I am trying to understand the intuition as well as the proof of the following theorem:

Let $A$ be be an algebra over a field $K$. If $A$ is a field and is contained in an affine $K$-domain, then $A$ is algebraic.

The proof that I am looking at is on pages 8 and 9 of this book.

Some of my questions are:

  1. Why do we need to consider the maximal K-algebraically independent subset?
  2. "For b ∈ $Quot(B)$, multiplication by b gives an $L$-linear endomorphism of $Quot(B)$"- Why is it $L$-linear?
  3. "Choosing an $L$-basis of $Quot(B)$, we obtain a map φ: $Quot(B)$ → $L^{m×m}$ assigning to each b ∈ $Quot(B)$ the representation matrix of this endomorphism." - How do we find this matrix? Do we just multiply the $L$-basis elements in $Quot(B)$ by $b$ and write what we get as columns? Where are the $m$'s coming from?

The rest of the proof also is unclear to me but thought I'd ask these questions first. If there is another proof of this that you could direct me to, that would also be great.


1 Answer 1


(1) Actually, it's A maximal $K$-algebraically independent subset - there can certainly be more than one. "Maximal" is used with respect to the partial ordering by inclusion. The set is chosen to be maximal so that $a_{r + 1}$, $\ldots$, $a_n$ will be algebraic over $L = K(a_1, \ldots a_r)$. This is what guarantees that $\textrm{Quot}(B) = L(a_{r + 1}, \ldots, a_n)$ is finite-dimensional over $L$.

(2) Automatically, if $R \subseteq S$ are commutative rings, and $s_0 \in S$, the mapping $u \colon S \to S$ defined by $u(s) = s_0 s$ is $R$-linear.

(3) $m$ is the dimension of $\textrm{Quot(B)}$ over $L$. Your description is correct. If a basis of $\textrm{Quot(B)}$ over $L$ is $\{b_1, \ldots, b_m \}$, then the first column will be the coordinates of $bb_1$ in thhat basis.

Prop. 7.9 in Atiyah-Macdonald is close, but not identical.


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