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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.

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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 $I=\langle h\rangle$.

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    $\begingroup$ You are right. Thanks for pointing it out。 $\endgroup$ – Pew Feb 23 '16 at 4:53
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    $\begingroup$ Nice! I think I take this as saying that, while truncated ideals make sense and can be very useful, the notion of a truncated ideal being radical isn't really a good "truncated" version of an ideal being radical... $\endgroup$ – Joshua Grochow Feb 26 '16 at 4:38
  • $\begingroup$ What do you think is a good "truncated" version of an ideal being real radical? mathoverflow.net/q/239946 $\endgroup$ – Pew May 28 '16 at 18:01

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