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.




2 Answers 2


Hint: $x^2 + y^2 = (x-iy)(x+iy)$ implies that there are zero divisors in the quotient ring.

  • $\begingroup$ 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? $\endgroup$ Oct 15, 2019 at 20:16
  • 4
    $\begingroup$ @MoBehzadKang: Yes, precisely. $\endgroup$ Oct 15, 2019 at 20:22

$R/I$ is a field $\iff$ $I$ is maximal.

So $R/I$ is not a field $\iff$ $I$ is not maximal.

As in the answer above, $I=\left<x^2+y^2\right>$ is not maximal since there is a proper ideal of $\mathbb{C}[x,y]$, $\left<x+iy\right>$ that properly contains $\left<x^2+y^2\right>$, that is, $\left<x^2+y^2\right>\subset\left<x+iy\right>$ but $\left<x^2+y^2\right>\neq\left<x+iy\right>$.


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