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

Question

I am reading Lemma 1.1 in Kemper's book A Course in Commutative Algebra. I am stuck in understanding the proof of part (b) of this lemma.

Some definitions in the books: By Affine $K$-algebra we mean a finitely generated algebra over a field $K$. By affine $K$-domain we mean an affine $K$-algebra that is an integral domain. The book's proof:

By way of contradiction, assume that $A$ has an element $a_1$ that is not algebraic. By hypothesis, $A$ is contained in an affine $K$-domain $B=K\left[a_1, \ldots, a_n\right]$ (we may include $a_1$ in the set of generators). We can reorder $a_2, \ldots, a_n$ in such a way that $\left\{a_1, \ldots, a_r\right\}$ forms a maximal $K$-algebraically independent subset of $\left\{a_1, \ldots, a_n\right\}$. Then the field of fractions $\operatorname{Quot}(B)$ of $B$ is a finite field extension of the subfield $L:=$ $K\left(a_1, \ldots, a_r\right)$. For $b \in \operatorname{Quot}(B)$, multiplication by $b$ gives an $L$-linear endomorphism of $\operatorname{Quot}(B)$. Choosing an $L$-basis of $\operatorname{Quot}(B)$, we obtain a map $\varphi$ : $\operatorname{Quot}(B) \rightarrow L^{m \times m}$ assigning to each $b \in \operatorname{Quot}(B)$ the representation matrix of this endomorphism. Let $g \in K\left[a_1, \ldots, a_r\right] \backslash\{0\}$ be a common denominator of all the matrix entries of all $\varphi\left(a_i\right), i=1, \ldots, n$. So $\varphi\left(a_i\right) \in K\left[a_1, \ldots, a_r, g^{-1}\right]^{m \times m}$ for all $i$. Since $\varphi$ preserves addition and multiplication, we obtain $$ \varphi(B) \subseteq K\left[a_1, \ldots, a_r, g^{-1}\right]^{m \times m} $$ $K\left[a_1, \ldots, a_r\right]$ is isomorphic to a polynomial ring and therefore factorial (see, for example, Lang [33, Chapter V, Corollary 6.3]). Take a factorization of $g$, and let $p_1, \ldots, p_k$ be those irreducible factors of $g$ that happen to lie in $K\left[a_1\right]$. Let $p \in K\left[a_1\right]$ be an arbitrary irreducible element. Then $p^{-1} \in A \subseteq B$ since $K\left[a_1\right] \subseteq A$ and $A$ is a field. Applying $\varphi$ to $p^{-1}$ yields a diagonal matrix with all entries equal to $p^{-1}$, so there exists a nonnegative integer $s$ and an $f \in K\left[a_1, \ldots, a_r\right]$ with $p^{-1}=g^{-s} \cdot f$, so $g^s=p \cdot f$. By the irreducibility of $p$, it follows that $p$ is a $K$-multiple of one of the $p_i$. Since this holds for all irreducible elements $p \in K\left[a_1\right]$, every element from $K\left[a_1\right] \backslash K$ is divisible by at least one of the $p_i$. But none of the $p_i$ divides $\prod_{i=1}^k p_i+1$. This is a contradiction, so all elements of $A$ are algebraic.

My questions are below:

• Why are there any irreducible factors that happen to lie in $k[a_1]$?

As far as I understand it, the common dominator of entries in all $\phi(a_i)$ means a common multiple of all $\phi(a_i)$. Could it happen that all factors of $g$ are in $K[a_2,...,a_r]$, and hence the set of such irreducible factors is an empty set? Are the irreducible factors of $g$ that are in $K[a_1]$ exist simply because we can multiply $g$ by $a_1 g$ to get a common multiple of dominators of entries in each $\phi(a_i)\, i=1,...,n$ that contains divisors in $k[a_1]$?

• Why applying $\phi$ to $p^{-1}$ yields a diagonal matrix with all entries equal to $p^{-1}$?

Answer 1

Why are there any irreducible factors that happen to lie in $k[a_1]$?

It doesn't matter. There doesn't need to be any.

Why applying $\phi$ to $p^{−1}$ yields a diagonal matrix with all entries equal to $p^{−1}$?

There is a field extension $Quot(B) / L$. $p^{-1}$ is in $K(a)$ which is a subfield of $L$. This is a special case of a more general phenomenon :

Let $E/F$ be a finite degree field extension, and let $v_1, ... v_n$ be a basis of $E$ as a $F$-vector space. Then, for any $a \in F$, the matrix representation of multiplication by $a$ is the diagonal matrix with value $a$.

This is because for any $x \in E$, write $x =\sum_{i=1}^n c_i v_i$, with $c_i \in F$. Then, $ax = \sum_{i=1}^n (a c_i) v_i$.