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If $R$ is an algebra without a unit, then the standard unitisation $R^\sharp$ can have maximal one-sided ideals other than $R$. Thus, it is natural to ask about the following. Let $R$ be an algebra without a unit (over a field with char 0 if it does matter).

Is there a unital algebra $A$ such that $R$ is the unique maximal right ideal in $A$?

EDIT: Assume $R$ has a faithful representation $\pi$ on a vector space $V$ (over $K$), what if we consider $A=\pi(R)+KI$?

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2 Answers 2

The nice examples given by Pace show that the answer is negative in general. There is also a natural condition on $R$ which gives a positive solution. It involves the Jacobson radical of a non-unital ring. (I'm following the first few exercises in section 4 of Lam's A First Course in Noncommutative Rings.)

The binary operation $a \circ b = a + b -ab$ on $R$ is associative and has unit element $0$. An element of $R$ is called left (right) quasi-regular if it has a left (right) inverse in the monoid $(R, \circ, 0)$ and is called quasi-regular if it has both a left and a right inverse, which are necessarily equal.

The Jacobson radical $J(R)$ is the sum of all left ideals of $R$ consisting of left quasi-regular elements. One can show that this is an ideal of $R$, and all elements in it are quasi-regular. So $J(R)$ is the largest left (right) ideal consisting of left (right) quasi-regular elements. One can show that this coincides with the usual Jacobson radical in case $R$ has an identity.

Claim: For a (not necessarily unital) ring $R$, the following are equivalent:

  1. $R$ is the unique maximal right (or left) ideal of its standard unitization $R^\sharp$;
  2. $R = J(R)$.

Proof: Assume 1 holds, so that $R^\sharp$ is local and $R = J(R^\sharp)$. Then for any $x \in R$, the element $1 - x = 1 \oplus (-x) \in K \oplus R = R^\sharp$ lies in $R^\sharp \setminus R$ and therefore is invertible. Its inverse is of the form $\alpha \oplus y \in R^\sharp$. Writing out the equation for $(1-x)(\alpha \oplus y) = 1 \oplus 0$ yields that $\alpha = 1$ and $x + y - xy = 0$, and similarly we obtain $y + x -yx = 0$. So every element of $R$ is quasi-regular and 2 holds.

Conversely, assume 2 holds. We know that $R$ is a maximal right (and maximal left) ideal of $R^\sharp$ simply because of the ring isomorphism $R/R^\sharp \cong K$; thus $J(R^\sharp) \subseteq R$. If we can show that $R$ is contained in $J(R^\sharp)$ then we will be done, for then $R \subseteq J(R^\sharp) \subseteq R$ yields that $R^\sharp/J(R^\sharp) = R^\sharp/R \cong K$ making $R$ local with unique maximal right ideal $R$.

So let $x \in R$. To show $0 \oplus x \in J(R^\sharp)$, let $\alpha \oplus r \in R$; we need to find a left inverse to $$1 \oplus 0 - (\alpha \oplus r)(0 \oplus x) = 1 \oplus-(\alpha x + rx) = 1 \oplus (-x')$$ where we set $x' = \alpha x + rx \in R$. To this end, fix a left quasi-inverse $y$ for $x$ in $R$. Then from $y+x'-yx' = 0$ we obtain $$(1 \oplus -y)(1 \oplus -x') = 1 \oplus 0$$ as desired. QED

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The answer to this question is no in general. Let $K$ be a field, and let $R=F\{s,t\ :\ st=s+t \}$, the non-unital algebra generated by the non-commuting variables $s,t$ subject to the single relation $st=s+t$.

Let $A$ be any unital $K$-algebra containing $R$. In $A$ we have $(1-s)(1-t)=1-s-t+st=1$. However, $(1-t)(1-s)=1-t-s+ts\neq 1$ since $ts\neq s+t$. Thus $(1-s)$ is only right invertible, and hence $A\setminus R$ does not consist only of units.

Edited to add: An easier example, which has the added benefit of being commutative is taking $R$ to be the non-unital ring generated by commuting variables $x,y$ subject only to the additional relation $x^2=x$. The element $1-x\in A\setminus R$ is not unit, since it is a non-trivial idempotent.

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