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I am self studying Commutative algebra out of Kemper's text. I came across an exercise in his text that is meant to better solidify the idea of mappings between affine varieties. As I am very new to the subject, I was hoping for some help in navigating through the question. It goes as follows:

Let $$X = \{ ( x_1 , x_2) \in \mathbb{C}^2 \: | \: x_1x_2 = 1 \}$$

Which of the following maps $\varphi_i : X \rightarrow X$ are morphisms, isomorphisms, or Zariski continuous?

(a) $\varphi_1(x_1,x_2) = (x_1^{-1} , x_2^{-1})$

(b) $\varphi_2(x_1,x_2) = (x_1^{2} , x_2^{2})$

(c) $\varphi_2(x_1,x_2) = (\bar{x_1} , \bar{x_2})$ (complex conjugation).

Here are some definitions to work with.

Let $K$ be a field and let $X \subseteq K^m$ and $Y \subseteq K^n$ be affine varieties. A map $f: X → Y $ is called a morphism (of varieties) if there exist polynomials $f_1,...,f_n \in K[x_1,...,x_m]$ such that $f$ is given by $f(P)=(f_1(P),...,f_n(P))$ for $P \in X$.

A morphism that has an inverse which is also a morphism is an isomorphism.

My question here is how do i approach these? Is there a technique I should think about. Any help would be appreciated.

An Isomorphic Teen
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1 Answers1

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I was not able to solve c), but I hope a) and b) will help you in your studies:

a)

First note, that neither $\{0\}\times\mathbb{C}$, nor $\mathbb{C}\times\{0\}$ is a subset of $X$, since the condition $x_1x_2=1$ is not fufilled. Therefore

$$\varphi_1:X\rightarrow X,\quad \varphi_1(x_1,x_2)=(x_1^{-1},x_2^{-1})$$

is well-defined. (Neither $x_1$, nor $x_2$ is equal to $0$.)

Using the condition of $X$, we can calculate as follows:

$$\varphi_1(x_1,x_2)=(x_1^{-1},x_2^{-1})=(1 x_1^{-1},1 x_2^{-1})=(x_1x_2x_1^{-1},x_1x_2x_2^{-1})=(x_2,x_1)=(f_1(x_1,x_2),f_2(x_1,x_2))$$

with polynomials $f_1,f_2\in\mathbb{C}[x_1,x_2]$, defined by

$$f_1(x_1,x_2)=x_2,\quad f_2(x_1,x_2)=x_1.$$

Therefore $\varphi_1$ is a morphism. Since $\varphi_1\circ\varphi_1=id_X$, it is even an isomorphism.

b)

$$\varphi_2:X\rightarrow X,\quad \varphi_2(x_1,x_2)=(x_1^2,x_2^2)$$

is well-defined, since any complex number can be squared. Again, both component functions are functions of $\mathbb{C}[x_1,x_2]$. Therefore $\varphi_2$ is a morphism. This is not an isomorphism, since for all $f\in \mathbb{C}[x_1,x_2]$ and all $(x_1,x_2)\in X$, there exists a $n\ge 0$:

$$f(\varphi_2(x_1,x_2))=f(x_1^2,x_2^2)=\sum_{k=0}^n a_k x_1^{2k}+b_k x_2^{2k}$$

This holds because for any $n_1\ge n_2\ge 0$, we find

$$(x_1^2)^{n_1}(x_2^2)^{n_2}=(x_1^2)^{n_1-n_2}$$

again by using the property of $X$. Especially we cannot find $f_1$ or $f_2$, such that

$$f_1(\varphi_2(x_1,x_2))=x_1\quad\text{and}\quad f_2(\varphi_2(x_1,x_2))=x_2$$

$$$$

c)

$$\varphi_3:X\rightarrow X,\quad \varphi_2(x_1,x_2)=(\overline{x_1},\overline{x_2})$$

is well-defined, since for any complex number exists a complex conjugate.

user408858
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