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.
1 Answer
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.
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$\begingroup$ Just, I don't really understand how $f_x^#$ acts on the element of $\mathcal O_{Y, y}$.. $\endgroup$– RamonMay 14, 2021 at 8:27
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$\begingroup$ Also, the fact that we have changed the sense of the morphism at the end it's because we dualize ? $\endgroup$– RamonMay 14, 2021 at 8:28
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$\begingroup$ 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 ? $\endgroup$– RamonMay 14, 2021 at 12:44