Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 24F (Algebraic Geometry)

A derivation of $A$ centred at the $k$-point $p:A\to k$ is a $k$-linear map $D:A\to k$ satisfying the twisted Leibniz ruleThe Zariski tangent space isSince $p(f)\ne 0$, $p$ extends uniquely to the localization $A[f^{-1}]$ by $p(a/f^n)=p(a)/p(f{)}^n$. Restriction gives a linear mapConversely, every $D\in T_{p}A$ has the unique extensionThe quotient rule verifies the Leibniz identity, and the assumption that $f$ is not a zero divisor makes the localization representation compatible with equality of fractions. Extension and restriction are inverse, so the natural map is an isomorphism.

Let $P(x)={\prod }_{i=1}^r(x-{\lambda }_i)$. Since both $x$ and $y$ are invertible on $X$, the formuladefines a regular map $F:X\to {\mathbb{A}}^1$.

Fix $t\in k$. To solve $F(x,y)=t$, setThe curve equation becomesThis is a nonconstant polynomial, so it has a root because $k$ is algebraically closed. At a root, neither $x$ nor $t-x$ can vanish, and the equation also forces $P(x)\ne 0$. The displayed value of $y$ is therefore defined and nonzero, satisfies $y^2=P(x)$, and obeys $F(x,y)=t$. Thus $F$ is surjective.