Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$ and $Y$ be varieties, and let $\phi:X\rightarrow Y$ be a regular map. Let $x\in X$, and $y=\phi(x)$. Write $\Theta_{X,x}$ and $\Theta_{Y,y}$ for the respective tangent spaces of $X$ and $Y$ at $x$ and $y$. Let $k[X]$ and $k[Y]$ denote respectively the ring of regular functions on $X$ and $Y$, and let $\mathfrak{m}_x$ and $\mathfrak{m}_y$ be the maximal ideals of $k[X]$ and $k[Y]$ at the points $x$ and $y$. Then $\phi$ induces a $k$-algebra homomorphism $\phi^*:k[Y]\rightarrow k[X]$ via right composition with $\phi$. This further induces a homomorphism $(d_x\phi)^*:\mathfrak{m}_y/\mathfrak{m}_y^2\rightarrow\mathfrak{m}_x/\mathfrak{m}_x^2$. Let $d_x:k[X]\rightarrow\Theta_{X,x}^*$ be the map that sends a regular function to the linear part of the Taylor expansion of that function at $x$. Then $d_x$ induces a map into $\mathfrak{m}_x/\mathfrak{m}_x^2$. Actually $d_x$ is an isomorphism from $\mathfrak{m}_x/\mathfrak{m}_x^2$ to $\Theta_{X,x}^*$. Let $d_x\phi$ be the dual map of $d_x\phi^*$. Then the above isomorphism implies that $d_x\phi:\Theta_{X,x}\rightarrow\Theta_{Y,y}$. The question here is: given any $f\in k[Y]$, why is $(d_x\phi)^*(d_yf)=d_x(\phi^*f)$?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

A fundamental property of $\phi ^*$is that for any $f\in k[Y]$ we have $\phi ^*(f)(x)=f(\phi (x))$.
In particular this implies that $\phi^*(\mathfrak m_y)\subset \mathfrak m_x$ and thus that $\phi^*(\mathfrak m_y^2)\subset \mathfrak m_x^2$.
There is thus an induced cotangent map $$d_x(\phi^*):\mathfrak m_y/\mathfrak m_y^2 \to \mathfrak m_x/\mathfrak m_x^2: \bar g\mapsto \overline {\phi^*(g)}\quad (g\in \mathfrak m_y) $$
On the other hand, given a function $f\in k[Y]$, we define $d_y(f)$ as the class $d_yf=\overline {f-f(y)}\in \mathfrak m_y/\mathfrak m_y^2$.
Following through we get
$$d_x(\phi^*)(d_yf)=d_x(\phi^*)(\overline {f-f(y)})=\overline {\phi^*(f-f(y))} =\overline {\phi^*(f)-\phi^*(f)(x)}=d_x(\phi^*(f))$$
which is exactly what you wanted .

Amusingly, I advised another user three hours ago not to use the tangent space but rather the more fundamental Zariski cotangent space.
Your question is another perfect illustration that the tangent space only confuses the issue in many foundational problems (and has nothing to do here)!

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.