Is there a reason why schemes are ordered pairs $(X,\mathcal O_X)$ rather than for example $(\mathcal O_X,X)$ or $\{X,\mathcal O_X\}$?
2026-04-11 17:55:30.1775930130
Why schemes are $(X,\mathcal O_X)$ rather than $(\mathcal O_X,X)$ or $\{X,\mathcal O_X\}$
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This is a general way to talk about structures that consist of several "parts". For example a field is a set $F$ together with two operations ($+$, $\cdot$) and two special elements ($0$ and $1$). So to unambiguosly specify a field, we could denote it with a pentuple $(F,+,\cdot,0,1)$. Then for example a field homomoprhism to another field $(F',+',\cdot',0',1')$ is a map $f\colon F\to F'$ such that $f(a+b)=f(a)+'f(b)$, $f(a\cdot b)=f(a)\cdot'f(b)$, $f(0)=0'$, $f(1)=1'$. It is necessary to explicitly mention the other parts $+,\cdot,0,1$ because we could define different field structures on the ery same set $F$. Then again, is this really less ambiguous? We might indeed employ a different convention and wirte the pentuple in different order, for example $(F,0,1,\cdot,+)$ or we might even consider it unnecessary to specifically mention $0$. But within a text (a course, a book), one should fix one such notation once and for all - later you will by abuse of language speak of "the field $\mathbb Q$" anyway.
So back to your original situation: One could certainly use the notation $(\mathcal O_X,X)$ instead of $(X,\mathcal O_X)$; but it is very convenient to always start with the underlying set (as I did with fields above), so here we start with $X$. (Incidentally, one could even consider $X$ redundant, but that is another story). And why don't we write $\{X,\mathcal O_X\}$? Well, in a set we cannot automatically distinguish between the elements by any implicit order. For example, if we did so for fields and said a dield is a set $\{F,+,\cdot,0,1\}$, then we would have no way to tell which of $+$, $\cdot$ is addition and which is multiplication, e.g., we could not tell if the distributive law should be $a\cdot(b+c)=(a\cdot b)+(a\cdot c)$ or $a+(b\cdot c)=(a+b)\cdot (a+c)$.