The proof of the weak Nullstellensatz in Qing Liu’s bookProof of Hilbert's NullstellensatzA proof of the Noether Normalization LemmaI need help in this proof in Lang's algebra bookQuotient of ring is flat gives an identity of idealsTheorem 1 in chapter II.4 of Mumford's Red BookAtiyah-MacDonald: proof of Proposition 7.9, weak Nullstellensatz.Lemma 4.1.7. in Bruns and Herzog, Cohen-Macaulay Ringsa proof in Matsumura's bookweak Bezout's theorem proofHilbert Nullstellensatz, Eisenbud's proof
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The proof of the weak Nullstellensatz in Qing Liu’s book
Proof of Hilbert's NullstellensatzA proof of the Noether Normalization LemmaI need help in this proof in Lang's algebra bookQuotient of ring is flat gives an identity of idealsTheorem 1 in chapter II.4 of Mumford's Red BookAtiyah-MacDonald: proof of Proposition 7.9, weak Nullstellensatz.Lemma 4.1.7. in Bruns and Herzog, Cohen-Macaulay Ringsa proof in Matsumura's bookweak Bezout's theorem proofHilbert Nullstellensatz, Eisenbud's proof
$begingroup$
I am reading Qing Liu’s book and the whole proof on pp 30 goes like this:
I do not understand the conclusion in the yellow-highlighted line. Let $f$ be the isomorphism highlighted in red. Why $f$ maps $alpha_i$ to $alpha_i$?
commutative-algebra
asked 2 days ago
user150248user150248
346110
$endgroup$
add a comment |
$begingroup$
I am reading Qing Liu’s book and the whole proof on pp 30 goes like this:
I do not understand the conclusion in the yellow-highlighted line. Let $f$ be the isomorphism highlighted in red. Why $f$ maps $alpha_i$ to $alpha_i$?
commutative-algebra
asked 2 days ago
user150248user150248
346110
$endgroup$
add a comment |
$begingroup$
I am reading Qing Liu’s book and the whole proof on pp 30 goes like this:
I do not understand the conclusion in the yellow-highlighted line. Let $f$ be the isomorphism highlighted in red. Why $f$ maps $alpha_i$ to $alpha_i$?
commutative-algebra
asked 2 days ago
user150248user150248
346110
$endgroup$
I am reading Qing Liu’s book and the whole proof on pp 30 goes like this:
I do not understand the conclusion in the yellow-highlighted line. Let $f$ be the isomorphism highlighted in red. Why $f$ maps $alpha_i$ to $alpha_i$?
commutative-algebra
commutative-algebra
asked 2 days ago
user150248user150248
346110
asked 2 days ago
user150248user150248
346110
edited 2 days ago
user150248
asked 2 days ago
user150248user150248
346110
asked 2 days ago
user150248user150248
346110
asked 2 days ago
user150248user150248
346110
346110
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
These corollaries (explicitly) talk about $k$-algebras, so every morphism is $k$-linear. This implies that $f(alpha_i)=alpha_i+mathfrakm$.
In the category of $k$-algebras $k$ is an initial object, so for any $k$-algebra $A$ there is exactly one $k$-algebra homomorphism $gcolon kto A$. This must be a ring morphism, so $g(1)=1_A$. And it must be $k$-linear, so $g(alpha)=alpha cdot 1_A$.
answered 2 days ago
PedroPedro
2,9291720
$endgroup$
$begingroup$
It may not be $k$-linear in the desired way. Perhaps $f^-1$ maps all the scalar elements to a proper subfield of $k$.
$endgroup$
– user150248
2 days ago
$begingroup$
Oh I think I get it. Qing has been always talking about k-algebra homomorphisms since the proof of the normalization lemma, although he did not give any definition of “finite homomorphism”.
$endgroup$
– user150248
2 days ago
add a comment |
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1 Answer
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1 Answer
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$begingroup$
These corollaries (explicitly) talk about $k$-algebras, so every morphism is $k$-linear. This implies that $f(alpha_i)=alpha_i+mathfrakm$.
In the category of $k$-algebras $k$ is an initial object, so for any $k$-algebra $A$ there is exactly one $k$-algebra homomorphism $gcolon kto A$. This must be a ring morphism, so $g(1)=1_A$. And it must be $k$-linear, so $g(alpha)=alpha cdot 1_A$.
answered 2 days ago
PedroPedro
2,9291720
$endgroup$
$begingroup$
It may not be $k$-linear in the desired way. Perhaps $f^-1$ maps all the scalar elements to a proper subfield of $k$.
$endgroup$
– user150248
2 days ago
$begingroup$
Oh I think I get it. Qing has been always talking about k-algebra homomorphisms since the proof of the normalization lemma, although he did not give any definition of “finite homomorphism”.
$endgroup$
– user150248
2 days ago
add a comment |
$begingroup$
These corollaries (explicitly) talk about $k$-algebras, so every morphism is $k$-linear. This implies that $f(alpha_i)=alpha_i+mathfrakm$.
In the category of $k$-algebras $k$ is an initial object, so for any $k$-algebra $A$ there is exactly one $k$-algebra homomorphism $gcolon kto A$. This must be a ring morphism, so $g(1)=1_A$. And it must be $k$-linear, so $g(alpha)=alpha cdot 1_A$.
answered 2 days ago
PedroPedro
2,9291720
$endgroup$
$begingroup$
It may not be $k$-linear in the desired way. Perhaps $f^-1$ maps all the scalar elements to a proper subfield of $k$.
$endgroup$
– user150248
2 days ago
$begingroup$
Oh I think I get it. Qing has been always talking about k-algebra homomorphisms since the proof of the normalization lemma, although he did not give any definition of “finite homomorphism”.
$endgroup$
– user150248
2 days ago
add a comment |
$begingroup$
These corollaries (explicitly) talk about $k$-algebras, so every morphism is $k$-linear. This implies that $f(alpha_i)=alpha_i+mathfrakm$.
In the category of $k$-algebras $k$ is an initial object, so for any $k$-algebra $A$ there is exactly one $k$-algebra homomorphism $gcolon kto A$. This must be a ring morphism, so $g(1)=1_A$. And it must be $k$-linear, so $g(alpha)=alpha cdot 1_A$.
answered 2 days ago
PedroPedro
2,9291720
$endgroup$
These corollaries (explicitly) talk about $k$-algebras, so every morphism is $k$-linear. This implies that $f(alpha_i)=alpha_i+mathfrakm$.
In the category of $k$-algebras $k$ is an initial object, so for any $k$-algebra $A$ there is exactly one $k$-algebra homomorphism $gcolon kto A$. This must be a ring morphism, so $g(1)=1_A$. And it must be $k$-linear, so $g(alpha)=alpha cdot 1_A$.
answered 2 days ago
PedroPedro
2,9291720
answered 2 days ago
PedroPedro
2,9291720
answered 2 days ago
PedroPedro
2,9291720
answered 2 days ago
PedroPedro
2,9291720
2,9291720
$begingroup$
It may not be $k$-linear in the desired way. Perhaps $f^-1$ maps all the scalar elements to a proper subfield of $k$.
$endgroup$
– user150248
2 days ago
$begingroup$
Oh I think I get it. Qing has been always talking about k-algebra homomorphisms since the proof of the normalization lemma, although he did not give any definition of “finite homomorphism”.
$endgroup$
– user150248
2 days ago
add a comment |
$begingroup$
It may not be $k$-linear in the desired way. Perhaps $f^-1$ maps all the scalar elements to a proper subfield of $k$.
$endgroup$
– user150248
2 days ago
$begingroup$
Oh I think I get it. Qing has been always talking about k-algebra homomorphisms since the proof of the normalization lemma, although he did not give any definition of “finite homomorphism”.
$endgroup$
– user150248
2 days ago
$begingroup$
It may not be $k$-linear in the desired way. Perhaps $f^-1$ maps all the scalar elements to a proper subfield of $k$.
$endgroup$
– user150248
2 days ago
$begingroup$
It may not be $k$-linear in the desired way. Perhaps $f^-1$ maps all the scalar elements to a proper subfield of $k$.
$endgroup$
– user150248
2 days ago
$begingroup$
Oh I think I get it. Qing has been always talking about k-algebra homomorphisms since the proof of the normalization lemma, although he did not give any definition of “finite homomorphism”.
$endgroup$
– user150248
2 days ago
$begingroup$
Oh I think I get it. Qing has been always talking about k-algebra homomorphisms since the proof of the normalization lemma, although he did not give any definition of “finite homomorphism”.
$endgroup$
– user150248
2 days ago
add a comment |
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