# Maximal ideals in a circular discrete convolution algebra

Let $G$ = $\mathbb{Z_n}$ and let $A$ = $\ell^1(G)$ with convolution, over $\mathbb{C}$. Let $A_{\mathbb{R}}$ denote the subring of real valued functions in $A$, so $A_{\mathbb{R}}$ is an algebra over $\mathbb{R}$.

I'm being asked to find the maximal ideals of $A_{\mathbb{R}}$, their dimensions, and the corresponding quotient fields.

I've found that $m$ = {$f \in A_{\mathbb{R}}$ : $\sum_{m=0}^{n-1} f([m])$ = 0} is a maximal ideal of dimension n-1 with corresponding quotient field $\mathbb{R}$. I also know that $m$ isn't the only the maximal ideal as the function which takes 1 on every element of G is not invertible and isn't in $m$. I don't really know where to go with this problem. I spent a fair bit of time trying to determine the invertibility of sums of delta functions, which seemed to depend on the parity of n, but that didn't lead anywhere. I'm way up the river on this, and any pointers in the right direction would be helpful.

For context, the previous two parts of the problem were to determine the maximal ideal space of $A$ and whether or not the Gelfand transform of $A$ was isometric.

Edit: With the help of the suggestions below, I've been able to identify a number of maximal ideals of $A_\mathbb{R}$. Namely, $\mathfrak{m} \cap A_\mathbb{R}$, where $\mathfrak{m}$ is a maximal ideal in $A$. I want to say that these are the only maximal ideals of $A_\mathbb{R}$ (or at least I think I do), but I am unsure as to how to approach this.

-
Via the Fourier transform, this convolution algebra is isomorphic to the algebra of functions on $G$ with pointwise addition and multiplication. From this point of view, some other surjective maps onto $\mathbb{R}$ are easier to see: the projections onto each coordinate. So there are these $n$ maximal ideals plus the one you found. I don't know how to find them all though. –  Ted Oct 2 '11 at 3:39
Hmmm, the Fourier transform will end up taking values in $\mathbb{C}$, not $\mathbb{R}$. So I'm not so sure about the above comment. But I'll leave it there in case it's useful. –  Ted Oct 2 '11 at 3:43
Here's some progress: The Fourier transform is given by $F(\chi) = \sum_{g \in G} f(g) \overline{\chi(g)}$ for $\chi \in G^*$. In general, this is $\mathbb{C}$-valued. But if $\chi$ is the trivial character, this projection $F \mapsto F(\chi)$ is real-valued, and corresponds to the maximal ideal you already found. And if $n$ is even, there is a non-trivial character onto $\{1,-1\}$ which gives you another maximal ideal. –  Ted Oct 2 '11 at 4:32
Hi Ted. Thanks for your input. What you’ve mentioned so far fits in with what I’ve been thinking i.e. trying to salvage some of the theory from the complex case. I’ve considered simply taking the intersection of a maximal ideal in $A$ with $A_\mathbb{R}$ but I don’t believe that this will be maximal in general. –  Grasshopper Oct 2 '11 at 4:53
Dear Grasshopper, Let $\mathfrak m$ be a maximal ideal in $A$. What is the quotient $A/\mathfrak m$? Now consider the quotient $A_{\mathbb R}/\mathfrak m \cap A_{\mathbb R}$. What are the possibilities for this quotient? Regards, –  Matt E Oct 2 '11 at 5:50

Note that $f \in A_{\mathbb R}$ iff $\hat{f}(\overline{\alpha}) = \overline{\hat{f}(\alpha)}$ for all $\alpha$. If $M_\alpha$ is the maximal ideal $\{f: \hat{f}(\alpha) = 0\}$ of $A$, then $M_{\alpha} \cap A_{\mathbb R} = M_{\overline{\alpha}} \cap A_{\mathbb R}$. For any $\alpha$, there is $h_\alpha \in A_{\mathbb R}$ such that $\widehat{h_\alpha}(\alpha) = \widehat{h_\alpha}(\bar{\alpha}) = 1$ and $\widehat{h_\alpha}(g) = 0$ for all other $g$.
Use this to show that $M_\alpha \cap A_{\mathbb R}$ is a maximal ideal in $A_{\mathbb R}$.
I decided to take one more look at the problem before I went to sleep and I think I've got it now. The map from $A_\mathbb{R}$ to $\mathbb{C}$ given by $f \mapsto \sum_{g\in G} f(g)\alpha(g)$ is an $\mathbb{R}$-algebra homomorphism. The function $h_\alpha$ that you mentioned ensures that the range of this map is either $\mathbb{R}$ or $\mathbb{C}$. In either case, the range is a field and so the kernel is a maximal ideal. Is this what you had in mind? I don't know how to show that this construction gives us every maximal ideal of $A_\mathbb{R}$, but maybe that'll be easier in the morning. –  Grasshopper Oct 2 '11 at 8:14
Is it the case that every maximal ideal of $A_\mathbb{R}$ is of this form? I tried assuming there is a maximal ideal that isn't equal to any of the $M_\alpha \cap A_\mathbb{R}$ but I'm getting nowhere with that. –  Grasshopper Oct 2 '11 at 23:43
@Grasshopper: To prove that you've found all maximal ideals, it's much easier to work in the Fourier-transformed space (with pointwise addition and multiplication) than the original space. First, figure out how to prove the complex equivalent: every maximal ideal of $A$ is one of the $M_{\alpha}$. Then adapt this proof for the real case. –  Ted Oct 3 '11 at 5:28
@Ted: I am unfamiliar with the Fourier-transformed space that you are referring to. I know that the $M_\alpha$ are the only maximal ideals of $A$ via the identification of the maximal ideals with the maximal ideal space. –  Grasshopper Oct 3 '11 at 6:28