Proving $R/J$ is local, where $R=k[\Gamma]$ and $J=(x^1)\unlhd R$.

Question

Let $\Gamma$ be the set of symbols of the type $x^q$, where $q\in\Bbb Q, \;q\ge0$. Setting $x^{q_1}\cdot x^{q_2}:=x^{q_1+q_2}$, $(\Gamma,\cdot)$ becomes a semigroup. Let then $k$ be a field. Let's then consider the semigroup algebra $k[\Gamma]$.

FIRST QUESTION: Is $k[\Gamma]=\{\sum_{i=1}^na_ix^{q_i}\;:\;n\ge1\;\mbox{integer}\;,a_i\in k,\;q_i\in\Bbb Q_{\ge0}\}$?

This last one is a $k$-algebra, with the usual sum, the usual product and product by elements of $k$. Thus in particular it is a ring which we call $R$. Writing down how multiplication works, it is immediate to see that $R$ is a commutative ring.

Let then consider $J:=(x^1)$ the ideal generated by $x^1$; thus $J=x^1R=\{\sum_{i=1}^na_ix^{q_i}\;:\;n\ge1\;\mbox{integer}\;,a_i\in k,\;q_i\in\Bbb Q_{\ge1}\}$.

So we can consider $\overline R:=R/J=\{\sum_{i=1}^na_ix^{q_i}+J\;:\;n\ge1\;\mbox{integer}\;,a_i\in k,\;q_i\in\Bbb Q\cap[0,1[\}$.

I must prove that $\overline R$ is a local ring. In order to see this, it would be enough to show that the sum of two non-invertible elements, is again not-invertible.

Thus, I began to investigate how invertible elements of $\overline R$ are made.

I conjectured that $\bar r=\sum_{i=1}^na_ix^{q_i}+J\in\overline R$ is invertible in $\overline R$ $\Longleftrightarrow$ $q_i=0$ for some $i=1,\dots,n$.

In this way we would see easily that $\overline R$ is a local ring.

I saw "$\Rightarrow$", but I have some problem in proving "$\Leftarrow$". I tried to build explicitly an inverse $\bar s=\sum_{j=1}^mb_jx^{p_j}+J$, but it seems really intricate.

SECOND QUESTION: Can someone help me in finding an inverse? Or tell me if there's a smarter way to prove that $\overline R$ is local.

Many thanks

Answer 1

1. Yes. This is the so called monoid ring.

2. A maximal ideal of $R$ is $M=(X^q:q\in\mathbb Q_{>0})$. Now let $I$ be an ideal which contains $X$. We want to show that $I\subseteq M$ or $I=R$. Let $f=a_0+a_{1}X^{q_1}+\cdots+a_{n}X^{q_n}\in I$, with $0<q_1<\cdots<q_n<1$, and suppose $a_0\ne0$. Now set $r_1=1-q_1$. Then $X^{r_1}f\in I$ and $X\in I$ implies $X^{r_1}\in I$. If $r_1\le q_1$ we get $a_0\in I$, and we are done. Otherwise, set $r_2=r_1-q_1$ and get $X^{r_2}\in I$, and so on. Notice that $r_2=1-2q_1$. Since there is a positive integer $m$ such that $1-mq_1>0$, and $1-(m+1)q_1\le0$, after $m$ steps we should stop. In conclusion, $a_0\in I$, so $I=R$.
   Thus we have shown that $M/(X)$ is the unique maximal ideal of $R/(X)$.