Examples of rings whose units $\cup\, \{0\}$ comprise a field

Question

Let $K$ denote a field. Then the polynomial ring $K[x]$ has the property that the sum of two units is either a unit, or zero. I'll bet there's heaps of other examples, though. So let a coolring be a commutative, unital ring $R$ in which the sum of two units is either a unit, or else zero. (Do these things have a name?) Thus every field is a coolring. Also, if $R$ is a both a coolring and an integral domain, then $R[X]$ is itself a coolring (and an integral domain), for any set of indeterminates $X$.

Question. What are some examples of coolrings that cannot be expressed in the form $R[X]$, where $X$ is a non-zero set of indeterminates?

I'd especially like to see a generic way of constructing such coolrings.

Answer 1

If $S$ is a coolring, then the set of units of $S$, together with zero, is closed under addition and multiplication. It contains $\pm 1$ and $0$, and so is a subring. Every non-zero element is a unit, so it is a subfield.

So coolrings can be characterized as being $k$-algebras $S$, for some field $k$, such that $S^{\times} = k^{\times}$.

First note that a coolring must be reduced (i.e. contain no non-zero nilpotents): if $x \in S$ is nilpotent, then $1+ x$ is invertible, and hence $1 + x$, and so $x$, lies in the field $k$, thus $x = 0$.

A basic source of interesting reduced $k$-algebras is the affine rings of (i.e. rings of regular functions on) affine algebraic sets over $k$. (If you want just integral domains, then restrict to irreducible affine alg. sets, i.e. affine varieties.) (All f.g. reduced $k$-algs. arise in this way.)

So you are looking for affine algebraic sets on which any nowhere zero regular function is necessarily constant.

As you observe, the affine spaces $\mathbb A^n$ have this property (these correspond to the coolrings $k[x_1,\ldots,x_n]$).

But lots of other affine algebraic sets do too.

E.g. any affine alg. set which admits a dominant morphism (i.e. a morphism with dense image) from an affine space does. In algebraic terms, this means that if the $k$-algebra $S$ embeds into $k[x_1,\ldots,x_n]$, then it is a coolring. This includes Hurkyl's examples $k[t^2,t^3]$ (which embeds in $k[t]$).

This can also be written as $k[x,y]/(y^2 - x^3)$, and the corresponding affine algebraic set is the cuspidal cubic $y^2 = x^3$.

E.g. any connected algebraic set which is a union of subsets each admitting a dominant morphism from some affine space has this property.

E.g. $k[x,y]/(xy)$ (which corresponds to the union of the $x$ and $y$ axes in the plane) is a coolring.

The property of having $S = R[t]$ just corresponds geometrically to the alg. set $Y$ attached to $S$ being the product $X \times \mathbb A^1$ for the algebraic set $X$ attached to $R$. So examples like the previous ones (the cuspidal cubic and the union of two lines) gives examples of coolrings which are not of the form $R[t]$ for some coolring $R$.

Another example is the coolring $k[x,y]/(y^2 - x^2(x-1) )$ corresponding to a nodal cubic.

There are examples which are not dominated by affine spaces too, e.g. the coolring $k[x,y]/(y^2 - x(x-1)(x+1))$ corresponding to the smooth affine cubic $y^2 = x(x-1)(x+1)$. (The point is that this is the complement of a single point in a projective curve.)

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Added: note that zcn's example of cones over projective varieties (i.e. homogeneous coordinate rings of proj. varieties) can also be explained by the ``covered by affine spaces'' rubric. Namely, any such cone is a union of lines through the origin (that's why it's called a cone!). On each line a nowhere zero function must be constant. Since these all these lines have a point in common (the origin) this constant must be independent of the line. Thus any nowhere zero function on such a cone is constant.

Answer 2

Any ring with trivial unit group is a coolring: if $1$ is the only unit, then the ring has characteristic $2$ (since $-1 \in R^\times \implies 1 = -1$), so the only sum of two units is $1 + 1 = 0$. The semilocal rings with trivial unit group are precisely finite products of $\mathbb{Z}/2\mathbb{Z}$, and if the product has more than one factor, these are not of the form $R[t]$ for any ring $R$ (since every nonunit is a zerodivisor).

For an infinite counterexample, take $S := (\mathbb{Z}/2\mathbb{Z})[x,y]/(xy)$, which has trivial unit group. If it were of the form $R[t]$ for some ring $R$, then $R$ would be Artinian with trivial unit group, hence $R = \prod_{i=1}^n \mathbb{Z}/2\mathbb{Z}$ is a finite product of $\mathbb{Z}/2\mathbb{Z}$'s. Then $n > 1$ since $S$ is not a domain, so $R[t]$ has idempotents, but $S$ does not (for proofs of some of these statements, see this answer).

Update: As other answers have given some geometric examples, let me mention a truly abundant source: any homogeneous coordinate ring of a projective variety is a coolring (over any field $k$). Indeed, such a ring is of the form $R = k[x_0, \ldots, x_n]/P$ for some homogeneous prime $P$, so the projection $k[x_0, \ldots, x_n] \twoheadrightarrow R$ induces a surjection $k^\times = k[x_0, \ldots, x_n]^\times \twoheadrightarrow R^\times$ (since $R$ is a postively graded domain, so $R^\times = (R_0)^\times$). Generalizing, a ring of the form $k[x_0, \ldots, x_n]/J$, where $J$ is a homogeneous radical ideal (and thus a finite intersection of homogeneous primes), is a coolring.

Answer 3

We have that $R$ satisfies your property if and only if $k = R^\times\cup \{0\}$ is a field. So we can state this problem inversely: if $k$ is a field, for which ring extensions $k\subset R$ is it the case that $R^\times = k^\times$?

A first observation: if $\operatorname{char} k\neq 2$, then every element of $R$ must be transcendental over $k$. This is because, if $x\in R$ is algebraic, then $k[x]$ is a product of field extensions of $k$. Each field extension must be trivial, to avoid increasing the unit group, and there can only be one field in the product, for the same reason (here we use $\operatorname{char} k\neq 2$). So $x\in k$.

Now, if we have $R^\times = k^\times$, then $S^\times = k^\times$ for any $k\subset S\subset R$. So we should begin our search by considering the case where $R$ is finitely generated over $k$. Said differently: we know that $R$ is a quotient of $k[X_\alpha]$ for some indexing set, but we are interested in what relations can hold between the variables.

This problem now has a nice statement in algebraic geometry: for what affine varieties $X$ over $k$ do we have $k[X]^\times = k^\times$? Equivalently: when are all non-vanishing regular functions on $X$ constant?

A simple example is zcn's $k[x,y]/(xy)$, the coordinate ring of two intersecting lines, which is a coolring for any $k$. guy-in-seoul just put this in his answer, so see his post for details. In short, joining two coolvarieties at a point will yield another coolvariety, so two lines is the natural non-polynomial-ring example.

As another example, if $X$ is a non-singular cubic plane curve given by $y^2 = x^3 + ax+b$ (or any affine curve with only one point at infinity), then all non-vanishing regular functions on $X$ are constant. For if $X\to\mathbb{A}^1_k \subset \mathbb{P}^1_k$ is a morphism, then it can be extended to $\tilde{X} = X\cup\{\infty\}\to \mathbb{P}^1_k$, which must be surjective (impossible) or constant.