Let $A = \mathbb R[x,y]$ where $x^2 + y^2 = 1$
Show that $A$ is an integral domain.
My thoughts are to show that $x^2 + y^2 = 1$ is irreducible over $\mathbb R[x,y] = \mathbb R[x][y],$ I am guessing that I can use Einstein criterion, is that correct? if so how?
Could someone push me in the right direction please?
EDIT:
I also know that:
1-Irreducibles are primes in a UFD.
2-$\mathbb R[x], \mathbb R[y]$ are UFD because $R$ is a field. And $<x^2 + y^2 -1>$ is a prime ideal in $B[y]$(where B = R[x] and where $B[y]$ is a UFD) iff $<x^2 + y^2 -1>$ is irreducible in $B[y]$
3- $I$ is prime iff $R/I$ is an integral domain.
I may as well make this an answer. The ring $A$ is simply $\mathbb R[X,Y]/(X^2+Y^2-1)$ which is an integral domain because the ideal $(X^2+Y^2-1)$ is prime. It's prime because $\mathbb R[X,Y]$ is an integral domain, so an ideal $(f(X,Y))$ is prime iff $f(X,Y)$ is irreducible. Indeed, $X^2+Y^2-1 = X^2+(Y+1)(Y-1)$ which is Eisenstein at the primes $Y\pm 1$.
A very similar argument shows that the ring $\mathbb R[X,Y]/(X^n+Y^n-1)$ is an integral domain. If you've seen any algebraic geometry, this is the same as the fact that the Fermat curve $V(X^n+Y^n-1)$ is irreducible, i.e. an algebraic variety. This time, write $X^n+(Y-1)(1+Y+\cdots+Y^{n-1})$ which is still Eisenstein at $Y-1$.