I am reading Theorem 2.26 from this lecture note that I randomly found from Google search. The last part of the proof makes me feel I am missing something.
I can understand the proof of (1) and $(\Rightarrow )$ part of (2), but it goes through the $(\Leftarrow )$ part with lots of detail missing (at least, I think).
Here is how much I can understand:
Suppose $I$ is not radical so that $I\subset \sqrt{I}$ (Here I use $\subset$ for STRICT containment, just because it makes much more sense). The $(\Rightarrow)$ part of (2) implies that $$|V(I)|=|V(\sqrt{I})|=\dim_k k[x_1,\cdots ,x_n]/\sqrt{I}$$ Then it suffices to show that $$\dim_k k[x_1,\cdots ,x_n]/\sqrt{I} < \dim_k k[x_1,\cdots ,x_n]/I$$ and the proof is done by taking contrapositive.
Since a vector space dimension is defined to be the number of elements in its basis, what I am thinking is somehow to show that a basis of $k[x_1,\cdots ,x_n]/\sqrt{I}$ is smaller than a basis of $k[x_1,\cdots ,x_n]/I$. However, I don't see any way to achieve this from the assumption that $I\subset \sqrt{I}$.
Any assistance is appreciated.