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

Question

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:

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