# Is $\mathbb{C}[x,y]/\langle x^2+y^2\rangle$ a field?

I'm interested in knowing whether or not $$\mathbb{C}[x,y]/\langle x^2+y^2\rangle$$ is a field, where $$\langle x^2+y^2\rangle$$ denotes the ideal generated by the polynomial $$x^2+y^2\in\mathbb{C}[x,y]$$ and $$\mathbb{C}$$ denotes the field of complex numbers.

I know the following:

1) For $$R$$ a commutative ring and $$I$$ an ideal of $$R$$, $$R/I$$ is a field if and only if $$I$$ is maximal.

2) For $$R$$ a principal ideal domain, the ideal $$I$$ of $$R$$ is maximal if and only if $$I$$ is generated by an irreducible element.

Putting these together, since $$x^2+y^2$$ is not irreducible in $$\mathbb{C}[x,y]$$ (as $$x^2+y^2=(x-iy)(x+iy)$$), one would think that the ideal $$\langle x^2+y^2\rangle$$ is not maximal in $$\mathbb{C}[x,y]$$ by 2), and thus, by 1), $$\mathbb{C}[x,y]/\langle x^2+y^2\rangle$$ is not a field.

However, this does not hold, because $$\mathbb{C}[x,y]$$ is not a principal ideal domain — in fact, for any commutative ring $$R$$ with $$1$$, any polynomial ring in more than one variable over $$R$$ is not a P.I.D.

Is there a way to refine my logic? I suspect the ring in question is not a field.

Thanks!

~Mo

Hint: $$x^2 + y^2 = (x-iy)(x+iy)$$ implies that there are zero divisors in the quotient ring.
• Hi Ihf! So, in the quotient ring of interest, both the elements $(x-iy)$ and $(x+iy)$ are zero divisors, correct? Thus, the quotient ring of interest is not even an integral domain, and thus cannot be a field? – Mo Behzad Kang Oct 15 '19 at 20:16
$$R/I$$ is a field $$\iff$$ $$I$$ is maximal.
So $$R/I$$ is not a field $$\iff$$ $$I$$ is not maximal.
As in the answer above, $$I=\left$$ is not maximal since there is a proper ideal of $$\mathbb{C}[x,y]$$, $$\left$$ that properly contains $$\left$$, that is, $$\left\subset\left$$ but $$\left\neq\left$$.