Consider $\mathbb{R}^2$ with the polynomial $P(x,w):=w^2+xw+x$. The zero set $N\left(P\right)$ of $P$ is a real analytic submanifold of $\mathbb{R}^2$. That is, the stalks of $\mathcal{A}_D/J_{N(P)}$ are regular analytic $\mathbb{R}$-algebras, here $\mathcal{A}_D$ denotes the sheaf of real analytic functions and $J_{N(P)}$ the ideal of all functions vanishing on $N(P)$.
My question is: Is $J_{N(P)}= (P)?$
I.e. is the analytic space defined by the ideal $(P)$ generated by $P$ smooth or is it non reduced?
I know that in real algebraic geometry the ideal $J_{N(P)}$ would be the real radical of $(P)$, but I do not know of a similar result in the analytic category.
I have tried to find elements of $\mathcal{A}_{D}/(P)$ that are nilpotent but with no success. How does one approach this in the real analytic category?
The answer is, no. After a change of coordinates around $0$, the polynomial may be assumed to be $P(x,w)=w^2+x$, as $w+1$ is invertible and positive around zero. Then by the Weierstraß division theorem one has
$\mathcal{A}_{D,0}/ (P)_0\cong \mathcal{A}_{D',0}^2.$
Here $D'$ is a domain in $\mathbb{R}$ containing $0$. Thus, the real analytic space defined by $P$ is not a real analytic manifold at $0$.