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In algebraic geometry, if $X$ is an affine variety than the tangent space $T_pX$ of $X$ at its point $p$ is the affine variety defined as the set of zeros of the ideal $J$ generated by the linear parts $df$ of the polynomials $f$ belonging to $I(X)$ (the ideal of $X$).

If $I(X)$ is generated by $f_1,\dotsc,f_m$, that is $I(X)=(f_1,f_2,\dotsc,f_m)$, then $J=(df_1,\dotsc,df_m)$. It's very common the notation $J=I(T_pX)$ which suggests that $J$ is the ideal of $T_pX$. But we know that the ideal of an affine variety is radical. My question is: how do we know that $J$ is radical?

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    $\begingroup$ On some level this is linear algebra: you can throw away $f_i$ until you get a linearly independent set inside the vector space of linear forms, then change coordinates so it's just $x_1, \dots, x_r$. Clearly that ideal is prime. $\endgroup$ – Hoot Jul 10 '16 at 22:10
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$J$ is generated by linear polynomials say $\{ l_i \}$ so $$ J = \bigcap_i \, (l_i). $$ The intersection of radical ideals is again radical so it suffices to show the ideals $(l_i)$ are radical. But $l_i$ is linear so must be irreducible so in fact $(l_i)$ is a prime ideal which is also radical.

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