Integral extension implies surjective map on spectra - Revision history

Vipul: 7 revisions

2008-05-12T16:23:54Z

7 revisions

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Vipul at 20:51, 9 March 2008

2008-03-09T20:51:30Z

← Older revision		Revision as of 20:51, 9 March 2008
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	{{ring map induces on spectrum fact}}		{{ring map induces on spectrum fact}}

			==Name==

			This result is sometimes termed '''lying over''', and is a precursor to the [[going up theorem]].
	==Statement==		==Statement==

Vipul at 02:05, 9 March 2008

2008-03-09T02:05:33Z

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			{{ring map induces on spectrum fact}}

	==Statement==		==Statement==

Vipul at 17:38, 3 March 2008

2008-03-03T17:38:39Z

← Older revision		Revision as of 17:38, 3 March 2008
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	==Statement==		==Statement==

	Suppose <math>S</math> is an [[integral extension]] of a ring <math>R</math>, in other words <math>f:R \to S</math> is an injective homomorphism of commutative unital rings with the property that every element of <math>S</math> is integral over the image of <math>R</math>. Then, the map:		Suppose <math>S</math> is an [[integral extension]] of a ring <math>R</math>, in other words <math>f:R \to S</math> is an injective homomorphism of commutative unital rings with the property that every element of <math>S</math> is integral over the image of <math>R</math>. Then, the following are true.

			The map:

	<math>f^*: Spec(S) \to Spec(R)</math>		<math>f^*: Spec(S) \to Spec(R)</math>

	from the [[spectrum]] of <math>S</math> to that of <math>R</math>, that sends a prime ideal of <math>S</math> to its [[contraction]] in <math>R</math>, is surjective. In other words, every prime ideal of <math>R</math> occurs as the contraction of a prime ideal of <math>S</math>.		from the [[spectrum]] of <math>S</math> to that of <math>R</math>, that sends a prime ideal of <math>S</math> to its [[contraction]] in <math>R</math>, is '''surjective'''. In other words, every prime ideal of <math>R</math> occurs as the contraction of a prime ideal of <math>S</math>.

	This result is sometimes termed the '''lying over theorem'''.		This result is sometimes termed the '''lying over theorem'''.
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	==Related facts==		==Related facts==

			* [[Integral extension implies inverse image of max-spectrum is max-spectrum]]
	* [[Going up theorem]]: The going up theorem is a somewhat stronger version of this result, and also follows as a corollary of this result.		* [[Going up theorem]]: The going up theorem is a somewhat stronger version of this result, and also follows as a corollary of this result.
	==Proof==		==Proof==

Vipul at 20:12, 2 February 2008

2008-02-02T20:12:53Z

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	This result is sometimes termed the '''lying over theorem'''.		This result is sometimes termed the '''lying over theorem'''.

			Note that injectivity of <math>f</math> is crucial for surjectivity of the map on spectra; this is analogous to the fact that surjective ring homomorphisms induce injective maps on spectra.

	==Related facts==		==Related facts==

Vipul at 20:04, 2 February 2008

2008-02-02T20:04:33Z

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	from the [[spectrum]] of <math>S</math> to that of <math>R</math>, that sends a prime ideal of <math>S</math> to its [[contraction]] in <math>R</math>, is surjective. In other words, every prime ideal of <math>R</math> occurs as the contraction of a prime ideal of <math>S</math>.		from the [[spectrum]] of <math>S</math> to that of <math>R</math>, that sends a prime ideal of <math>S</math> to its [[contraction]] in <math>R</math>, is surjective. In other words, every prime ideal of <math>R</math> occurs as the contraction of a prime ideal of <math>S</math>.

			This result is sometimes termed the '''lying over theorem'''.

			==Related facts==

			* [[Going up theorem]]: The going up theorem is a somewhat stronger version of this result, and also follows as a corollary of this result.
	==Proof==		==Proof==

Vipul: /* Proof */

2008-02-02T20:01:50Z

Proof

← Older revision		Revision as of 20:01, 2 February 2008
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	==Proof==		==Proof==

	{{~~fillin~~}}		The goal is to prove that starting with a prime ideal <math>P</math> of <math>R</math>, we can find a prime ideal <math>Q</math> of <math>S</math> such that <math>f^{-1}(Q) = P</math>.

			We localize <math>R</math> at <math>P</math>, and localize <math>S</math> at the image of <math>U = R \setminus P</math> to get <math>S'</math>. Then <math>R_P</math> is a local ring with unique maximal ideal <math>P' = PR_P</math>, and <math>f</math> induces a map <math>R_P \to S' = S[U^{-1}]</math>.

			We thus have an inclusion <math>f:R_P \to S'</math>. Consider the image <math>P'S'</math>. This is an ideal of <math>S'</math>. If <math>P'S'</math> is a proper ideal, it is contained in some maximal ideal <math>M</math>, and the contraction of that maximal ideal to <math>R_P</math> is precisely <math>P'</math>. Contracting back along the localization, we find a prime ideal of <math>S'</math>, whose contraction is exactly <math>P</math>. (we are using the fact that contracting a maximal ideal of <math>S'</math> yields a prime, though not necessarily maximal, ideal of <math>S</math>).

			Thus, the main goal is to show that <math>P'S' \ne S'</math> (this is where we need to use integrality). The idea is to construct a <math>R</math>-subalgebra of <math>S'</math>, called <math>S''</math>, that is ''finite'' over <math>R_P</math>, and use [[Nakayama's lemma]] to derive a contradiction. Here are the steps:

			* Since <math>S</math> is integral over <math>R</math>, <math>S'</math> is integral over <math>R_P</math>
			* If <math>P'S' = S'</math>, then the element <math>1 \in S'</math> can be written as a <math>P'</math>-linear combination of finitely many elements from <math>S'</math>
			* Let <math>S''</math> be the <math>R_P</math>-subalgebra generated by these finitely many elements. Then <math>S''</math> is finitely generated and integral over <math>R_P</math>, hence it is finitely generated as a module over <math>R_P</math>. {{proofat\|[[finitely generated and integral implies finite]]}}
			* We thus have <math>P'S'' = S''</math> (since <math>1 \in P'S''</math>). Since <math>P'</math> is the [[Jacobson radical]] of <math>R_P</math>, Nakayama's lemma tells us that <math>S'' = 0</math>, yielding a contradiction.

Vipul: New page: ==Statement== Suppose $S$ is an integral extension of a ring $R$ , in other words $f:R \to S$ is an injective homomorphism of commutative unital rings ...

2008-02-02T19:35:06Z

New page: ==Statement== Suppose <math>S</math> is an integral extension of a ring <math>R</math>, in other words <math>f:R \to S</math> is an injective homomorphism of commutative unital rings ...

New page

==Statement==

Suppose <math>S</math> is an [[integral extension]] of a ring <math>R</math>, in other words <math>f:R \to S</math> is an injective homomorphism of commutative unital rings with the property that every element of <math>S</math> is integral over the image of <math>R</math>. Then, the map:

<math>f^*: Spec(S) \to Spec(R)</math>

from the [[spectrum]] of <math>S</math> to that of <math>R</math>, that sends a prime ideal of <math>S</math> to its [[contraction]] in <math>R</math>, is surjective. In other words, every prime ideal of <math>R</math> occurs as the contraction of a prime ideal of <math>S</math>.

==Proof==

{{fillin}}