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2008-05-12T18:05:33Z

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Vipul at 22:13, 19 January 2008

2008-01-19T22:13:59Z

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==Statement==

Suppose <math>k</math> is an [[algebraically closed field]]. Then, the following equivalent statements hold true:

* Any field <math>K</math>, which is finitely generated as a <math>k</math>-algebra, must be <math>k</math> itself (i.e. isomorphic to <math>k</math> as a <math>k</math>-algebra)
* For any [[maximal ideal]] of the polynomial ring in finitely many variables over <math>k</math>, the quotient field is <math>k</math>
* The maximal ideals in the polynomial ring <math>k[x_1,x_2,\ldots,x_n]</math> are in bijection with the points in <math>k^n</math>, where the maximal ideal corresponding to a point <math>(a_1,a_2,\ldots,a_n)</math> is the ideal <math>(x_1 - a_1, x_2 - a_2, \ldots, x_n - a_n)</math>
* The [[max-spectrum of a commutative unital ring|max-spectrum]] of <math>k[x_1,x_2,\ldots,x_n]</math>, with the max-spec topology, is homeomorphic to <math>k^n</math> with the [[Zariski topology]], where the bijection is as described above
* Any proper ideal of <math>k[x_1,x_2,\ldots,x_n]</math> has a nonempty vanishing set in <math>k^n</math>
* If a system of polynomial equations over <math>k</math> (in <math>n</math> variables) is ''consistent'' (i.e. one cannot derive the equation <math>0 = 1</math> by manipulating them) then the system has a solution in <math>k^n</math>

Weak nullstellensatz for algebraically closed fields - Revision history

Vipul: 1 revision

Vipul at 22:13, 19 January 2008