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$ S_m/m S_m $ contains just one prime ideal

Questions on scheme morphismsQuestion about the proof of the going-up theorem of Cohen-Seidenberg.Spectrum of a ring is irreducible if and only if nilradical is prime (Atiyah-Macdonald, Exercise 1.19)$mathbbZ_(2)$ has one maximal idealQuestion on the existence of a prime ideal contained in the $ker$ of a homomorphism $mathbbC[x,y]rightarrowmathbbC[t]$.completion and heights of prime idealsFiber over a prime idealJacobson radical = nilradical iff every open set of $textSpecA$ contains a closed point.If $x$ is a prime ideal, then $overlinex=V(p_x)$, where $V(x)$ is the set of prime ideals that contains $x$.Reducing an approximation claim from a prime ideal to a maximal ideal

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I'm trying to read these lecture notes on my own for self studying. At the beginning of the second lecture, page 8, the author wants to deduce the geometric form of the Zariski's main theorem from the algebraic form. The setting is as follows: $ g: X to Y $ is map of irreducible affine varieties that is bijective on closed points such that $ K(X)/K(Y)$ is separable and $ Y $ is normal, then $ g $ is an isomorphism.

He let $ R = K[Y] subset K[X] = S $. We are to show that $ R = S $. If $ R neq S $ then $ S/R neq 0 $ as an $ R $-module, so there exists a maximal ideal $ m $ such that $ (S/R)_m neq 0 $, hence $ R_m neq S_m $. After some time, he says that the fiber $ S_m/mS_m $ contains just one prime ideal, so it is isolated in its fiber (We need this for the hypothesis of the algebraic form of ZMT).

How do we know that $ S_m/mS_m $ contains just one prime ideal, since the hypothesis is only that the map is bijective on closed points?

EDIT: I think the phrase "bijective on closed points" perhaps means that the map $ textSpec ,S to textSpec , R $ restricted to $ textSpec_m , S = X $ bijects onto $ textSpec_m , R = Y $, that is, there is a unique prime ideal of $ S $ that also happens to be maximal lying over any maximal ideal of $ R $. But I'm not sure if this interpretation is correct or not.

commutative-algebra

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edited Mar 15 at 22:36

calm

asked Mar 15 at 18:35

calmcalm

1387

$endgroup$

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  1
  

  
  
  
  
  
  

  
  $begingroup$
  You are correct in your interpretation. However, one can also note that the fiber over a closed point is closed (since closed immersions are stable under base change), so if the fiber contained a non-closed point, it would have to contain infinitely many closed points (since the closure of a non-closed point in an affine variety contains infinitely many closed points).
  $endgroup$
  – jgon
  Mar 16 at 0:23
  
  

  
  
  
  

add a comment | 

0

$begingroup$

I'm trying to read these lecture notes on my own for self studying. At the beginning of the second lecture, page 8, the author wants to deduce the geometric form of the Zariski's main theorem from the algebraic form. The setting is as follows: $ g: X to Y $ is map of irreducible affine varieties that is bijective on closed points such that $ K(X)/K(Y)$ is separable and $ Y $ is normal, then $ g $ is an isomorphism.

He let $ R = K[Y] subset K[X] = S $. We are to show that $ R = S $. If $ R neq S $ then $ S/R neq 0 $ as an $ R $-module, so there exists a maximal ideal $ m $ such that $ (S/R)_m neq 0 $, hence $ R_m neq S_m $. After some time, he says that the fiber $ S_m/mS_m $ contains just one prime ideal, so it is isolated in its fiber (We need this for the hypothesis of the algebraic form of ZMT).

How do we know that $ S_m/mS_m $ contains just one prime ideal, since the hypothesis is only that the map is bijective on closed points?

EDIT: I think the phrase "bijective on closed points" perhaps means that the map $ textSpec ,S to textSpec , R $ restricted to $ textSpec_m , S = X $ bijects onto $ textSpec_m , R = Y $, that is, there is a unique prime ideal of $ S $ that also happens to be maximal lying over any maximal ideal of $ R $. But I'm not sure if this interpretation is correct or not.

commutative-algebra

share|cite|improve this question

edited Mar 15 at 22:36

calm

asked Mar 15 at 18:35

calmcalm

1387

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You are correct in your interpretation. However, one can also note that the fiber over a closed point is closed (since closed immersions are stable under base change), so if the fiber contained a non-closed point, it would have to contain infinitely many closed points (since the closure of a non-closed point in an affine variety contains infinitely many closed points).
  $endgroup$
  – jgon
  Mar 16 at 0:23
  
  

  
  
  
  

add a comment | 

$begingroup$

I'm trying to read these lecture notes on my own for self studying. At the beginning of the second lecture, page 8, the author wants to deduce the geometric form of the Zariski's main theorem from the algebraic form. The setting is as follows: $ g: X to Y $ is map of irreducible affine varieties that is bijective on closed points such that $ K(X)/K(Y)$ is separable and $ Y $ is normal, then $ g $ is an isomorphism.

He let $ R = K[Y] subset K[X] = S $. We are to show that $ R = S $. If $ R neq S $ then $ S/R neq 0 $ as an $ R $-module, so there exists a maximal ideal $ m $ such that $ (S/R)_m neq 0 $, hence $ R_m neq S_m $. After some time, he says that the fiber $ S_m/mS_m $ contains just one prime ideal, so it is isolated in its fiber (We need this for the hypothesis of the algebraic form of ZMT).

How do we know that $ S_m/mS_m $ contains just one prime ideal, since the hypothesis is only that the map is bijective on closed points?

EDIT: I think the phrase "bijective on closed points" perhaps means that the map $ textSpec ,S to textSpec , R $ restricted to $ textSpec_m , S = X $ bijects onto $ textSpec_m , R = Y $, that is, there is a unique prime ideal of $ S $ that also happens to be maximal lying over any maximal ideal of $ R $. But I'm not sure if this interpretation is correct or not.

commutative-algebra

share|cite|improve this question

edited Mar 15 at 22:36

calm

asked Mar 15 at 18:35

calmcalm

1387

$endgroup$

share|cite|improve this question

edited Mar 15 at 22:36

calm

asked Mar 15 at 18:35

calmcalm

1387

share|cite|improve this question

edited Mar 15 at 22:36

calm

asked Mar 15 at 18:35

calmcalm

1387

asked Mar 15 at 18:35

calmcalm

1387

asked Mar 15 at 18:35

calmcalm

1387