Irreducible element in $k[X,Y]/(X^2+Y^2)$ Let $k$ be a field and $I=(X^2+Y^2)$ the ideal generated by $X^2+Y^2$ in the polynomial ring $k[X,Y]$. Denote by $\bar{X}$ the natural image of $X$ in the quotient ring $R=k[X,Y]/I$.
Is $\bar{X}$ an irreducible element in $R$ if $k=\mathbb{R}$? What if $k=\mathbb{C}$?
How should I approach this problem? My initial attempt was to write $\bar{X} = \bar{f}\bar{g}$ in $R$ (which means $X-fg = h(X^2+Y^2)$ in $k[X,Y]$) and try to deduce that $\bar{f}$ or $\bar{g}$ is a unit in $R$ but I did not make progress. Any hints and general advice on solving this type of a problem are appreciated.

When $k=\mathbb R$ the ring $R=\mathbb R[X,Y]/(X^2+Y^2)$ is an integral domain. We show that $x=X\bmod (X^2+Y^2)$ is irreducible. Suppose that $x=f(x,y)g(x,y)$ in $R$. Now define a norm function $N:R\to k[X]$ by $N(h)=a_0^2(X)+X^2a_1^2(X)$, where $a_0(X)+Ya_1(X)$ is the remainder of $h(X,Y)$ on division by $Y^2+X^2$. It's not hard to show that $N(h_1h_2)=N(h_1)N(h_2)$. It follows that the invertible elements of $R$ are only those from $\mathbb R^\times$. Moreover, from $x=f(x,y)g(x,y)$ we get $N(x)=N(f)N(g)$, that is, $X^2=[a_0^2(X)+X^2a_1^2(X)][b_0^2(X)+X^2b_1^2(X)]$. We get $a_0^2(X)+X^2a_1^2(X)=b_0^2(X)+X^2b_1^2(X)=X$ (which is not possible since $a_0(0)=0$ and $b_0(0)=0$), or $a_0^2(X)+X^2a_1^2(X)=X^2$ and $b_0^2(X)+X^2b_1^2(X)=1$ (or vice versa). In the last case $\deg b_0=0$ and $b_1=0$, so $g(x,y)\in\mathbb R^\times$. 
(Thought not explicit, I've used that $u^2+v^2=0$ iff $u=v=0$, where $u,v\in\mathbb R$.)
When $k=\mathbb C$ the ring $R=\mathbb C[X,Y]/(X^2+Y^2)$ is not an integral domain, and I don't know what irreducible means.
Edit. One can show that $a_0^2(X)+X^2a_1^2(X)=1$ in $\mathbb C[X]$ iff $a_1=0$ and $a_0=\pm1$.
From $a_0^2(X)+X^2a_1^2(X)=1$ we get $[a_0(X)+iXa_1(X)][a_0(X)-iXa_1(X)]=1$, so $a_0(X)+iXa_1(X)=c$ and $a_0(X)-iXa_1(X)=c^{-1}$ for some $c\in\mathbb C^\times$. Subtracting the two equations we get $2iXa_1(X)=c-c^{-1}$, so $a_1=0$ and $a_0=c=\pm1$. This shows that the invertible elements of $R$ are only those from $\mathbb C^\times$. As before we can show that $x$ can't be written as a product of two non-invertible elements.
