Timeline for To prove the Nullstellensatz, how can the general case of an arbitrary algebraically closed field be reduced to the easily-proved case of an uncountable algebraically closed field?

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Nov 14, 2017 at 16:42	comment	added	ACL		@users: $K$ has the power of the continuum. Fix a surjective map $\phi\colon k\to\mathbf Z$. For every $(x_n)\in A$, the sequence $\phi(x_n)$ of elements in~$\mathbf Z$ has a limit in~$\widehat{\mathbf Z}$ (in a compact space, every ultrafilter converges). By density of $\mathbf Z$ in $\widehat{\mathbf Z}$ (and surjectivity of $\phi$), this map is surjective. This induces a surjective map $\widehat\phi\colon K\to\widehat{\mathbf Z}$, hence $K$ has at least the cardinality of $\widehat{\mathbf Z}$, which is the continuum.
Nov 14, 2017 at 16:39	comment	added	ACL		@users: $K=A/m$ is algebraically closed. Take a non-constant polynomial $P\in K[T]$, and lift it to a family $(P_n)$ where $P_n\in k[T]$. For almost every $n$, $P_n$ is not constant, hence there exists $x_n\in k$ such that $P(x_n)=0$; for the other $n$, set $x_n=0$. Then the class of $x=(x_n)$ satisfies $P(x)=0$.
Aug 15, 2017 at 6:30	comment	added	user111524		@ACL , Martin Brandenburg : why is $A/m$ algebraically closed and has at least the power of the continuum ?
Jan 6, 2013 at 18:14	comment	added	Martin Brandenburg		I like this proof, it is completely elementary (and remark that also Brian's proof contains a choice of a maximal ideal).
Jan 6, 2013 at 18:08	history	edited	Martin Brandenburg	CC BY-SA 3.0	added 7 characters in body
Jan 6, 2013 at 16:34	history	answered	ACL	CC BY-SA 3.0