1

The following lemma is from Qing Liu's "Algebraic Geometry and Arithmetic Curves" p.61. In line 4 of the proof, why is $k\subseteq A/\mathfrak p\subseteq k(x)$? I think it means that there are injective ring homomorphisms $k\to A/\mathfrak p$ and $A/\mathfrak p\to k(x)$, but I don't know how to construct such homomorphisms.

$k(x)$ is the residue field $\mathcal O_{X,x}/\mathfrak m_x$, where $\mathfrak m_x$ is the maximal ideal of $\mathcal O_{X,x}$.

enter image description here

zxcv
  • 1,445

1 Answers1

2

$X$ is a variety over $k$, hence $A$ is a $k$-algebra and so must be $A/\mathfrak p$. On the other hand, $k(x)$ is the fraction field of $A/\mathfrak p$, so this gives you the other inclusion.

asdq
  • 3,900
  • Thanks. But I don't fully understand your answer. I have two questions: 1. Are we restricting $V$ to be an affine variety, and not general affine scheme? I think we should, in order to say that $A$ is a $k$-algebra. – zxcv Apr 25 '19 at 14:04
  • Accepting that $A$ is a $k$-algebra, I still don't understand why $A$ is isomorphic to $A/\mathfrak p$. Could you explain why?
    • – zxcv Apr 25 '19 at 14:08
  • Any scheme $X$ over $k$ (i.e. a morphism $X\to \operatorname{Spec}k$) has the property that $\mathcal O_X(U)$ is a $k$-algebra for any open $U\subset X$. This means that we have a structure morphism $k\to A$, postcomposing with the projection $A\to A/\mathfrak p$ then gives the quotient the structure of a $k$- algebra. I don't claim that $A$ is isomorphic to $A/\mathfrak p$. – asdq Apr 25 '19 at 14:40