Show, using a specific approach, that $\dim \Bbb P^n=\dim\Bbb A^n=n$.

Question

This is Exercise 1.8.4(1) of Springer's, "Linear Algebraic Groups (Second Edition)". It is not a duplicate of

The dimension of $\mathbb P^n$ is $n$

because I'm after a particular perspective; namely, the approach Springer takes (using transcendence degree and not Krull dimension).

The Question:

Prove (using transcendence degree) that

$$\dim\Bbb P^n=\dim\Bbb A^n=n.$$

Here $\Bbb P^n$ is projective $n$-space.

The Details:

For a definition of irreducible, see

The components of a Noetherian space are its maximal irreducible closed subsets.

For a definition of affine, see

A morphism of affine varieties $\phi: X\to Y$ is an isomorphism iff the algebra homomorphism $\phi^*$ is an isomorphism.

Recall from the above the definition of the principal open set $D(f)$. Define $k[X]_f$ as $k[X][T]/(1-fT)$.

On page 16 of my copy of Springer's book, we have:

Let $X$ be an irreducible variety. First assume that $X$ is affine. Then $k[X]$ is an integral domain. Let $k(X)$ be its quotient field. If $U=D(f)$ is a principal open subset of $X$, then $k[U]=k[X]_f$ from which it follows that the quotient field $k(U)$ is isomorphic to $k(X)$. [. . .] The same holds for any affine open subset $U$.

Let $X$ be arbitrary. [. . .] For any two affine open sets $U,V$ of $X$, the quotient fields $k(U),k(V)$ can be canonically identified. It follows that we can speak of the quotient field $k(X)$ of $X$. If $X$ is an irreducible $F$-variety, we define similarly the $F$-quotient field $F(X)$. The dimension $\dim X$ of $X$ is the transcendence degree of $k(X)$ over $k$. If $X$ is reducible and if $(X_i)_{1\le i\le m}$ is the set of its components, we define $\dim X=\max_i(\dim X_i)$.

If $X$ is affine and $k[X]=k[x_1,\dots, x_r]$ then $\dim X$ is the maximal number of elements among the $x_i$ that are algebraically independent over $k$.

Thoughts:

I have a (very) rough understanding of transcendence degree.

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My academic supervisor explained to me in some detail most of a proof (from the basics here) that the circle $\{(X,Y)\in k^2\mid X^2+Y^2-1=0\}$ is one dimensional; however, certain parts were glossed over, we ran out of time to complete the proof, and all I have are pictures of his blackboard to go by; and he's at a conference for the next few days. It makes sense to me that it would have one dimension, since there is one "degree of freedom" and the circle can be parameterised by a single variable.

He said that, in practice, we rarely if ever calculate these things directly and that there is a bunch of handy theorems at a higher level for such computations, like comparing the set to another of known dimension. A proof using said theorems would be anachronistic from my point of view.

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My understanding of $\Bbb P^n$ is that it is "the set of $k$-lines in $k^{n+1}$"; that is, it is defined by quotienting $k^{n+1}\setminus \{0\}$ out by the equivalence relation $x\sim y$ given by $x=\lambda y$ for some $\lambda\in k\setminus \{0_k\}$.

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I don't think I could answer this myself any time soon. I am looking for a proof involving the definitions given above. Krull dimension is not covered yet, although I am aware of the equivalence of the two definitions. To use the Krull dimension would take me too far afield for my purposes.

This question was selected by my academic supervisor. He found it helped him understand things when he was a PhD student.

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I hope that is enough context.

Please help :)

Answer 1

The quotient field of $k[x_1, \dots, x_n]$ is $k(x_1, \dots, x_n)$ which has a transcendence basis over $k$ given by $\{x_1, \dots, x_n\}$. It follows that $\Bbb A^n$ is $n$-dimensional. $\Bbb P^n$ has an open affine subset isomorphic to $\Bbb A^n$, so it has the same quotient field and hence the same dimension.

As mentioned in the comments, it's not entirely obvious to see that the transcendence degree of $k(x_1, \dots, x_n)$ over $k$ is $n$. One possible definition of the transcendence degree is the maximum number of algebraically independent elements. But with this definition, it's hard to compute. More useful is the definition as the cardinality of a transcendence basis. A subset $X \subset K$ for a field extension $K/k$ is called a transcendence basis over $k$ if $X$ is algebraically independent (so that the subextension $k(X)/k$ is purely transcendental, i.e. isomorphic to the function field over $k$ with variables in $X$) and $K/k(X)$ is algebraic. With this definition, it's easy to see that the transcendence degree of $k(x_1, \dots, x_n)$ is $n$.

The situation with the transcendence degree is a lot like with dimension in linear algebra: if you define it as the supremum over the cardinality of all linear independent subsets, it's hard to compute in examples. But you can show that this is the same as the size of a basis, and this makes it easier to compute. It is proven for example in Lang's Algebra (Theorem VIII.1.1) that the cardinality of a transcendence basis does not depend on the transcendence basis and hence, the cardinality of a transcendence basis is the transcendence degree.

Answer 2

Well, if you use the transcendence degree over $k$ to define the dimension, then the result is pretty much immediate. First compute that $K(\mathbb P^n)=k(t_1,...,t_n)$. It is the field of fractions of $k[t_1,...,t_n]$. The natural map $k[t_1,...,t_n]\to k(t_1,...,t_n)$ is injective, because the polynomial ring is an integral domain. This tells us that $t_1,...,t_n$ are algebraically independent in the field extension $k\to k(t_1,...,t_n)$. Now let $L$ be the smallest subfield of $k(t_1,...,t_n)$ which contains all the $t_i$. $L$ is the whole field itself, so $L\to k(t_1,...,t_n)$ is an algebraic extension and $\{t_1,...,t_n\}$ is a transcendence basis of the field extension $k\to k(t_1,...,t_n)$. So by definition $\operatorname{trdeg}_kK(\mathbb P^n)=n$.