Characterization of invertible elements of a ring

Question

As invertible matrices are characterized by having non zero determinant, I was wondering if there's a similar characterization of invertible elements for more general rings with identity (perhaps something like a generalization of the determinant).

Thanks in advance

Answer 1

Here's one thing you can do in some examples. Suppose $A$ is a $k$-algebra for some field $k$, i.e. $A$ is a ring and a $k$-vector space such that the multiplication $A \times A \to A$ is $k$-bilinear. Now $A$ acts on itself by multiplication: this gives an injective ring homomorphism

$$f = a \mapsto (b \mapsto ab) : A \to \operatorname{End}_k(A).$$

You should check this if it's not obvious! Now, suppose $\dim_k A$ is finite. By choosing a $k$-basis for $A$ we get an isomorphism of $k$-vector spaces $r : A \to k^n$ (where $n$ is some natural number), and this induces a ring isomorphism $\rho : \operatorname{End}_k(A) \to M_n(k)$ such that $\rho(m)(r(a)) = r(m(a))$ for all $m \in \operatorname{End}_k(A)$ and $a \in A$. This gives us the map $\varphi := \det \circ \rho \circ f : A \to k$. Notably, if $\varphi(a) = 0$ then $f(a)$ is not invertible in $M_n(k)$ so $a$ is not invertible in $A$! The converse need not hold, but we can still determine when elements of $A$ are invertible through $\varphi$ (as follows).

If $X \in M_n(k)$ is in the image of $\rho \circ f$ we have $X = \rho(f(a))$ for some $a \in A$ and thus $$X(r(1)) = \rho(f(a))(r(1)) = r(f(a)(1)) = r(a) \implies X = \rho(f(r^{-1}(X(r(1))).$$ Once we've constructed the map $r$ by choosing a $k$-basis for $A$, the right-hand condition is easy to check for any specific matrix $X$. Therefore, we have an easy way to tell if an element of $M_n(k)$ is in the image of $\rho \circ f$.

Now, if $\varphi(a) \neq 0$, we know $\rho(f(a))$ has some matrix inverse $X \in M_n(k)$. It follows that $a$ is invertible in $A$ if and only if $X \in \operatorname{img}(\rho \circ f)$, which we now know how to check.

Of course, this isn't even close to fully general, but it's a useful technique: not only does this give an algorithm for determining if elements of $A$ are invertible, but it gives you an algorithm for inverting those elements – if $a \in A$ is invertible then $a^{-1} = r^{-1}(\rho(f(a))^{-1}(r(1)))$! This is useful, for example, in field theory: if $A = k[x]/(p)$ for some irreducible $p \in k[x]$ our construction gives us an algorithm for computing the inverses of elements in the field $A$.

Answer 2

In general, this is a very hard problem. Let me start with giving some basic examples of rings where you can characterise the units nicely:

• $F$ is a skew field, then $U(F) = F\setminus \{0\}$
• $\mathbb{Z}/n \mathbb{Z}$, then $U(\mathbb{Z}/n \mathbb{Z}) = \{a: \gcd(a,n) = 1\}$.
• Let $R$ be a commutative ring. Then $U(R[X])$ is the set of all polynomials of the form

$$a_0 + a_1 X +\dots+ a_n X^n$$

where $a_0$ is a unit in $R$ and $a_1, \dots, a_n$ are nilpotent.

• The units of the formal power series $R[[X]]$ over a commutative ring $R$ are exactly those power series

$$\sum_{n=0}^\infty a_n X^n$$

where $a_0$ is a unit of $R$.

• Generalising the previous example, in a local ring, the units are exactly those elements that are not contained in the unique maximal ideal.

• Let $X$ be a compact Hausdorff space. Then the units of the ring of continuous functions $C(X, \mathbb{R})$ are exactly the elements the functions that vanish nowhere on $X$.

• If $\{R_i:i \in I\}$ is a collection of rings, then $$U\left(\prod_{i \in I} R_i\right) = \prod_{i \in I} U(R_i)$$

But the quest for finding units is sometimes very difficult. For example, finding the units for the group ring $RG$ where $R$ is a ring and $G$ is a (finite) group is currently active research.