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Is an ideal generated by linear homogeneous polynomials a radical ideal?

I know that a linear variety in $\mathbb{P}^n$ is always isomorphic to an algebraic set of form $V_+(x_0,\ldots,x_r)$. (Can this be useful?)

user26857
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Bruno Rodrigues dos Santos
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  • Polynomials over what base ring? A field? If so is the field algebraically closed? Also note that if your polynomial ring is $R[x_1, \cdots, x_n]$, then your ideal is just $(x_1, \cdots, x_n)$. – Hamed Jun 22 '18 at 22:19
  • In general if the base ring $R$ has no nilpotent elements, then your ideal is radical. – Hamed Jun 22 '18 at 22:24
  • @Hamed: Are you saying that $(x^2)$ is radical in $\mathbb{C}[x]$? Because this is false. – RghtHndSd Jun 23 '18 at 01:59
  • @RghtHndSd, No I'm not saying that (by "your" ideal, I meant the ideal the OP is asking about). Concretely, I'm saying that: the ideal $(x_1, \cdots, x_n)$ (the ideal generated by all linear homogeneous polynomials) in $R[x_1, \cdots, x_n]$ is radical iff $R$ has no nilpotent elements. – Hamed Jun 24 '18 at 01:21
  • Over a algebraic closed field. Thanks for ask – Bruno Rodrigues dos Santos Jun 25 '18 at 13:42
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    Related: https://math.stackexchange.com/questions/2496792/ideals-in-polynomial-rings-generated-by-linear-polynomials – user26857 Jun 25 '18 at 17:34
  • $x^2$ is not linear – Bruno Rodrigues dos Santos Jun 27 '18 at 02:08

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Let $h_1,\ldots,h_r\in k[x_0,\ldots,x_n]$ be homogeneous linear polynomials. It's easy to see that we can write $I=(h_1,\ldots,h_r)=(f_1,\ldots,f_s)$ where the $f_i$'s are linear homogeneous polynomials such that if $x_i$ is a variable in $f_j$ so $x_i$ does not appear in $f_l$ for all $l\neq j$ (to see this just scale the matrix obtained from the coefficients of $h_1,\ldots,h_r$). Hence $$k[x_0,\ldots,x_n]/I$$ is a ring of polynomials in $k$, showing that $I$ is a prime ideal.

user26857
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Bruno Rodrigues dos Santos
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