Derive the weak Nullstellensatz from the strong one

Question

I'd like to derive the weak Nullstellensatz

An ideal $J\subset K[x_1,\dots,x_n]$ has a common zero exactly if it is a proper ideal.

from the strong one

$\sqrt{J} = I(V(J))$

This seems pretty easy:

\begin{align} J \text{ has no common zero} & \Longleftrightarrow V(J) \text{ is empty } \\ & \Longleftrightarrow 1\in I(V(J)) = \sqrt{J} \\ & \Longleftrightarrow \sqrt{J} = K[x_1,\dots,x_n] \\ & \overset{(*)}{\Longleftarrow} J=K[x_1,\dots,x_n] \end{align}

The missing part is $(*)$. Obviously $J\subset \sqrt{J}$ for all ideals.

But why does $\sqrt{J} = K[x_1,\dots,x_n]$ imply $J = K[x_1,\dots,x_n]$?

If $1\in rad\ J $ then $1\in J $. Or not? – Blumer Jul 25 '17 at 11:51 — Blumer, Jul 25 '17 at 11:51
@Blumer aaww of course. Thx. – Andy Miles Jul 25 '17 at 11:58 — Andy Miles, Jul 25 '17 at 11:58

score 4 · Accepted Answer · answered Jul 25 '17 at 11:58

4

If $1\in rad\ J =\{x\in \ |\ \exists n\in \mathbb N :\ x^n\in J\}$. Thus $\exists n\in \mathbb N$ such that $1=1^n\in J$.

answered Jul 25 '17 at 11:58

Blumer

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Derive the weak Nullstellensatz from the strong one

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