The question I was posed was to construct a commutative finite dimensional $\mathbb{Q}$-algebra $A$ say, with a simple $A$-module S say, such that $\text{dim}_{\mathbb{Q}}S = n$ with $n$ arbitrary in the naturals.

This question was posed to me in Representation Theory, but since I've been doing a bit of Galois Theory recently, my head jumped to the following (potential?) solution:

Let $f_{n}(x) = x^{n} - 2$, then by Eisenstein's criterion with respect to the prime $p=2$, $ f_{n}$ is irreducible in $\mathbb{Q}$[x] for every $n$. Then let $I_{n} = \left<f_{n}\right>$ be the ideal in $\mathbb{Q}[x]$ generated by $f_{n}$. Then since $\mathbb{Q}[x]$ is a PID, prime ideals are maximal, and further since $f_{n}$ is irreducible and so prime, $I_{n}$ is a prime and hence maximal ideal of $\mathbb{Q}[x]$. Hence $E_{n} = \mathbb{Q}[x] / I_{n}$ is a finite degree field extension of $\mathbb{Q}$. In particular if $\alpha _{n} $ is a roof of $f_{n}$ in $\mathbb{C}$, then $E_{n} \cong \mathbb{Q}(\alpha _ {n})$. Further $| E_{n} : \mathbb{Q} | = \text{deg}(f_{n}) = n$. So $E_{n}$ is a field extension of $\mathbb{Q}$ of degree $n$. Then if we view $\mathbb{Q}$ as a commutative $\mathbb{Q}$ algebra, then $E_{n}$ is a $\mathbb{Q}$ module in the obvious way, and it is simple, since it is a field, and further $\text{dim}_{\mathbb{Q}}S = \left| E_{n} : \mathbb{Q} \right| = n$.

My query is two-fold, firstly is this a valid solution to the problem? And secondly, I suspect that the author of the question was expecting a construction more rooted in Representation Theory than Galois Theory. Could any give me a hint of where I should start in this regard?

$\textbf{Correction: }$ In the above construction I stated that $E_{n}$ is a simple $\mathbb{Q}$-module since it is a field. This is incorrect since $E_{n}$ has a composition series of length $n$ as a $\mathbb{Q}$-module. I had meant to say that $E_{n}$ is a simple $E_{n}$-module since it is a field, where $E_{n}$ is the finite-dimensional commutative $\mathbb{Q}$-algebra.

  • $\begingroup$ This is really clever! Thanks so much! It's a nice commutative algebra approach to the question. But thanks for the verification and neat example. I was wondering if there was some simple example rooted in representation theory (usually matrices, but then they're not commutative, in my course) that I was just somehow missing. $\endgroup$ – Adam Higgins May 14 '18 at 13:09
  • $\begingroup$ Only I'm having doubts now :) I think I fooled myself with ideas for univariate polynomial rings. Now it seems like in the example I gave both $R/M$ and $M/M^2$ are both $1$ dimensional. $\endgroup$ – rschwieb May 14 '18 at 13:14
  • $\begingroup$ No worries. Were they 1-dimensional as vector spaces over $\mathbb{Q}$? I thought they were n-dimensional. But I might be wrong. $\endgroup$ – Adam Higgins May 14 '18 at 13:28
  • 1
    $\begingroup$ But, yes, your construction also fits the bill. $\endgroup$ – Jyrki Lahtonen May 15 '18 at 9:48
  • 1
    $\begingroup$ @JyrkiLahtonen no worries. I was thinking about the group algebra for quite a long time. I managed to find lots of examples for a given $n$ by looking at $\mathbb{Q}$-algebras over cyclic groups and considering cyclotomic polynomials. But this can only give either one-dimensional simple modules, or simple modules of even dimension as vector spaces. Worse still, it can’t even give all even dimensions, since the possible dimensions of simple modules of the $\mathbb{Q}[C_{n}]$ Algebra are determined by the irreducible factors of $x^{n} - 1$, which are all cyclotomic polynomials. Then.. $\endgroup$ – Adam Higgins May 16 '18 at 9:28

I had a class today, in which I finally managed to resolve this question in my own head, with the help of others. Not only was my suggested example the one they were expecting, but I think you can show that any such example is similar, in a specific way, to the one I gave as follows

$\textbf{Claim 1: }$ Let $A$ be a finite dimensional commutative algebra over a field $k$, and let $S$ be a simple $A$-module such that $\operatorname{dim}_{k}(S) = n$. Then $S$ is a field extension of $k$ of degree $n$.

$\textbf{Proof: }$ $S$ is simple so there exists $s\in S$ non zero. For such an $s$, define $\phi_{s}: A \rightarrow S$ such that $ a \mapsto a \cdot s$. Then $\phi_{s}$ is a non-identically zero left-$A$-module homomorphism in to a simple $A$-module $S$, so in particular $\phi_{s}$ is surjective. Then $\operatorname{ker}(\phi_{s})$ is a sub-module of the $A$-module $A$, such that $A/ \operatorname{ker}(\phi_{s}) \cong S$ with $S$ simple, and so $\operatorname{ker}(\phi_{s})$ is in fact a maximal $A$-sub-module of $A$. Then $A$-sub-modules of $A$ are left-ideals, but since $A$ is commutative, in fact $\operatorname{ker}(\phi_{s})$ in a two-sided maximal ideal of $A$, and so $A / \operatorname{ker}(\phi_{s})$ is a field. Hence $S$ is a field, and so since $k \hookrightarrow S$ we see that $S$ is a field extension of $k$ with $\left|S : k \right| = \operatorname{dim}_{k}(S) = n$.

$\textbf{Claim 2: }$ Let $A$ be a finite-dimensional commutative $\mathbb{Q}$-algebra, with $S$ a simple $A$-module such that $\operatorname{dim}_{\mathbb{Q}}(S) = n$. Then $S \cong \mathbb{Q}[x] / \left(f(x)\right) $ as $\mathbb{Q}$-algebras, for $f$ some irreducible polynomial of degree $n$ in $\mathbb{Q}[x]$.

$\textit{Remark 1: }$ The isomorphism is only with respect to the induced $\mathbb{Q}$-algebra structure on $S$ (from it's $A$-algebra structure). The $A$-algebra structure of $S$ and $\mathbb{Q}[x] / \left(f(x)\right)$ (if it even admits one), need not be the same.

$\textbf{Proof: }$ By the above claim, $S$ is a field extension of $\mathbb{Q}$ of degree $n$. Then in characteristic zero, all finite degree field extensions are separable. Furthermore, by the primitive-element theorem, all finite and separable field extensions are simple (and hence algebraic) extensions. That is there is an $s \in S$ algebraic over $\mathbb{Q}$ such that $S = \mathbb{Q}(s)$. Then let $f(x) = m_{s}(x) \in \mathbb{Q}[x]$ be the (monic) minimum polynomial of $s$ over $\mathbb{Q}$. Then $f$ is irreducible of degree no greater than $n$, $m$ say. Then, as fields (that is $\mathbb{Q}$-algebras) $S = \mathbb{Q}(s) \cong \mathbb{Q}[x] / \left(f(x)\right)$. Moreover, $n = \left| S : \mathbb{Q} \right| = \operatorname{deg}(f) = m$, so $\operatorname{deg}(f) = n$.

Hence if the above two proofs are valid (which I believe they are), we have shown that any example satisfying the original criteria will have the be (as a $\mathbb{Q}$-algebra) of the form $\mathbb{Q}[x] / \left(f(x)\right)$ for some irreducible polynomial $f$. Which explains why these were the only examples I could come up with.

$\textit{Remark 2: }$The only specific fact about $\mathbb{Q}$ that I have used in the above proofs is that any finite degree field extension of $\mathbb{Q}$ is separable. Hence, if $k$ is any field such that all finite degree field extensions of $k$ are separable, then the analogous result applies for $k$. We know that any field of characteristic zero has the property that all finite degree field extensions are separable, and so the analogous result holds for all fields of characteristic zero.

$\textit{Remark 3:}$I have been told that the class of fields for which all finite fields extensions are separable are exactly the perfect fields. So we conclude that this result is in fact just a special case of the analogous result for perfect fields. Stated below as a proposition.

$\textbf{Proposition: }$ Let $k$ be a be a perfect field, with $A$ a finite dimensional commutative $k$-algebra and $S$ a simple $A$-module with $\operatorname{dim}_{k}S = n$. Then, as $k$-algebras, $S \cong k[x] / \left( f(x) \right)$, for some irreducible $f \in k[x]$ with $\operatorname{deg}(f) = n$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.