Suppose $\mathbb{R}[X]$ is the normal multivariate polynomial ring where $X = x\_1...x\_n$. $\mathbb{R}[X]\_t$ is the truncated ring such that $\mathbb{R}[X]\_t =\left\{f: f \in \mathbb{R}[X], deg(f) \leq t \right \}$. 


Suppose $I\_t$ is a subspace of $\mathbb{R}[X]\_t$ that satisfies the "real radical-like" properties:


(1) If $f \in I\_t, g \in I\_t$, then $f + g \in I\_t$.


(2) If $f \in I\_t, g \in \mathbb{R}[X]\_t, deg(fg)\leq t$, then $fg \in I\_t.$ 


(3) If $\displaystyle \sum\_i p\_i^2 \in I\_t$, then $p\_i \in I\_t$


Now suppose $I = \langle I\_t \rangle$ ($I$ is the ideal generated by $I\_t$), my question is: Is $I$ still a real radical ideal?


In other words, I want to prove: $\displaystyle \sum\_i p\_i^2 \in I \Rightarrow p\_i \in I$ or find a counter example which I don't know how. 



ideal generated from a truncated "real radical-like" set are still real radical?
I think the answer is no. Let us take $t=3$ and $I\_3$ the $1$-dimensional subspace of $\mathbf R[X]\_3$ generated by $h=x(x^2+y^2)$. Then $I\_3$ satisfies the conditions 1-3. The ideal $I$ in $\mathbf R[X]$ generated by $I\_3$ contains the element
$$
x\cdot h=(x^2)^2+(xy)^2,
$$
but neither $x^2$ nor $xy$ belong to $I=\langle h\rangle$.


