Adjoining an inverse to a central element of an algebra

Question

Given a (not necessarily commutative) algebra $A$, and a central element $c \in A$, is it always possible to enlarge $A$ to a an algebra $A'$ in which $c$ is invertible? I guess one can take a set of generators and relations for $A$, and add a formal inverse $c^{-1}$ along with extra relations making $c^{-1}$ central and satisfying $c^{-1}c = c^{-1}c = 1$. But I guess this is a specific example of a general formal process. What could this process be?

Answer 1

Let $R$ be a ring and let $c$ be a central element of $R$. We can form the polynomial ring $R[x]$ and its quotient ring $$ R_c := R[x] / (cx - 1) \,. $$ We have a homomorphism of rings $i$ from $R$ to $R_c$ given by the composite of the inclusion from $R$ to $R[X]$ and the quotient map from $R[x]$ to $R_c$. The element $i(c) = [c]$ is a unit in $R_c$ with inverse given by $[x]$ because $$ [c] \cdot [x] = [cx] = [1] = 1_{R_c} $$ and similarly $$ [x] \cdot [c] = [xc] = [cx] = [1] = 1_{R_c} . $$

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The pair $(R_c, i)$ is universal with this property, in the following sense: Let $(S, j)$ be another pair consisting of a ring $S$ and a homomorphism of rings $j$ from $R$ to $S$ such that $j(c)$ is a unit in $S$. Then there exists a unique homomorphism of rings $f$ from $R_c$ to $S$ with $f \circ i = j$. In other words, the homomorphism $j$ factors uniquely through the homomorphism $i$.

The ring $R_c$ is in this sense the universal way of making the element $c$ invertible.

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So under what conditions can we embed $R$ into a larger ring $S$ such that $c$ becomes a unit in $S$? Well, we need a pair $(S, j)$ as above such that $j$ is injective. (We refer to this condition as $(*)$.)

If on the one hand such a pair $(S, j)$ exists, then follows from the factorization $j = f \circ i$ that $i$ also needs to be injective. If on the other hand $i$ is injective, then one can choose the pair $(S, j)$ as $(R_c, i)$.

We hence see that $(*)$ is possible if and only if the homomorphism $i$ is injective.

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We now claim that $i$ is injective if and only if $c$ is no zero divisor in $R$.

Suppose first that $i$ is injective. If $c$ were a zero divisor in $R$, then it would also be a zero divisor in $R_c$. Indeed, there would exist some nonzero element $r$ of $R$ with $cr = 0$. But then $i(r)$ would be a nonzero element of $R_c$ (by the injectivity of $i$) such that $i(c) \cdot i(r) = i(cr) = i(0) = 0$, which would mean that $i(c)$ would be a zero divisor. But this would contradict $i(c)$ being a unit. This shows that $c$ cannot be a zero divisor if $i$ is injective.

Suppose on the other hand that $c$ is not a zero divisor. The element $cx - 1$ of $R[x]$ is central, whence the two-sided ideal $(cx - 1)$ is given by $$ (cx - 1) = \{ p \cdot (cx - 1) \mid p \in R[x] \} \,. $$ Let $r$ be an element of the kernel of $i$. Then $0 = i(r) = [r]$, which means that $r$ is contained in the two-sided ideal $(cx - 1)$, and is thus of the form $$ r = p (cx - 1) $$ for some polynomial $p$ in $R[x]$. The leading coefficient of the polynomial $cx - 1$ is $c$, which is not a zero divisor in $R[x]$. It follows that $$ \deg( p (cx - 1) ) = \deg(p) + \deg( cx - 1 ) = \deg(p) + 1 \,. $$ But $r$ is an element of $R$, and therefore cannot be of degree $\geq 1$ in $R[x]$. We hence find that $r = 0$ and $p = 0$. This shows that $\ker(i) = 0$, so that $i$ is injective.

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We have overall shown that $R$ can be embedded into a larger ring in which $c$ is invertible if and only if $c$ is not a zero divisor in $R$. Such a larger ring can then be constructed as the quotient $R[x] / (cx - 1)$, and this is the universal construction for making $c$ invertible.

Answer 2

Let $S=\{a^n \mid n \in \Bbb N\}$. Then $S$ satisfies the Ore condition, so one can form the Ore localization $AS^{-1}$. This is constructed in the same way that localizations of commutative rings work. This is treated in many non-commutative algebra books, e.g. in Lam's "Lectures on Modules and Rings".

In this special case, there's another contruction yielding the same result: we can form the polynomial ring $A[x]$ and divide by the two-sided ideal $(ax-1)A[x]=A[x](ax-1)=A[x](ax-1)A[x]$ to get $A[x]/(ax-1)A[x]$. The class of $x$ will be a formal inverse of $a$.