# Polynomial Ring Quotient of $\mathbb{R}[x,y]$ by $x^2 + y^2 -1$

I want to find a simpler ring that is isomorphic to this polynomial ring quotient: $$\mathbb{R}[x,y]/(x^2 + y^2 -1)$$.

I've tried to use the isomorphism theorem by defining $$\phi$$ a ring homomorphism so that I can show the quotient ring is isomorphic to $$Im(\phi)/\ker(\phi)$$. But I'm stuck and not sure if this is the right approach.

$$\phi: \mathbb{R}[x,y] \longrightarrow \mathbb{R}[x]\diagup(x^2-1) \,\times\,\mathbb{R}[y]\diagup(y^2 -1) \,\,\,\,$$ the ring homomorphism

$$\phi\big(p(x,y)\big) := \big(p(x,0),p(0,y)\big) \in \mathbb{R}[x]\times\mathbb{R}[y]\,\,\,\,$$ with each entry mod its respective equation

where

• $$\phi(0)=(0,0)$$
• $$\phi(1)=(1,1)$$
• $$\phi\big(p(x,y)q(x,y)\big)=\big(p(x,0)q(x,0),p(0,y)q(0,y)\big)=\big(p(x,0),p(0,y)\big)\big(q(x,0),q(0,y)\big)=\phi\big(p(x,y)\big)\phi\big(q(x,y)\big)$$
• $$\phi\big(p(x,y)+q(x,y)\big)=\big(p(x,0)+q(x,0),p(0,y)+q(0,y)\big)=\big(p(x,0),p(0,y)\big)+\big(q(x,0),q(0,y)\big)=\phi\big(p(x,y)\big)+\phi\big(q(x,y)\big)$$

Now if $$p(x,y)\in \ker\phi$$ then

• $$p(x,0) \mod (x^2 -1) = 0 \,\,\,\Longrightarrow \,\,\,p(\pm1,0)=0$$
• $$p(0,y) \mod (y^2 -1) = 0 \,\,\,\Longrightarrow \,\,\,p(0,\pm1)=0$$

Would we then have:

$$\mathbb{R}[x,y]\diagup\ker(\phi) \cong \mathbb{R}\times\mathbb{R}\times\mathbb{R}\times\mathbb{R}$$

• The polynomial $xy$ is in the kernel but not in the ideal generated by $x^2+y^2-1$.
– user497429
Commented Apr 23, 2020 at 7:56
• The quotient ring is an integral domain, see this post. Note that $\Bbb R\times \cdots \times \Bbb R$ has zero divisors. Commented Apr 23, 2020 at 10:38
• I wonder what you mean by computing a quotient ring? Anyway, not every quotient ring is isomorphic to some rings you previously know, and this is the case here. Commented Apr 23, 2020 at 10:38
• @user26857 I want to find a simpler ring that’s isomorphic and easier to work with. Commented Apr 23, 2020 at 20:59
• Bad luck! There is no such "simpler" ring. Commented Apr 23, 2020 at 21:00