Is there an example of a ring where every left-ideal is two-sided but not every right ideal?
WHAT FOLLOWS IS A FAILED EXAMPLE
As an argument but not a proof that the example fails, consider:
$ f(a) \;a\; f(a) = 1 f(a) = f(a) $
Thus $f(a) (a \; f(a)) = f(a) \times 1$
This suggests that $(a \; f(a))$ must also be $1$ (it certainly can't equal $f(a)$ since $f(a)$ can't be idempotent), but I'm not sure how to make this an actual argument.
Let $F$ be the completely free algebra in the signature $(+, \times, f, \mathbb{Z}[w])$ where $+$ and $\times$ are binary functions, $f$ is a unary, and $\mathbb{Z}$ is a set of constants.
Additionally, let the function $d : F \to \mathbb{N}$ be defined as follows:
- $d(\mathbb{Z})$ is 0
- $d(f(x))$ is $1 + d(x)$ for all $x$.
- $d(x+y)$ is the maximum of $d(x)$ and $d(y)$ for all $x$ and $y$.
- $d(x\times y) = d(x+y)$
Next, I will build a set of equations, $\Delta$. The intent of this construction is to make $f(x)$ be a left inverse but not a right inverse of $x$ for every $x$.
- Let $\Sigma_0$ consist of the axioms of non-commutative rings, all variable-free true equations (i.e. of the form $t_1(\varnothing) = t_2(\varnothing)$) in $\mathbb{Z}$, and the variable-free equation $f(0) = 0$. Note that $\Delta_0$ is $f$-free except for a single sentence.
- Let $\Delta_0$ be the equational deductive closure of $\Sigma_0$. Note that $\Delta_0$ does not include the non-theorem $xy = yx$ because we can't conclude that it's true. We know that $xy = yx$ holds of all integers $\mathbb{Z}[w]$, but can't conclude that only integers exist.
- Let $\Sigma_1$ be the union of $\Delta_0$, sentences $f(x)x = 1$ for each $x$ such that $d(x) = 0$ and $x \neq 0$ is in $\Delta_0$, and sentences $f(x)x = 0$ for each $x$ such that $d(x) = 0$ and $x = 0$ is in $\Delta_0$.
- Let $\Delta_1$ be the equational deductive closure of $\Sigma_1$.
- Analogously, let $\Sigma_{n+1}$ be the union of $\Delta_n$, sentences $f(x)x = 1$ for each $x$ such that $d(x) = n$ and $x \neq 0$ is in $\Delta_n$, and sentences $f(x)x = 0$ for each $x$ such that $d(x) = n$ and $x = 0$ is in $\Delta_n$.
- Let $\Delta_{n+1}$ be the equational deductive closure of $\Sigma_{n+1}$.
I define $\Delta$ as $\cup_{n \in \mathbb{N}}\Delta_n$.
I define $A$ as $F/\Delta$, which is well-defined since $F$ is a completely-free algebra and $\Delta$ is a set of equations.
I claim that $A$ has precisely two left ideals.
Let $I$ be a left ideal of $A$. Suppose $I$ is $\{0\}$, then $I$ is closed under left multiplication by arbitrary elements of $A$. Suppose $I$ is not $\{0\}$. It follows that $I$ has a nonzero element $c$. $f(c)$ is a left inverse of $c$ and thus left multiplication by $rf(c)$ will hit the arbitrarily chosen ring element $r$ since $f(c)c = 1$ and thus $(rf(c))c = r(f(c)c) = r1 = r$. Thus, if $I$ is not $(0)$, then it is $(1)$.
Every left ideal of $A$ is two-sided since $(0)$ and $(1)$ are both two-sided ideals.
This is called being a left-not-right duo ring.
Consider the twisted polynomial ring $k[x;\sigma]/(x^2)$ where $k$ is a field and $\sigma$ is a ring endomorphism of $k$ such that $1<[k:\sigma(k)]$. This means it's like the normal polynomial ring except you have a rule to commute coefficients like this:
$x\lambda = \sigma(\lambda)x$
We want our ring $R$ to be that ring modulo the ideal $(x^2)$.
It's easy to see that $Rx=kx$ is the only nontrivial left ideal, and that it is a right ideal too.
Now consider two different $\sigma(k)$ basis elements of $k$, call them $b_1$ and $b_2$. It is not hard to show that $b_1xR\cap b_2xR=\{0\}$, and of course given that the only nontrivial left ideal is $Rx$, they cannot both be left ideals (because then they would be equal.)