# About transcendence degree of an affine $K$-domain

Exercise 5.6 (G. Kemper) If $$A$$ is an affine $$K$$-domain, then the transcendence degree of $$A$$ is the size of a maximal algebraically independent subset of $$A$$.

Kemper's Proof: Let $$T\subseteq A$$ be a maximal algebraically independent subset. Then $$T$$ is a maximal algebraically independent subset of $$Quot(A)$$, so it is a transcendence basis. Since any two transcendence basis has the same size, then $$trdeg(Quot(A))=|T|=trdeg(A)$$.

I'm trying to proof (unsuccessfully) the affirmation "$$T$$ is a maximal algebraically independent subset of $$Quot(A)$$" because it's not obvious to me, but a more elementary question arrises (I'm learning): when we say "maximal" we mean among all other algebraically independent subsets or it is related to the size of the set? For example, the set $$\{\overline{x_1}\}$$ is a maximal algebraically independent subset of $$K[x_1,x_2,x_3]/(x_1x_2,x_1x_3)$$, but is not "maximal in size", since $$\{\overline{x_2},\overline{x_3}\}$$ is algebraically independent too.

If someone could give an explanation about these terms and proof of the affirmation I'm really glad. Thanks!

## 1 Answer

Maximal in this context always means with respect to inclusion, not size. Formally, a maximal element of a poset $$(P, \le)$$ is an element $$m$$ such that if $$m \le n$$ then $$m = n$$ ("maximal with respect to $$\le$$"). A poset can have many maximal elements and maximal elements of a collection of subsets ordered with respect to inclusion can have many different sizes.

To show that $$T$$ is a maximal algebraically independent subset of $$\text{Quot}(A)$$ we can argue as follows. Let $$t = \frac{f}{g} \in \text{Quot}(A)$$ be arbitrary, where $$f, g \in A$$. By maximality $$T \cup \{ f \}$$ is algebraically dependent in $$A$$, so there exists some polynomial relation

$$\sum a_n f^n = 0$$

where $$a_n \in k[T] \subseteq A$$. Similarly $$T \cup \{ g \}$$ is algebraically dependent in $$A$$, so there exists some polynomial relation

$$\sum b_n g^n = 0$$

where $$b_n \in k[T]$$. So $$\frac{f}{g}$$ is a quotient of two elements algebraic over $$\text{Quot}(k[T])$$ and hence is itself algebraic over $$\text{Quot}(k[T])$$, meaning there exists some polynomial relation $$\sum c_n t^n = 0$$ where $$c_n \in \text{Quot}(k[T])$$. This means $$T \cup \{ t \}$$ is algebraically dependent as desired, for any $$t$$. So $$T$$ is maximal.

• Very clear! Thanks! Commented Aug 31, 2020 at 11:56