A finitely presentable real algebra without morphisms to the real field and without morphism from complex field?

Question

I guess there must standard examples but I can't seem to find them. I tried ${A = \mathbb{R}[x,y]/(x^2 -1, y^2 x + x)}$, which is not the domain of a homomorphism to $\mathbb{R}$ because it would have to send $x$ to $1$, and so $y^2$ to $i$. But then I have trouble checking that there is no homomorphism ${\mathbb{C} \rightarrow A}$. So I tried ${\mathbb{R}[x,y]/(x^2 -1, y^2 x + x, y^3 + y)}$ hoping it would be easier but no luck. Do you know a simple example?

Answer 1

What about $$B=\Bbb{R}[x,y]/(x^2+y^2+1)$$

• There is no morphism to $\Bbb{R}$ (the image of $x^2$ would be strictly negative)

• And $B$ is an integral domain so a morphism from $\Bbb{C}$ would send $i^2$ to $-1$, but $-1$ is not a square in $B$ (since $B[z]/(z^2+1)\cong \Bbb{C}[x,y]/(x^2+y^2+1)$ is an integral domain)

Answer 2

There is no simpler example despite the lack of an objective standard.

We can show that if $A$ is generated by a single element over $\mathbb R$, i.e. $A\simeq \mathbb R[x]/(f(x))$, then it cannot be an example. Indeed, since there is no $\mathbb R$-homomorphism to $\mathbb R$, $f(x)$ has no real root, hence it's a product of irreducible factors of degree $2$, say $f=\prod_i f_i^{n_i}$ where $f_i$'s are distinct. By CRT, $$\mathbb R[x]/(f(x))\simeq \prod_i\mathbb R[x]/(f_i(x))^{n_i}\simeq \prod_i\mathbb C[y]/(y^{n_i})$$ which admits an embedding of $\mathbb C$. In the language of algebraic geometry, there is a morphism from $\text{Spec}(A)$ to $\text{Spec}(\mathbb C)$ over $\text{Spec}(\mathbb R)$, which is not surprising, given $\text{Spec}(A)$ is a discrete set full of $\mathbb C$-points.

Therefore one would need at least two variables. Argubly, there is not an example simpler than $\mathbb R[x, y]/(x^2+y^2=-1)$, as if $x$ or $y$ has degree $1$ in $f(x,y)$, then $f(x,y)$ must have real roots by solving $x$ in terms of $y$ (or the other way around).