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

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 $$p^{-1}$$?
• It's better if questions here are self-contained. Please write out Lemma 1.1, complete with definitions of all the necessary notations, e.g., what is $B$, what is $L$, what is $K$, what are $a_1,\dots,a_r$? Mar 20 at 0:43

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, $$ax = \sum_{i=1}^n (a c_i) v_i$$.

• Thanks for your explanation. I understood the second question now. May you explain also why the first question doesn't matter? In particular, if all matrices are just in $K[a_1,\cdots, a_r]^{m\times m}$ hence the dominators of entries are all 1, how would this proof work? Mar 20 at 16:53
• There is no step in the proof that requires any of the $p_i$'s to actually exist. The proof works just fine if there aren't any irreducible factors of $g$ that are in $K[a_i]$. Mar 20 at 16:59