# Lemma 5.2.5 in Springer's Linear Algebraic Groups

I'm stuck trying to understand the first paragraph of this proof.

Let $X\rightarrow Y$ be a dominant morphism of affine varieties and denote $B=k[X]$,$A=k[Y]$. Assume there exists $b\in B$ such that $B=A[b]$, so $B=A[T]/I$ for some ideal $I$. Pick $x\in X$ and let $\epsilon:A\rightarrow k$ be the homomorphism defined by $\phi(x)$. Extend it to a homomorphism $\epsilon:A[T]\rightarrow k[T]$.

He claims that if $\epsilon(I)=0$, then $k[\phi^{-1}(\phi(x))]=k[T]$, but I don't quite see it...

## 1 Answer

Let $\varphi:A\rightarrow B$ be the morphism associated to $\phi:X\rightarrow Y$ and let $\pi:A[T]\rightarrow B$ be the surjection such that $\varphi=\pi\circ\iota$, where $\iota:A\rightarrow A[T]$ is the inclusion map. Let $\tilde{\epsilon}:A[T]\rightarrow K[T]$ be the extension of $\epsilon$ which is a surjection whose kernel is the ideal of $A[T]$ generated by $\ker(\epsilon)$. The hypothesis on $\tilde{\epsilon}$ implies that there exists a surjection $\eta:B\rightarrow k[T]$ such that $\eta\circ\pi=\tilde{\epsilon}$.

Note that $\mathfrak{m}:=\ker(\epsilon)=\{f\in A:f(\phi(x))=0\}$ is the ideal of $\{\phi(x)\}$. Therefore $\varphi(\mathfrak{m})$ generates an ideal ( $\varphi(\mathfrak{m})B$ ) whose zero zet is $\phi^{-1}(\phi(x))$ (in fact $Z(\varphi(\mathfrak{m}))=\{x'\in X:\varphi(f)(x')=0\ \forall f\in\mathfrak{m}\}=\{x'\in X:f(\phi(x'))=0\ \forall f\in\mathfrak{m}\}=\phi^{-1}(Z(\mathfrak{m}))$. ).

Then it is enough to prove that $B/\varphi(\mathfrak{m})B\cong k[T]$. Now, $\ker(\eta)=\pi(\ker(\tilde{\epsilon}))=\pi((\ker(\epsilon)))=(\varphi(\ker(\epsilon)))=\varphi(\mathfrak{m})B$ and therefore $\eta$ induces an isomorphism $B/\varphi(\mathfrak{m})B\cong k[T]$.

Sorry if this is too long!.