Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $A$ be the algebra of continuous functions $\mathbb{R} \to \mathbb{R}$ which are periodic of period $1$, and write $M$ for the $A$-module of antiperiodic functions of period $1$ (meaning $f \in M$ satisfies $f(x+1) = -f(x)$ for all $x \in \mathbb{R}$). In Etingof's Introduction to Representation Theory, he asks the reader to show that $M$ is indecomposable and that $A \oplus A \cong M \oplus M$ as $A$-modules. How does one prove either of these things? I would be happy with hints.

Note that since $A$ and $M$ are not isomorphic as $A$-modules (clearly $M$ is not cyclic) this illustrates the failure of the Krull-Schmidt theorem without finiteness conditions.

share|cite|improve this question
up vote 1 down vote accepted

Let's suppose $M$ can be decomposed into a direct sum of non-zero submodules, $M = M_1 \oplus M_2$, and define $$ X_i := \{x \in \mathbb R : f(x) = 0\quad\forall f\in M_i \} $$ $X_i$ is closed. If $x_n$ is a sequence of elements of $X_i$ convergent to $x_0$ then for each $f\in M_i$, $f(x_n) = 0$ and, since $f$ is continuous, $f(x_0) = 0$. Hence $x_0 \in X_i$.

The sets $$ Y_i := \mathbb R \setminus X_i\quad i = 1, 2 $$ are not empty and open.
For each $x\in \mathbb R$, it exists a function $f\in M$ such that $f(x) \neq 0$. Writing $f$ as $f_1 + f_2$, with $f_1\in M_1$ and $f_2\in M_2$, we can deduce that $x\in Y_1 \cup Y_2$ and therefore $Y_1 \cup Y_2 = \mathbb R$.
Since $\mathbb R$ is connected, we must conclude that $Y_1 \cap Y_2 \neq \emptyset$.

Let $y_0$ be an element of $Y_1 \cap Y_2$ and $f_i\in M_i$ such that $f_i(y_0) \neq 0$, $i = 1, 2$.
We can choose an interval $I$ centered on $y_0$ such that $$ f_i(x) \neq 0 \quad \forall x \in I, i = 1, 2 $$ Let's take the set $$ J = \{ x + n \in \mathbb R : x\in I, n\in \mathbb Z \} $$ the function defined by $$ d(x) = \begin{cases} f_1(x)/f_2(x) &\text{if } x\in J\\ 0 & \text{if } x\notin J \end{cases} $$ and a map $0 \neq h\in A$ with support in $J$.

$f^*_1 := h^2 f_1$ is a non-zero element of $M_1$, $f^*_2 := h f_2$ is an element of $M_2$ and $d^* := h d$ is an element of $A$. Moreover $$ d^* f^*_2 = h^2 d f_2 = h^2 f_1 = f^*_1 $$ Therefore $f^*_1$ belongs also to $M_2$. That contradicts the made assumption $M_1\cap M_2 = \emptyset$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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