# How is the induced morphism $(Z,\mathcal{O}_Z)\to (X,\mathcal{O}_X)$ defined?

I'm reading Algebraic Geometry I by Görtz and Wedhorn and have a question concerning the proof of Lemma 1.55. $X$ is an irreducible affine algebraic set and $Z\subseteq X$ is an irreducible closed set. I don't understand the following sentence:

As the inclusion $Z\to X$ is a morphism of affine algebraic sets it induces a morphism $(Z,\mathcal{O}_Z)\to (X,\mathcal{O}_X)$.

Can you explain how that induced morphism is defined?

Edit: Here is my approach to prove $\mathcal{O}'_Z(U)\subseteq\mathcal{O}_Z(U)$.

Let $f\in\mathcal{O}'_Z(U)$. For each $x\in U$ we find by definition of $\mathcal{O}'_Z(U)$ an open set $V_x\subseteq X$ ($V$ for short) with $x\in V$ and $g\in\mathcal{O}_X(V)$ such that $f_{\mid U\cap V}=g_{\mid U\cap V}$. The morphism $(Z,\mathcal{O}_Z)\to (X,\mathcal{O}_X)$ implies that we have $g_{\mid Z\cap V}\in\mathcal{O}_Z(Z\cap V)$. By the first axiom of spaces with functions we get $g_{\mid U\cap V}\in\mathcal{O}_Z(U\cap V)$, hence $f_{\mid U\cap V}\in\mathcal{O}_Z(U\cap V)$. Finally $f$ can be obtained by gluing $f_{\mid U\cap V_x}$ for all $x\in U$. So by the axiom of gluing we get $f\in\mathcal{O}_Z(U)$.

Is that proof correct? If yes, is that the way the authors of the book probably wanted the reader to verify the claim or is there an easier way?

Let $i: Z \to X$ denote the canonical inclusion map given by $i(z) = z$ for $z \in Z$. For any open subset $U \subset X$ one has a $k$-algebra homomorphism $i^*_U : \mathcal{O}_X(U) \to \mathcal{O}_Z(i^{-1}U) = \mathcal{O}_Z(U \cap Z)$ defined by $i^*_U(f) = f \circ i|_{U \cap Z}$ for any function $f \in \mathcal{O}_X(U)$ (cf. Definition 1.35(2) on p. 20). The map $i$ together with the collection of the homomorphisms $i^*_U$ for all open $U \subset X$ are the data that define the morphism $(Z,\mathcal{O}_Z) \to (X,\mathcal{O}_X)$. Note that while $i : Z \to X$ is injective, the maps $i^*_U : \mathcal{O}_X(U) \to \mathcal{O}_Z(U\cap Z)$ are usually not injective. If a function $f \in \mathcal{O}_X(U)$ vanishes at all points $z \in U \cap Z$ then $i^*_U(f)=0$ even though $f$ may not be identically $0$ on $U$. For instance, if $A = k[T_1,\ldots,T_n]$ is the coordinate ring of $\mathbb{A}^n(k)$ and $I \subset A$ is the defining ideal of $X$, while $J$ is the defining ideal of $Z$, then the map $i^*_X : \mathcal{O}_X(X) \to \mathcal{O}_Z(Z)$ is simply the quotient map $A/I \to A/J$ with kernel the ideal $J/I$ of $A/I$.
• Yeah, that's exactly the textbook I'm talking about. I think you meant $i^*_U : \mathcal{O}_X(U) \to \mathcal{O}_Z(i^{-1}U) = \mathcal{O}_Z(U \cap Z)$. But nevertheless, the morphism you are presenting me is a morphism $(X,\mathcal{O}_X)\to (Z,\mathcal{O}_Z)$ instead of $(Z,\mathcal{O}_Z)\to (X,\mathcal{O}_X)$, isn't it? I already thought about that one, but it's a morphism in the wrong direction. That's why I'm so confused and posted the question. Anyway, thanks a lot for your extensive answer. Commented Jun 1, 2014 at 6:14
• Thanks for noticing the typo in $\mathcal{O}_Z(U\cap Z)$, I corrected it in my original answer. The morphism is in the right direction, from $Z$ to $X$. Whenever we have a map of sets $f : S \to T$, and a third set $R$ is given, then we have a natural map in the opposite direction $f^* : \operatorname{Map}(T,R) \to \operatorname{Map}(S,R)$ defined by $f^*(\phi) = \phi \circ f$ for $\phi \in \operatorname{Map}(T,R)$. There is no obvious way to obtain from $f$ a map $\operatorname{Map}(S,R) \to \operatorname{Map}(T,R)$ unless $f$ is a bijection (in which case we would take $(f^{-1})^*$). Commented Jun 1, 2014 at 7:23
• This is of course true, but actually you need the implication (i) $\Rightarrow$ (iii). According to Definition 1.28 on p. 16, a morphism between affine algebraic sets is a map given by polynomials, and in our case, we have $i(z) = z = (T_1(z),\ldots,T_n(z))$ for $z \in Z$, so $i$ is already known to be a morphism of affine algebraic sets. In order to deduce that it defines a morphism of spaces with functions $(Z,\mathcal{O}_Z) \to (X,\mathcal{O}_X)$, we need (iii). Alternatively, you can use your observation about the map $\Gamma(X) \to \Gamma(Z)$ and the implication (ii) $\Rightarrow$ (iii). Commented Jun 2, 2014 at 11:36
• If we can choose $h=1$, this would mean that $f \in \mathcal{O}_Z(U)$ extends to a globally defined function $g \in \Gamma(Z)$, which need not be the case (e.g. if $X = \mathbb{A}^2(k)$, $Z = \{T_1^2-T_2=0\}$, $U = Z\setminus\{(0,0)\}$, then $f = T_1/T_2 \in \mathcal{O}_Z(U)$ does not extend to a function in $\Gamma(Z) \simeq k[T_1]$). As $D(h) = D(h^n)$ for $n\in \mathbb{Z}_{\geq0}$, one could assume $n=1$, but that is hardly a simplification in this case. Commented Jun 8, 2014 at 6:26
• This is true, but the real reason, IMHO, is that we want to represent $f$ as a quotient of functions defined on the whole of $Z$, in order to take advantage of the surjectivity of the homomorphism $\Gamma(X)\to\Gamma(Z)$. Luckily, this can be done on open sets of the form $D(h)$, and we can squeeze such a set between any point $x\in Z$ and any neighbourhood $U$ of $x$ in $Z$. So, the desired representation exists locally around every point. If we could have achieved the same with a larger set, i.e. if $f$ could have been extended beyond $U$, we would have done that, but we can't in general. Commented Jun 8, 2014 at 13:52