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I'm stuck with the following:

Let $R=\left\{x\dfrac{f(x)}{g(x)}\,:\,f(x),g(x)\in\mathbb{R}[x],\,g(0)\neq 0\right\}$. Show that $R$ has not maximal ideals.

My principal problem is that $R$ is not unital, so if $M$ is an ideal, I can't use that $R/M$ is a field.

user26857
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sinbadh
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  • so this is the ring of rational functions on $\mathbb{R}$ except it's not allowed to have a singularity at 0? – Zoe Allen Mar 05 '23 at 01:16
  • I don't think so. g(x) has no x as divisor. So in the numerator we never have constant functions. Don't you think? – sinbadh Mar 05 '23 at 01:53
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    @ZoeAllen I believe it is the ideal generated by $x$ of that ring, viewed as a ring. – Arturo Magidin Mar 05 '23 at 03:25
  • As shown here a commutative ring with no proper nontrivial ideals is either the cyclic group of order $p$ with zero multiplication for some prime $p$, or a field. If $M$ is a maximal ideal of a commutative ring, with or without unity, then $R/M$ is a commutative ring with no proper nontrivial ideals. – Arturo Magidin Mar 05 '23 at 03:28

1 Answers1

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Let $A=\{x^2\frac{f(x)}{g(x)}:\;f(x)\in\mathbb{R}[x],\,g(x)\in\mathbb{R}[x],\,g(0)\neq0\}$.

Note that for all $r_1,r_2\in R$ you have $r_1 r_2\in A$.

Note also that for any $p(x),q(x)\in\mathbb{R}[x]$ such that neither $p(x)$ nor $q(x)$ has $x$ as a factor, we have that $A\subseteq \big<x\frac{p(x)}{q(x)}\big>$. For such a choice of $p(x)$ and $q(x)$, note that $\big<x\frac{p(x)}{q(x)}\big>-A=\{nx\frac{p(x)}{q(x)}:n\text{ is a non-zero integer}\}$.

Now let $I$ be a proper ideal, i.e. $I\neq0$ and $I\neq R$. We are going to prove that $I$ is not maximal.

Case one: $I\subseteq A$.

For this case, we need only prove that $A$ is not maximal. Just observe that $\big<A,x\big>-A=\{nx:n\text{ is a non-zero integer}\}\neq R-A$. Therefore, $A$ is not maximal.

Case two: $I\not\subseteq A$.

That means that there exists an element of $I$ that is not in $A$. Such an element must be of the form $x\frac{p(x)}{q(x)}$, where neither $p(x)$ nor $q(x)$ has $x$ as a factor. We immediately have $A\subseteq \big<x\frac{p(x)}{q(x)}\big>\subseteq I$. But since $I\neq R$, there exist $r(x),t(x)$ such that neither $r(x)$ nor $t(x)$ has $x$ has a factor and such that $x\frac{r(x)}{t(x)}\notin I$.

Now, the key question is this. Under what conditions is it the case that $cx\frac{r(x)}{t(x)}\in\big<x\frac{r(x)}{t(x)}\big>+I$, where $c$ ranges over the rational numbers?

If for a particular $c$ we have $cx\frac{r(x)}{t(x)}\in\big<x\frac{r(x)}{t(x)}\big>+I$, we can let $cx\frac{r(x)}{t(x)}=\zeta+\beta$, where $\alpha\in\big<x\frac{r(x)}{t(x)}\big>$ and $\beta\in I$. We can write $\zeta=zx\frac{r(x)}{t(x)}+a$, where $z$ is an integer and $a\in A$. Since $A\subseteq I$, we have $(c-z)x\frac{r(x)}{t(x)}=a+\beta\in I$.

We can conclude that the set of rational values $c$ such that $cx\frac{r(x)}{t(x)}\in\big<x\frac{r(x)}{t(x)}\big>+I$ is precisely the set $\mathbb{Z}+Q$ viewed as an additive subgroup of $\mathbb{Q}$, where $Q=\{f\in\mathbb{Q}:\;fx\frac{r(x)}{t(x)}\in I\}$. Recall that $x\frac{r(x)}{t(x)}\notin I$, so of course $1$ is not an element of $Q$.

Now, look at the following link

$(\mathbb{Q},+)$ has no maximal subgroups

whence it follows that $\mathbb{Z}+Q$ cannot be all of $\mathbb{Q}$.

There therefore exists a rational number $c_0$ such that $c_0 x\frac{r(x)}{t(x)}\notin\big<x\frac{r(x)}{t(x)}\big>+I$.

So $I$ is not maximal.

user26857
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Chris Sanders
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