# Which commutative rings admit exactly one ring homomorphism from $\mathbb{R}$?

Which commutative rings $$A$$ have the property that there is exactly one ring homomorphism $$\mathbb{R} \to A$$?

For the sake of a better name, let's call these rings "really unique" (pun intended). Clearly, $$0$$ is really unique. It is well-known that $$\mathbb{R}$$ is really unique. The class of really unique rings is closed under limits in $$\mathbf{CRing}$$. So for example, $$\mathbb{R}^I$$ for every index set $$I$$ is really unique.

The polynomial ring $$\mathbb{R}[X]$$ is really unique: If $$f : \mathbb{R} \to \mathbb{R}[X]$$ is a ring homomorphism, then $$f(r)$$ is a unit for every $$r \in \mathbb{R}^*$$, so $$f(r)$$ is constant. But then $$f$$ is just coming from a ring homomorphism $$\mathbb{R} \to \mathbb{R}$$ and we are done. The same holds for every polynomial ring over $$\mathbb{R}$$ (in any number of variables).

With limits we can then build more really unique rings, such as the "axis ring" $$\mathbb{R}[X] \times_{\mathbb{R}} \mathbb{R}[Y] \cong \mathbb{R}[X,Y]/\langle XY \rangle$$.

Of course every subring of a really unique commutative ring $$A$$ that contains the copy of $$\mathbb{R}$$ is also really unique. So for example, $$\mathbb{R}[X^2,X^3] \cong \mathbb{R}[X,Y]/\langle X^3-Y^2 \rangle$$ is really unique. For every topological space $$S$$ the ring $$C(S)$$ of real-valued continuous functions on $$S$$ is really unique (it is a subring of $$\mathbb{R}^S$$).

The question is partially motivated by this one, where $$A = M_2(\mathbb{R})$$ (not commutative) and uniqueness fails. In his answer, Qiaochu also reminded us that $$\mathbb{C} \cong \mathbb{R}[X]/\langle X^2+1 \rangle$$ is not really unique (in particular, being finitely generated over $$\mathbb{R}$$ is not sufficient).

Since $$\mathbb{C}$$ is not really unique, it follows that for a really unique $$\mathbb{R} \hookrightarrow A$$ the only intermediate field $$F$$ which is finitely generated over $$\mathbb{R}$$ must be $$\mathbb{R}$$ (using Zariski's Lemma). But I don't know what happens when we drop finite generation.

The "universal" example of a non-really unique ring is $$\mathbb{R} \otimes \mathbb{R}$$ (tensor product over the integers or rationals, doesn't matter). In general, a commutative ring $$A$$ is really unique if it admits at least one homomorphism from $$\mathbb{R}$$ and every homomorphism $$\mathbb{R} \otimes \mathbb{R} \to A$$ factors over the codiagonal $$\mathbb{R} \otimes \mathbb{R} \to \mathbb{R}$$. But this just a reformulation of the definition.

If a full classification is not possible, answers with interesting examples are also appreciated. We may also restrict to finitely generated $$\mathbb{R}$$-algebras if necessary.

Here is the proof that $$\mathbb{R}$$ is really unique. I will write it down here so that we don't need to switch between tabs, but also since I suspect that the method may be generalized. So let $$f : \mathbb{R} \to \mathbb{R}$$ be a ring homomorphism. Clearly, $$f(q)=q$$ for all $$q \in \mathbb{Q}$$, and since $$x \geq 0 \iff x \text{ is a square}$$ we see that $$f$$ preserves the order. Also, $$f$$ is injective since $$\mathbb{R}$$ is a field, and hence reflects the order. Now $$f(r)$$ is the supremum of all rational numbers $$q$$ with $$q \leq f(r)$$, i.e. $$f(q) \leq f(r)$$. This is equivalent to $$q \leq r$$, so the supremum is $$r$$. $$\checkmark$$

• Do you know any examples of commutative rings that contain a copy of $\mathbb{R}$, are not really unique, but don't contain a copy of $\mathbb{C}$? In particular, does $\mathbb{R}\otimes_\mathbb{Q}\mathbb{R}$ contain a copy of $\mathbb{C}$? Maybe that's obvious? Commented Sep 5 at 8:09
• To answer my own question: yes, $\mathbb{R}\otimes_\mathbb{Q}\mathbb{R}$ does contain a copy of $\mathbb{C}$. Commented Sep 5 at 10:30
• @JeremyRickard: $\mathbb{R}\otimes_\mathbb{Q}\mathbb{R}$ does not contain a copy of $\mathbb{C}$, since it has a homomorphism to $\mathbb{R}$. Commented Sep 5 at 14:21
• @EricWofsey Yes, you're right. I was confused about what I meant by "contains a copy of $\mathbb{C}$". What is true is that it contains a copy of $\mathbb{C}\times\mathbb{R}$, but I'm not sure that helps with anything. Commented Sep 5 at 14:38
• @JeremyRickard A real closed field extending $\mathbb R$ should be a counterexample since you can map transcendental elements of $\mathbb R$ to elements that differ from them by an infinitesimal. Commented Sep 5 at 15:23