# Morphism of tangent space induce from a morphism of varieties.

I am trying to show that a morphism of varieties $$f: X \to Y$$ induces a linear map between the tangent spaces $$\tilde{f}: T_aX \to T_{b} Y$$, for $$a \in X$$ and $$b = f(a) \in Y$$. My idea was to use the intrinsic caracterization of the tangent $$T_aX \cong (m_a/m_a^2)^*$$ where $$m_a$$ is the unique maximal ideal of the local ring $$\mathcal O_{X, a}$$ (which is the stalk at $$a$$ of the regular functions). I know that $$f$$ comes with a $$k$$-algebra morphism $$f^*: \mathcal O_Y(U) \to \mathcal O_X(f^{-1}(U)): \varphi\mapsto f^*\varphi = \varphi\circ f$$ for $$U \subset Y$$ an open set. But from that I don't really see how should I construct $$\tilde f$$ explicitly. Could one of you give me an explanation ? I am not very comfortable with the notion of $$\text{Spec}$$ etc. so it would be nice to answer the question without using these tools.

Question: "Could one of you give me an explanation ? I am not very comfortable with the notion of Spec etc. so it would be nice to answer the question without using these tools."

Answer: If you are familiar with the "stalk" of a sheaf you can use the following method: The map $$f$$ induce for any point $$x\in X$$ a map of local rings ($$y:=f(x)$$)

$$f_x:=f^{\#}_x: \mathcal{O}_{Y,y} \rightarrow \mathcal{O}_{X,x}$$

mapping maximal ideal to the maximal ideal: $$f_x(\mathfrak{m}_y) \subseteq \mathfrak{m}_x$$. This gives a canonical map

$$\mathfrak{m}_y/\mathfrak{m}_y^2 \rightarrow \mathfrak{m}/\mathfrak{m}^2$$

inducing a map at tangent spaces

$$(df)_x: T_x(X) \rightarrow \kappa(x)\otimes_{\kappa(y)} T_y(Y).$$

Note: You must dualize to get the tangent space:

$$T_x(X):= Hom_{\kappa(x)}(\mathfrak{m}_x/\mathfrak{m}_x^2, \kappa(x)).$$

Note: Locally it $$\phi: A \rightarrow B$$ and if $$\mathfrak{p}\subseteq B$$ is a prime ideal with $$\mathfrak{q}:=\phi^{-1}(\mathfrak{p})$$, you get using localization an induced map

$$L1.\text{ }\phi_{\mathfrak{q}}: A_{\mathfrak{q}} \rightarrow B_{\mathfrak{p}}.$$

This is the map corresponding to the map $$f_x$$ with $$x$$ the point corresponding to the prime ideal $$\mathfrak{q}$$. The map in $$L1$$ maps the maximal ideal to the maximal ideal.

In Hartshorne, Chapter I.3 they define for any "algebraic variety" $$X$$ and any open set $$U$$ the ring of regular functions $$\mathcal{O}_X(U)$$. This defines a sheaf of rings on $$X$$ - the structure sheaf. In Theorem I.3.2 in the same book they prove the isomorphism

$$\mathcal{O}_{X,x} \cong A(X)_{\mathfrak{m}_x}.$$

With this definition any map $$f: X \rightarrow Y$$ of algebraic varieties induced a map of local rings

$$f^{\#}_x: \mathcal{O}_{Y,f(x)} \rightarrow \mathcal{O}_{X,x}$$

with $$f^{\#}_x(\mathfrak{m}_y) \subseteq \mathfrak{m}_x$$. Hence you get an induced map on cotangent and tangent spaces. The sheaf $$\mathcal{O}_X$$ has the property that there is an isomorphism

$$H^0(X, \mathcal{O}_X) \cong A(X)$$

where $$A(X)$$ is the coordinate ring of $$X$$. There is a 1-1 correspondence between points in $$X$$ and maximal ideals in $$A(X)$$.

This construction is "more elementary" than the $$Spec$$-construction. If you understand this construction it is not a difficult thing to understand the "more general" $$Spec$$-construction.

Remark: If you want to read about the $$Spec$$-construction there are many books available: Hartshorne "Algebraic geometry", Fulton "Algebraic curves" etc. The Fulton book is the "more elementary" and is a good introductory book.

• Just, I don't really understand how $f_x^#$ acts on the element of $\mathcal O_{Y, y}$.. May 14, 2021 at 8:27
• Also, the fact that we have changed the sense of the morphism at the end it's because we dualize ? May 14, 2021 at 8:28
• Thank you for your nice answer. Could you just tell me where can I find details about what you said on localization at a prime ideal ? May 14, 2021 at 12:44