Minimal set of generators of an ideal
$begingroup$
Let $R$ be a commutative local ring with maximal ideal $mathfrak{m}$. Let $I$ be an ideal and $xin R$ such that $x$ is not a zero divisor on $R/I$. Then a minimal set of generators for $I$ is sent to a minimal set of generators for $(I +(x))/(x)$ in $R/(x)$.
My idea was to try to prove that $I/mathfrak{m}Icong (I + (x))/(mathfrak{m}I +(x))$, but my attempts have been unsuccessful.
commutative-algebra
asked Jan 20 at 9:10
user09127user09127
1846
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add a comment |
$begingroup$
Let $R$ be a commutative local ring with maximal ideal $mathfrak{m}$. Let $I$ be an ideal and $xin R$ such that $x$ is not a zero divisor on $R/I$. Then a minimal set of generators for $I$ is sent to a minimal set of generators for $(I +(x))/(x)$ in $R/(x)$.
My idea was to try to prove that $I/mathfrak{m}Icong (I + (x))/(mathfrak{m}I +(x))$, but my attempts have been unsuccessful.
commutative-algebra
asked Jan 20 at 9:10
user09127user09127
1846
$endgroup$
add a comment |
$begingroup$
Let $R$ be a commutative local ring with maximal ideal $mathfrak{m}$. Let $I$ be an ideal and $xin R$ such that $x$ is not a zero divisor on $R/I$. Then a minimal set of generators for $I$ is sent to a minimal set of generators for $(I +(x))/(x)$ in $R/(x)$.
My idea was to try to prove that $I/mathfrak{m}Icong (I + (x))/(mathfrak{m}I +(x))$, but my attempts have been unsuccessful.
commutative-algebra
asked Jan 20 at 9:10
user09127user09127
1846
$endgroup$
Let $R$ be a commutative local ring with maximal ideal $mathfrak{m}$. Let $I$ be an ideal and $xin R$ such that $x$ is not a zero divisor on $R/I$. Then a minimal set of generators for $I$ is sent to a minimal set of generators for $(I +(x))/(x)$ in $R/(x)$.
My idea was to try to prove that $I/mathfrak{m}Icong (I + (x))/(mathfrak{m}I +(x))$, but my attempts have been unsuccessful.
commutative-algebra
commutative-algebra
asked Jan 20 at 9:10
user09127user09127
1846
asked Jan 20 at 9:10
user09127user09127
1846
asked Jan 20 at 9:10
user09127user09127
1846
asked Jan 20 at 9:10
user09127user09127
1846
asked Jan 20 at 9:10
user09127user09127
1846
1846
add a comment |
add a comment |
1 Answer
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$begingroup$
Lemma: For any ideal $I$ and $x$ in $R$, one has $I cap (x) = x(I:_R x)$. In addition if $x$ is a nonzerodivior on $R/I$, $I:x = I$, so $I cap (x) = xI$.
Now, apply the 3rd and 2nd isomorphism theorems to have
$$
frac{I+(x)/(x)}{m(I+(x)) +(x) / (x)} cong frac{I + (x)}{m(I+(x)) + (x)} cong frac{I+(x)}{mI + (x)} = frac{I}{ mI + I cap (x)}.
$$
By the lemma above,
$$
I/ (mI + I cap (x)) cong I / (mI + xI) = I/mI.
$$
answered Jan 20 at 19:02
YoungsuYoungsu
1,813715
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1 Answer
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1 Answer
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$begingroup$
Lemma: For any ideal $I$ and $x$ in $R$, one has $I cap (x) = x(I:_R x)$. In addition if $x$ is a nonzerodivior on $R/I$, $I:x = I$, so $I cap (x) = xI$.
Now, apply the 3rd and 2nd isomorphism theorems to have
$$
frac{I+(x)/(x)}{m(I+(x)) +(x) / (x)} cong frac{I + (x)}{m(I+(x)) + (x)} cong frac{I+(x)}{mI + (x)} = frac{I}{ mI + I cap (x)}.
$$
By the lemma above,
$$
I/ (mI + I cap (x)) cong I / (mI + xI) = I/mI.
$$
answered Jan 20 at 19:02
YoungsuYoungsu
1,813715
$endgroup$
add a comment |
$begingroup$
Lemma: For any ideal $I$ and $x$ in $R$, one has $I cap (x) = x(I:_R x)$. In addition if $x$ is a nonzerodivior on $R/I$, $I:x = I$, so $I cap (x) = xI$.
Now, apply the 3rd and 2nd isomorphism theorems to have
$$
frac{I+(x)/(x)}{m(I+(x)) +(x) / (x)} cong frac{I + (x)}{m(I+(x)) + (x)} cong frac{I+(x)}{mI + (x)} = frac{I}{ mI + I cap (x)}.
$$
By the lemma above,
$$
I/ (mI + I cap (x)) cong I / (mI + xI) = I/mI.
$$
answered Jan 20 at 19:02
YoungsuYoungsu
1,813715
$endgroup$
add a comment |
$begingroup$
Lemma: For any ideal $I$ and $x$ in $R$, one has $I cap (x) = x(I:_R x)$. In addition if $x$ is a nonzerodivior on $R/I$, $I:x = I$, so $I cap (x) = xI$.
Now, apply the 3rd and 2nd isomorphism theorems to have
$$
frac{I+(x)/(x)}{m(I+(x)) +(x) / (x)} cong frac{I + (x)}{m(I+(x)) + (x)} cong frac{I+(x)}{mI + (x)} = frac{I}{ mI + I cap (x)}.
$$
By the lemma above,
$$
I/ (mI + I cap (x)) cong I / (mI + xI) = I/mI.
$$
answered Jan 20 at 19:02
YoungsuYoungsu
1,813715
$endgroup$
Lemma: For any ideal $I$ and $x$ in $R$, one has $I cap (x) = x(I:_R x)$. In addition if $x$ is a nonzerodivior on $R/I$, $I:x = I$, so $I cap (x) = xI$.
Now, apply the 3rd and 2nd isomorphism theorems to have
$$
frac{I+(x)/(x)}{m(I+(x)) +(x) / (x)} cong frac{I + (x)}{m(I+(x)) + (x)} cong frac{I+(x)}{mI + (x)} = frac{I}{ mI + I cap (x)}.
$$
By the lemma above,
$$
I/ (mI + I cap (x)) cong I / (mI + xI) = I/mI.
$$
answered Jan 20 at 19:02
YoungsuYoungsu
1,813715
answered Jan 20 at 19:02
YoungsuYoungsu
1,813715
answered Jan 20 at 19:02
YoungsuYoungsu
1,813715
answered Jan 20 at 19:02
YoungsuYoungsu
1,813715
1,813715
add a comment |
add a comment |
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