I have the following problem I'm working on: Prove or disprove that $\mathbb{R}[x,y]/\langle{x^2+y^2}\rangle$ is a field.
I know that this amounts to asking whether or not the polynomial $x^2 + y^2$ is irreducible in $\mathbb{R}[x,y]$ – if it is irreducible, then the ideal generated by the polynomial $x^2 + y^2$ is maximal in $\mathbb{R}[x,y]$, yielding that the quotient ring $\mathbb{R}[x,y]/\langle{x^2+y^2}\rangle$ is a field, whereas if it is reducible, the quotient ring has zero divisors and is not even an integral domain ($\Rightarrow$ it cannot be a field).
To this point, suppose it is in fact reducible. Then it must be the product of two linear polynomials in the variables $x$ and $y$
$\Rightarrow$ $x^2 + y^2 = (ax + by + c)(dx + ey + f)$ , where $a,b,c,d,e,f \in \mathbb{R}$
$\Rightarrow$ $x^2 + y^2 = adx^2 + dexy + afx + bdxy + bey^2 + bfy + cdx + cey + cf$
$\Rightarrow$ $ad = 1, be = 1, ae+bd = 0, af+cd = 0, bf + ce = 0, cf=0$.
From the resulting equations, one can try and get a contradiction to $x^2 + y^2$ being a product of two linear polynomials in the variables $x$ and $y$, or try and get a solution for the real coefficients that will make $x^2 + y^2$ factor.
I tried to achieve the former. If $cf = 0$, then $c = 0$ or $f = 0$ (using that $\mathbb{R}$ is an integral domain). If $c = 0$ and $f \neq 0$, then $bf = 0$ $\Rightarrow$ $b = 0$, yielding a contradiction to $be = 1$. If $c \neq 0$ and $f = 0$, $ce = 0$ $\Rightarrow$ $e = 0$, yielding a contradiction to $be = 1$. However, in the case that both $c = 0$ and $f = 0$, I'm not able to reach such a contradiction. Is my logic so far sound? How can I refine my argument from here?
Thanks!
Best Answer
You don't have to show (ir)reducibility because we work over a polynomial ring in strictly more than one variable.
Observe that $$(X^2 + Y^2) \subsetneq (X,Y)$$
and hence your ideal is not maximal. It follows that the quotient ring is not a field.