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Postmodern Algebra

Sam Macdonald

Section 8.3 Algebra and Module Finite

“All poetry is putting the infinite within the finite.”
―Robert Browning

Definition 8.26. Module Finite.

Given an \(R\)-algebra \(S\text{,}\) we say \(S\) is module-finite over \(R\) if it is finitely generated as an \(R\)-module.

Remark 8.27.

Instead of asking about how \(S\) is generated as an algebra over \(R\text{,}\) we can ask how it is generated as a module over \(R\text{.}\) The notion of module-finite is much stronger than algebra-finite, since a linear combination is a very special type of polynomial expression.

Proof.

Let \(\Gamma\) be a generating set for \(M\) as an \(R\)-module. If \(M\) is a finitely generated \(R\)-module, then we can find elements \(f_1 \ldots, f_r\) that generate \(M\) as an \(R\)-module. Since \(\Gamma\) generates \(M\text{,}\) for each \(i\) we can find finitely many elements \(\gamma_{i,1}, \ldots, \gamma_{i,n_i} \in \Gamma\) and \(R\)-coefficients \(r_{i,1}, \ldots, r_{i,n_i}\) such
\begin{equation*} f_i = r_{i,1} \gamma_{i,1} + \cdots + r_{i,n_i} \gamma_{i,n_i}. \end{equation*}
The submodule \(N\) of \(M\) generated by all the \(\gamma_{i,j}\) contains the elements \(f_1, \ldots, f_n\text{,}\) but since \(M = Rf_1 + \cdots + Rf_n\text{,}\) we conclude that \(M\) is generated by those finitely many \(\gamma_{i,j}\text{,}\) and thus by a finite subset of \(\Gamma\text{.}\)
The other proof is essentially the same, with the appropriate replacements: whenever we talk about a set that generates \(M\) as an \(R\)-module, we should instead consider a set that generates \(S\) as an \(R\)-algebra, and instead of taking linear combinations of elements we should consider polynomials in those elements with \(R\)-coefficients.

Proof.

  1. Suppose \(R\subseteq S\) is algebra-finite, i.e., \(S\) is a finitely generated \(R\)-algebra. Then, there exist \(f_1,\ldots,f_n\in S\) such that \(S=R[f_1,\ldots,f_n]\text{.}\) To see this, note that since \(S\) is finitely generated over \(R\text{,}\) we can write \(S=R[x_1,\ldots,x_m]\) for some \(x_1,\ldots,x_m\in S\text{.}\) Then, we can express each \(x_i\) as a polynomial in \(f_1,\ldots,f_n\) with coefficients in \(R\text{,}\) so \(S\subseteq R[f_1,\ldots,f_n]\text{.}\) Conversely, if \(S=R[f_1,\ldots,f_n]\) for some \(f_1,\ldots,f_n\in S\text{,}\) then \(S\) is generated over \(R\) by \(f_1,\ldots,f_n\text{,}\) and so \(S\) is algebra-finite.
  2. Suppose \(R\subseteq S\) is module-finite, i.e., \(S\) is a finitely generated \(R\)-module. Then, there exist \(f_1,\ldots,f_n\in S\) such that \(S=Rf_1+\cdots+Rf_n\text{.}\) To see this, note that since \(S\) is finitely generated over \(R\text{,}\) we can write \(S=Rx_1+\cdots+Rx_m\) for some \(x_1,\ldots,x_m\in S\text{.}\) Then, we can express each \(x_i\) as a linear combination of \(f_1,\ldots,f_n\) with coefficients in \(R\text{,}\) so \(S\subseteq R[f_1,\ldots,f_n]\text{.}\) Conversely, if \(S=Rf_1+\cdots+Rf_n\) for some \(f_1,\ldots,f_n\in S\text{,}\) then \(S\) is generated over \(R\) by \(f_1,\ldots,f_n\text{,}\) and so \(S\) is module-finite.

Example 8.30. \(\Q[x]/(x^{n})\) Not Free Algebra.

Given \(n \geq 2\text{,}\) the \(\Q\)-algebra \(S=\Q[x]/(x^n)\) is generated as an algebra by the element \(x\text{.}\) Note, however, that this is not a free \(\Q\)-algebra.

Proof.

Let \(S = R f_1 + \cdots + R f_n\text{,}\) meaning that \(f_1, \ldots, f_n\) is a set of module generators for \(S\) over \(R\text{.}\) Note that every \(R\)-linear combination of \(f_1, \ldots, f_n\) is also an element of \(R[f_1, \ldots, f_n]\text{,}\) and thus \(S\) is a subalgebra of \(R[f_1, \ldots, f_n]\text{.}\)
On the other hand, since \(f_1, \ldots, f_n \in S\) and \(S\) is an \(R\)-algebra, every polynomial in \(f_1, \ldots, f_n\) with coefficients in \(R\) is also in \(S\text{,}\) and thus \(S = R[f_1, \ldots, f_n]\text{,}\) so that \(S\) is algebra-finite over \(R\text{.}\)
The converse, however, is false: it is harder to be module-finite than algebra-finite.

Example 8.32. .

  1. The Gaussian integers \(\Z[i]\) satisfy the well-known property (or definition, depending on your source) that any element \(z\in\Z[i]\) admits a unique expression \(z=a+bi\) with \(a,b\in \Z\text{.}\) That is, \(\Z[i]\) is generated as a \(\Z\)-module by \(\{1,i\}\text{;}\) moreover, \(\{1,i\}\) is a free module basis! As a \(\Z\)-algebra, \(\Z[i]\) is generated by \(i\text{,}\) but it is not a free \(\Z\)-algebra, since \(i^2 -1 = 0\text{.}\)
  2. If \(R\) is a ring and \(x\) an indeterminate, the algebra-finite extension \(R\subseteq R[x]\) is not module-finite. Indeed, \(R[x]\) is a free \(R\)-module on the basis \(\{1,x,x^2,x^3,\dots\}\text{.}\)
  3. Another map that is not module-finite is the inclusion \(R := k[x] \subseteq k[x,\frac{1}{x}] =: S\text{.}\) First, note that any element of \(k[x,\frac{1}{x}]\) can be written in the form \(\frac{f(x)}{x^n}\) for some \(f \in k[x]\) and some \(n \geqslant 0\text{.}\) Now any finitely generated \(R\)-submodule of \(S\) is of the form
    \begin{equation*} M = \sum_i R \cdot \frac{f_i(x)}{x^{n_i}} = \sum_i k[x] \cdot \frac{f_i(x)}{x^{n_i}}. \end{equation*}
    If \(n := \max \{ n_i \}_i\text{,}\) then \(M \subseteq \frac{1}{x^n} k[x] \neq k[x,\frac{1}{x}] = S\text{.}\)
  4. The extension \(\Z \subseteq \Q\) is neither module-finite nor algebra-finite. To see that, we first claim that the set
    \begin{equation*} P = \left\lbrace \frac{1}{p} \mid p \textrm{ prime integer} \right\rbrace \end{equation*}
    generates \(\mathbb{Q}\) as a \(\Z\)-algebra. The key point here is the Fundamental Theorem of Arithmetic: since any positive integer \(n\) can be written as a product \(n = p_1^{a_1} \cdots p_s^{a_s}\) where the \(p_i\) are all prime and the \(a_i \geqslant 0\) are nonnegative integers, we see that the rational number \(\frac{m}{n} \neq 0\) can be written as
    \begin{equation*} \frac{m}{n} = m \left( \frac{1}{p_1} \right)^{a_1} \cdots \left( \frac{1}{p_s} \right)^{a_s} \in \Z \left[ \frac{1}{p_1}, \ldots, \frac{1}{p_s} \right] \subseteq \Z[P]. \end{equation*}
    On the other hand, note that any finite subset of \(P\) is contained in
    \begin{equation*} \left\lbrace \frac{1}{p} \mid p \leqslant q \textrm{ prime integer} \right\rbrace \end{equation*}
    for some fixed prime \(q\text{,}\) and that
    \begin{equation*} \Z \left[ \frac{1}{p} \mid p \leqslant q \textrm{ is prime} \right] \end{equation*}
    contains only rational numbers whose denominator is a product of primes smaller than \(q\text{.}\) But there are infinitely many primes, and thus this cannot be all of \(\Q\text{.}\) By this [provisional cross-reference: lem-subsets-of-generating-sets], we can conclude that \(\Q\) is not a algebra-finite over \(\Z\text{.}\) But then \(\Q\) cannot be module-finite over \(\Z\) by [provisional cross-reference: lem-module-finite-implies-algebra-finite].

Proof.

Let \(S = R a_1 + \cdots + R a_r\) and \(N = S b_1 + \cdots + S b_s\text{.}\) Then we claim
\begin{equation*} N=\sum_{i=1}^{r}\sum_{j=1}^{s} R a_i b_j. \end{equation*}
Indeed, given \(n = \sum_{j=1}^{s} s_j b_j\text{,}\) rewrite each \(s_j=\sum_{i=1}^{r} r_{ij} a_{i}\) and substitute to get
\begin{equation*} n=\sum_{i=1}^{r}\sum_{j=1}^{s} r_{ij} a_i b_j \end{equation*}
as an \(R\)-linear combination of the \(a_i b_j\text{.}\)

Example 8.34. Algebra Finite and Transitivity.

Let \(A\subseteq B \subseteq C\) be rings. It follows from the definitions that
  1. If \(A\subseteq B\) algebra-finite and \(B\subseteq C\) algebra finite, then \(A\subseteq C\) algebra finite.
  2. \(\displaystyle A\subseteq C \,\text{algebra-finite} \Longrightarrow B\subseteq C \,\text{algebra-finite}.\)

Example 8.35. Infinitely Generated Subalgebra.

Let \(k\) be a field and
\begin{equation*} B=k[x,xy,xy^2,xy^3,\cdots] \subseteq C=k[x,y], \end{equation*}
where \(x\) and \(y\) are indeterminates. While \(B\) and \(C\) are both \(k\)-algebras, \(C\) is a finitely generated \(k\)-algebra, while \(B\) is not.
Solution.
To see this, first note by this [provisional cross-reference: lem-subsets-of-generating-sets] it is sufficient to show that no finite subset of \(\{ xy^n \mid n \geqslant 1 \}\) generates \(B\) over \(k\text{.}\) Since any such subset is contained in \(\{ xy^n \mid 1 \leqslant n \leqslant m \}\) for some fixed \(m\) it is sufficient to show that \(B\) is not \(k[x,xy,\dots,xy^m]\) for any \(m\text{.}\) Now note that every element of \(k[x,xy,\dots,xy^m]\) is a \(k\)-linear combination of monomials \(x^iy^j\) with \(j \leqslant mi\text{,}\) so this ring does not contain \(xy^{m+1}\text{.}\) Therefore, \(B\) is not a finitely generated \(A\)-algebra.

Remark 8.36.

Let \(A\subseteq B \subseteq C\) be rings. It follows from the definitions that
  1. If \(A\subseteq B\) module-finite and \(B\subseteq C\) module-finite, then \(A\subseteq C\) module-finite.
  2. \(\displaystyle A\subseteq C \,\text{module-finite} \Longrightarrow B\subseteq C \,\text{module-finite}.\)
However, we will see that \(A\subseteq C\) module-finite \(\not\implies A\subseteq B\) module-finite. This construction is a bit more involved, so we will leave it for the problem sets.

Remark 8.37.

Any surjective ring homomorphism \(\varphi\!: R \to S\) is both algebra-finite and module-finite, since \(S\) must then be generated over \(R\) by \(1\text{.}\) Moreover, we can always factor \(\varphi\) as the surjection \(R\into R/\ker(\varphi)\) followed by the inclusion \(R/\ker (\varphi) \hookrightarrow S\text{,}\) so to understand algebra-finiteness or module-finiteness it suffices to restrict our attention to injective homomorphisms.
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