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I (naively) believed that the Zariski tangent space was the tangent space of an algebraic manifold, i.e. of the manifold described by the solution of a polynomial set of equations.

However, I see some definitions in which it is declared that the Zariski space corresponds to the space of solutions of the linearized equations in a point. It is quite clear that such a space is not really the tangent space of the solution of the polynomial system. Take for example a system with two dimensions, $x$ and $y$, with equation: $$ x^2=0 $$ The solution is the manifold $M$ with $x=0$ and any value of $y$. At the point $P=(0,0)$, the solution of the linearized system is the whole 2d plane, that is not the tangent space of $M$.

So, my first question is: is the Zariski tangent space the tangent space of the algebraic manifold?

If not, how is it possible to calculate the true tangent space of the algebraic manifold?

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Algebraic geometry has learned over the years that it's beneficial to remember more about the solutions to a system of equations than just the underlying set - there are lots of fun/cool/useful technical tools which rely on this sort of thing and prove interesting and beautiful results. For instance, it makes sense to keep track of the difference between the solutions to the equation $x^2=0$ and the solutions to the equation $x=0$. (See for example here for recent discussion, and there's lots more where that came from.)

What you're running in to here is that the construction of the Zariski tangent space via linearization at a point will care about this extra structure, while your conception of the tangent space to an algebraic manifold doesn't care about it at all (i.e. you're only concerned with the underlying set). To bridge these gaps, if you want to adapt the Zariski tangent space to only care about the underlying set, given an algebraic manifold cut out by an ideal $I$, you would want to use the radical $\sqrt{I}$ as the defining ideal to work with the Zariski tangent space construction.

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  • $\begingroup$ Actually, in my question, I am only focusing on the set of solutions (although I understand that sometimes it is useful to know something more). I understand the suggestion of taking the radical of the ideal: it solves my specific example. Does it always work? If not, is there a generalization? $\endgroup$ Commented Nov 11, 2022 at 20:24
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    $\begingroup$ Yes, it always works. The radical of an ideal is the largest ideal with the same underlying set of solutions. $\endgroup$
    – KReiser
    Commented Nov 11, 2022 at 20:39

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