. Otherwise, dividing all the generators by $b$ would yield a larger ideal in the same ideal class.) I suppose that may be why the terminology is being used.

• I haven't seen "relatively prime" used this way before, but I'm not a commutative algebraist, so I don't know. (Personally, I would have chosen some other terminology, perhaps "normalized".) – Dave Witte Morris Apr 2 '15 at 22:46
• I agree this is a strange use of the term "relatively prime." – KConrad Apr 2 '15 at 22:53
• Your explanation of why the terminology might have been chosen makes sense. But as you say it is not necessarily a good terminology. This is exactly what beginners often confuse with the correct definition of relative primeness as generating the unit ideal. – Colin McLarty Apr 2 '15 at 23:39