What happens to a variety when we take radical of its defining ideal? Let $V$ be an affine variety in $K^n$ with ideal $I=I(V)$, where $K$ is an algebraically closed field. Let $V'$ be the variety with defining ideal $Radical(I)$. Usually $K[x_1,\ldots,x_n]/I$ and $K[x_1,\ldots,x_n]/Radical(I)$ have different Hilbert series. Does $V'$ consist of several components of $V$? Which part of $V$ is not in $V'$? Thank you very much.
 A: As varieties they are the same. Usually one considers varieties defined by radical ideals for the correspondence of Hilbert's Nullstellensatz.
Here you can see one reason why one wants to work with schemes. If for example $n = 1$, $I = (x^2)$ and hence  $\sqrt{I}= (x)$, you get the origin in $\mathbb{A}^1$ in both cases, but one of them is actually a bit "thicker" (which the classical varieties cannot really see at first). That is, because this origin has multiplicities coming from $x^2 = 0$. In scheme-theoretic language this is called non-reduced. This amounts to having nilpotency in your coordinate ring (or in your sheaf).
A: If $V$ is an affine variety then $I(V)$ is radical because $f(x)^n = 0$ implies $f(x) = 0$ over a field.
In particular (which is probably what you intended to ask), $I(V)$ is not the ideal generated by just any defining polynomials for $V$; in general you have to take the radical. For example, if $V$ is $x^2=0$ then $I(V) = \langle x \rangle$.
In this setting, $x^2=0$ and $x=0$ are the same as varieties. If you want to see a difference, consider them as schemes.
