# When do the units of a ring together with 0 form a field?

Let $$R$$ be a commutative ring with identity. Let $$U(R)$$ be the multiplicative group of units of $$R$$. For what rings $$R$$ does $$U(R)\cup\{0\}$$ form a field? This is equivalent to asking: for what commutative rings is it true that the sum of two units is a unit or 0?

Fields trivially have this property. A nontrivial example is a polynomial ring over a field, since units look like $$a_0+a_1x+\ldots+a_nx^n$$ with $$a_0$$ a unit and $$a_1,\ldots,a_n$$ nilpotent. Edit: fields have no nontrivial nilpotent elements, so this example is also pretty trivial. I would appreciate any interesting examples of rings with the described property.

Relevant, but without conclusive answers: When does sum of two units give a unit?

• Mar 20, 2020 at 23:47
• Nilpotent elements in a field, Simple? Mar 21, 2020 at 0:57
• I would also appreciate any interesting examples of such rings. Mar 21, 2020 at 2:13
• $k[x,y,\bigcup_{n\ge 0} y/x^n]$ satisfies the condition so I doubt there is a simple classification even for integral domains of finite dimension. Mar 21, 2020 at 3:08

There might not be a thorough classification of such rings.

You may give it a name, though, say a "good ring" is a ring in which the sum of two units is either $$0$$ or a unit.

It is then possible to deduce some properties and equivalent definitions of good rings.

For example (all the rings I talk about are commutative):

A ring $$R$$ is good if and only if it contains a subfield $$k$$ such that $$R^\times = k^\times$$.

A ring $$R$$ is good if and only if for any $$1 \neq u\in R^\times$$, one has $$u - 1 \in R^\times$$.

If $$R$$ is a good ring, then the polynomial ring $$R[x]$$ is also good.

Note that the last construction can be iterated, thus e.g. $$k[x_1, \dotsc, x_n]$$ is good.

The realistic question is how useful this definition is. Usually, notions in commutative algebra serve as tools for other subjects such as algebraic geometry, algebraic number theory, etc.

That said, of course one can always study something just for fun, and may even find great applications later.

• @reuns This looks very complicated, as there are hardly any assumption on $R_n$... Mar 21, 2020 at 3:10
• Yes the point is to prove the classification problem is complicated even for integral domains of finite transcendental degree. If $R_\infty$ is $k$-good then any sub $k$-algebra is $k$-good so we have such a sequence $R_0=k, R_{n+1}=R_n[a_n]$, say for $n\le m$ they are algebraic over $k[x_1,\ldots,x_m]$, it remains to check if it is possible to parametrize $R_n$. Mar 21, 2020 at 3:18