Mixing
Hilbert Nullstellensatz, Eisenbud's proof
$begingroup$
I am trying to understand the proof on Eisenbud, Commutative Algebra with a View Toward Algebraic Geometry.
The theorem I am trying to prove is:
Let $k$ be an algebraically closed field. If $I subset k[x_{1},...x_{n}]$ is an ideal, then $I(Z(I))=mathrm{rad}(I)$.
"The proof goes like follows: we know that points of $X subset A^{r}(k)$ are in 1-1 correspondence to maximal ideals in $A(X)$, thus to maximal ideals in $k[x_{1},...,x_{n}]$ containing $I(X)$. Thus points in $Z(I)$ corresponds to the maximal ideals in $k[x_{1}, ...x_{n}]$ containing $I$. Thus $I(Z(I))$ is the intersection of all maximal ideals containing $I$."
I can't see "Thus points in Z(I) corresponds to the ideals in $k[x_{1}, ...x_{n}]$ containing I. Thus $I(Z(I))$ is the intersection of all maximal ideals containing I." Any help will be appreciated.
commutative-algebra
asked Jan 13 at 16:01
IvanIvan
113
$endgroup$
add a comment |
$begingroup$
I am trying to understand the proof on Eisenbud, Commutative Algebra with a View Toward Algebraic Geometry.
The theorem I am trying to prove is:
Let $k$ be an algebraically closed field. If $I subset k[x_{1},...x_{n}]$ is an ideal, then $I(Z(I))=mathrm{rad}(I)$.
"The proof goes like follows: we know that points of $X subset A^{r}(k)$ are in 1-1 correspondence to maximal ideals in $A(X)$, thus to maximal ideals in $k[x_{1},...,x_{n}]$ containing $I(X)$. Thus points in $Z(I)$ corresponds to the maximal ideals in $k[x_{1}, ...x_{n}]$ containing $I$. Thus $I(Z(I))$ is the intersection of all maximal ideals containing $I$."
I can't see "Thus points in Z(I) corresponds to the ideals in $k[x_{1}, ...x_{n}]$ containing I. Thus $I(Z(I))$ is the intersection of all maximal ideals containing I." Any help will be appreciated.
commutative-algebra
asked Jan 13 at 16:01
IvanIvan
113
$endgroup$
$begingroup$
You have two sentences in your question. Which part is your question? I think "it" should be "is".
$endgroup$
– Youngsu
Jan 13 at 19:11
$begingroup$
I can't see both conclusions. ( I have corrected the typo)
$endgroup$
– Ivan
Jan 14 at 9:10
$begingroup$
The first conclusion is the previous statement with $X = Z(I)$. (I think you are missing "maximal" here though.) For the second conclusion, one way to do this is showing they are contained in each other using the first sentence of your question.
$endgroup$
– Youngsu
Jan 14 at 16:16
$begingroup$
Thanks I get it now!
$endgroup$
– Ivan
Jan 14 at 16:40
$begingroup$
Great! Good luck.
$endgroup$
– Youngsu
Jan 14 at 17:37
add a comment |
$begingroup$
I am trying to understand the proof on Eisenbud, Commutative Algebra with a View Toward Algebraic Geometry.
The theorem I am trying to prove is:
Let $k$ be an algebraically closed field. If $I subset k[x_{1},...x_{n}]$ is an ideal, then $I(Z(I))=mathrm{rad}(I)$.
"The proof goes like follows: we know that points of $X subset A^{r}(k)$ are in 1-1 correspondence to maximal ideals in $A(X)$, thus to maximal ideals in $k[x_{1},...,x_{n}]$ containing $I(X)$. Thus points in $Z(I)$ corresponds to the maximal ideals in $k[x_{1}, ...x_{n}]$ containing $I$. Thus $I(Z(I))$ is the intersection of all maximal ideals containing $I$."
I can't see "Thus points in Z(I) corresponds to the ideals in $k[x_{1}, ...x_{n}]$ containing I. Thus $I(Z(I))$ is the intersection of all maximal ideals containing I." Any help will be appreciated.
commutative-algebra
asked Jan 13 at 16:01
IvanIvan
113
$endgroup$
I am trying to understand the proof on Eisenbud, Commutative Algebra with a View Toward Algebraic Geometry.
The theorem I am trying to prove is:
Let $k$ be an algebraically closed field. If $I subset k[x_{1},...x_{n}]$ is an ideal, then $I(Z(I))=mathrm{rad}(I)$.
"The proof goes like follows: we know that points of $X subset A^{r}(k)$ are in 1-1 correspondence to maximal ideals in $A(X)$, thus to maximal ideals in $k[x_{1},...,x_{n}]$ containing $I(X)$. Thus points in $Z(I)$ corresponds to the maximal ideals in $k[x_{1}, ...x_{n}]$ containing $I$. Thus $I(Z(I))$ is the intersection of all maximal ideals containing $I$."
I can't see "Thus points in Z(I) corresponds to the ideals in $k[x_{1}, ...x_{n}]$ containing I. Thus $I(Z(I))$ is the intersection of all maximal ideals containing I." Any help will be appreciated.
commutative-algebra
commutative-algebra
asked Jan 13 at 16:01
IvanIvan
113
asked Jan 13 at 16:01
IvanIvan
113
edited Jan 14 at 16:39
Ivan
asked Jan 13 at 16:01
IvanIvan
113
asked Jan 13 at 16:01
IvanIvan
113
asked Jan 13 at 16:01
IvanIvan
113
113
$begingroup$
You have two sentences in your question. Which part is your question? I think "it" should be "is".
$endgroup$
– Youngsu
Jan 13 at 19:11
$begingroup$
I can't see both conclusions. ( I have corrected the typo)
$endgroup$
– Ivan
Jan 14 at 9:10
$begingroup$
The first conclusion is the previous statement with $X = Z(I)$. (I think you are missing "maximal" here though.) For the second conclusion, one way to do this is showing they are contained in each other using the first sentence of your question.
$endgroup$
– Youngsu
Jan 14 at 16:16
$begingroup$
Thanks I get it now!
$endgroup$
– Ivan
Jan 14 at 16:40
$begingroup$
Great! Good luck.
$endgroup$
– Youngsu
Jan 14 at 17:37
add a comment |
$begingroup$
You have two sentences in your question. Which part is your question? I think "it" should be "is".
$endgroup$
– Youngsu
Jan 13 at 19:11
$begingroup$
I can't see both conclusions. ( I have corrected the typo)
$endgroup$
– Ivan
Jan 14 at 9:10
$begingroup$
The first conclusion is the previous statement with $X = Z(I)$. (I think you are missing "maximal" here though.) For the second conclusion, one way to do this is showing they are contained in each other using the first sentence of your question.
$endgroup$
– Youngsu
Jan 14 at 16:16
$begingroup$
Thanks I get it now!
$endgroup$
– Ivan
Jan 14 at 16:40
$begingroup$
Great! Good luck.
$endgroup$
– Youngsu
Jan 14 at 17:37
$begingroup$
You have two sentences in your question. Which part is your question? I think "it" should be "is".
$endgroup$
– Youngsu
Jan 13 at 19:11
$begingroup$
You have two sentences in your question. Which part is your question? I think "it" should be "is".
$endgroup$
– Youngsu
Jan 13 at 19:11
$begingroup$
I can't see both conclusions. ( I have corrected the typo)
$endgroup$
– Ivan
Jan 14 at 9:10
$begingroup$
I can't see both conclusions. ( I have corrected the typo)
$endgroup$
– Ivan
Jan 14 at 9:10
$begingroup$
The first conclusion is the previous statement with $X = Z(I)$. (I think you are missing "maximal" here though.) For the second conclusion, one way to do this is showing they are contained in each other using the first sentence of your question.
$endgroup$
– Youngsu
Jan 14 at 16:16
$begingroup$
The first conclusion is the previous statement with $X = Z(I)$. (I think you are missing "maximal" here though.) For the second conclusion, one way to do this is showing they are contained in each other using the first sentence of your question.
$endgroup$
– Youngsu
Jan 14 at 16:16
$begingroup$
Thanks I get it now!
$endgroup$
– Ivan
Jan 14 at 16:40
$begingroup$
Thanks I get it now!
$endgroup$
– Ivan
Jan 14 at 16:40
$begingroup$
Great! Good luck.
$endgroup$
– Youngsu
Jan 14 at 17:37
$begingroup$
Great! Good luck.
$endgroup$
– Youngsu
Jan 14 at 17:37
add a comment |
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$begingroup$
You have two sentences in your question. Which part is your question? I think "it" should be "is".
$endgroup$
– Youngsu
Jan 13 at 19:11
$begingroup$
I can't see both conclusions. ( I have corrected the typo)
$endgroup$
– Ivan
Jan 14 at 9:10
$begingroup$
The first conclusion is the previous statement with $X = Z(I)$. (I think you are missing "maximal" here though.) For the second conclusion, one way to do this is showing they are contained in each other using the first sentence of your question.
$endgroup$
– Youngsu
Jan 14 at 16:16
$begingroup$
Thanks I get it now!
$endgroup$
– Ivan
Jan 14 at 16:40
$begingroup$
Great! Good luck.
$endgroup$
– Youngsu
Jan 14 at 17:37