Some subfield of ring $\mathbb{R}[x,y]/(x^2+y^2+1)$

As an $\mathbb R$-vector space we know that $\mathbb R[x,y]/(x^2+y^2+1)$ has basis $x^m,x^ny$ for $m,n\geq0$. Thus every element can be written uniquely as $a+by$ with $a,b\in\mathbb R[x]$.

Now compute $(a+by)^2+1=(a^2-b^2(1+x^2)+1)+2aby$. If this vanishes, then $ab=0$, so either $a=0$ or $b=0$. Since the first term also vanishes, we must have either $a^2+1=0$ or $b^2(1+x^2)=1$. These are equations in $\mathbb R[x]$, which have no solutions. Thus there is no element $x\in\mathbb R[x,y]/(x^2+y^2+1)$ satisfying $x^2+1=0$, and hence no homomorphism $\mathbb C\to\mathbb R[x,y]/(x^2+y^2+1)$.

Second try: Here’s a partial answer showing that there is no $ℝ$-algebra homomorphism $ℂ → ℝ[x,y]/(x^2 + y^2 + 1)$ at least. I hope this is correct …

Let $f = x^2 + y^2 + 1$ in $ℝ[x,y]$. Note that $f$ is prime in $ℝ[x,y]$ and in $ℂ[x,y]$ (as an Eisenstein polynomial, for example). If there was an $ℝ$-algebra homomorphism $ℂ → ℝ[x,y]/(f)$, then there would be a surjective $ℝ$-algebra homomorphism $$α\colon ℂ[x,y]/(f) → ℝ[x,y]/(f)$$ with $α(x) = x$ and $α(y) = y$ that is not injective:

• For $h ∈ ℝ[x,y]$ with $α(i) = [h]_{(f)}$, we have $\deg h ≠ 0$ (else $α(i)$ would be real), so $h - i$ would be nonzero with $h - i ∈ \ker α$.

Since $\dim ℂ[x,y] = 2$ and $\operatorname{height} (f) = 1$, the kernel of $α$ would be maximal and hence, by Hilbert’s Nullstellensatz, $α$ would factor as an isomorphism $ℂ → ℝ[x,y]/(f)$, but $(f)$ is not maximal in $ℝ[x,y]$.