Proving that the radical of an ideal is an ideal [duplicate]
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Commutative Ring: Nilpotent elements closed under addition? [duplicate]
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I'm trying to solve problem 1.18 in Fulton's 'Algebraic Curves', which is illustrated in the attached image, but I'm having some difficulties understanding the first part.
The ring R is assumed to be commutative and unital, so the first thing that came to mind was to consider the binomial expansion of $(a+b)^{n+m}$, but I don't see why powers of $a$ which are less than n, or powers of $b$ which are less than m should be contained in the ideal.
An explanation as to how I can show that $Rad(I) $ contains $a+b$ would be appreciated, particularly I'd like to know if one can indeed be guaranteed the containment (in I) of the aforementioned terms from the binomial expansion.
I suspect that I may be missing or forgetting some ring theoretic fact.
ring-theory ideals radicals
asked Jan 2 at 15:08
ToricTorusToricTorus
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Jan 2 at 16:06
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This question already has an answer here:
Commutative Ring: Nilpotent elements closed under addition? [duplicate]
1 answer
I'm trying to solve problem 1.18 in Fulton's 'Algebraic Curves', which is illustrated in the attached image, but I'm having some difficulties understanding the first part.
The ring R is assumed to be commutative and unital, so the first thing that came to mind was to consider the binomial expansion of $(a+b)^{n+m}$, but I don't see why powers of $a$ which are less than n, or powers of $b$ which are less than m should be contained in the ideal.
An explanation as to how I can show that $Rad(I) $ contains $a+b$ would be appreciated, particularly I'd like to know if one can indeed be guaranteed the containment (in I) of the aforementioned terms from the binomial expansion.
I suspect that I may be missing or forgetting some ring theoretic fact.
ring-theory ideals radicals
asked Jan 2 at 15:08
ToricTorusToricTorus
124
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marked as duplicate by rschwieb ring-theory
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Jan 2 at 16:06
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This question already has an answer here:
Commutative Ring: Nilpotent elements closed under addition? [duplicate]
1 answer
I'm trying to solve problem 1.18 in Fulton's 'Algebraic Curves', which is illustrated in the attached image, but I'm having some difficulties understanding the first part.
The ring R is assumed to be commutative and unital, so the first thing that came to mind was to consider the binomial expansion of $(a+b)^{n+m}$, but I don't see why powers of $a$ which are less than n, or powers of $b$ which are less than m should be contained in the ideal.
An explanation as to how I can show that $Rad(I) $ contains $a+b$ would be appreciated, particularly I'd like to know if one can indeed be guaranteed the containment (in I) of the aforementioned terms from the binomial expansion.
I suspect that I may be missing or forgetting some ring theoretic fact.
ring-theory ideals radicals
asked Jan 2 at 15:08
ToricTorusToricTorus
124
$endgroup$
This question already has an answer here:
Commutative Ring: Nilpotent elements closed under addition? [duplicate]
1 answer
I'm trying to solve problem 1.18 in Fulton's 'Algebraic Curves', which is illustrated in the attached image, but I'm having some difficulties understanding the first part.
The ring R is assumed to be commutative and unital, so the first thing that came to mind was to consider the binomial expansion of $(a+b)^{n+m}$, but I don't see why powers of $a$ which are less than n, or powers of $b$ which are less than m should be contained in the ideal.
An explanation as to how I can show that $Rad(I) $ contains $a+b$ would be appreciated, particularly I'd like to know if one can indeed be guaranteed the containment (in I) of the aforementioned terms from the binomial expansion.
I suspect that I may be missing or forgetting some ring theoretic fact.
This question already has an answer here:
Commutative Ring: Nilpotent elements closed under addition? [duplicate]
1 answer
ring-theory ideals radicals
ring-theory ideals radicals
asked Jan 2 at 15:08
ToricTorusToricTorus
124
asked Jan 2 at 15:08
ToricTorusToricTorus
124
edited Jan 2 at 15:34
ToricTorus
asked Jan 2 at 15:08
ToricTorusToricTorus
124
asked Jan 2 at 15:08
ToricTorusToricTorus
124
asked Jan 2 at 15:08
ToricTorusToricTorus
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Jan 2 at 16:06
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2 Answers
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The fact is that in the binomial expansion you have a sum of products $a^ib^{n+m-i}$ with $0 le i le n+m$. So either $ige n$ in which case $a^i in I$ or $n+m-ige m$ in which case $b^{n+m-i}$. In both cases $a^ib^{n+m-i} in I$.
answered Jan 2 at 15:17
mathcounterexamples.netmathcounterexamples.net
27k22158
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I'm reminded yet again that one shouldn't be lazy in ones thinking. Thank you.
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– ToricTorus
Jan 2 at 15:29
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Suppose, for instance, that $m=3$ and that $n=2$. Then$$(a+b)^{m+n}=(a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5.$$Then $a^5$, $a^4b$, $a^3b^2$ and $a^2b^3$ all belong to $I$, since $a^2$ does. And $ab^4$ and $b^5$ also belong to $I$, since $b^3$ does. Therefore, $(a+b)^5in I$.
answered Jan 2 at 15:14
José Carlos SantosJosé Carlos Santos
171k23132240
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2 Answers
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2 Answers
2
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oldest
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active
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active
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votes
$begingroup$
The fact is that in the binomial expansion you have a sum of products $a^ib^{n+m-i}$ with $0 le i le n+m$. So either $ige n$ in which case $a^i in I$ or $n+m-ige m$ in which case $b^{n+m-i}$. In both cases $a^ib^{n+m-i} in I$.
answered Jan 2 at 15:17
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
$begingroup$
I'm reminded yet again that one shouldn't be lazy in ones thinking. Thank you.
$endgroup$
– ToricTorus
Jan 2 at 15:29
add a comment |
$begingroup$
The fact is that in the binomial expansion you have a sum of products $a^ib^{n+m-i}$ with $0 le i le n+m$. So either $ige n$ in which case $a^i in I$ or $n+m-ige m$ in which case $b^{n+m-i}$. In both cases $a^ib^{n+m-i} in I$.
answered Jan 2 at 15:17
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
$begingroup$
I'm reminded yet again that one shouldn't be lazy in ones thinking. Thank you.
$endgroup$
– ToricTorus
Jan 2 at 15:29
add a comment |
$begingroup$
The fact is that in the binomial expansion you have a sum of products $a^ib^{n+m-i}$ with $0 le i le n+m$. So either $ige n$ in which case $a^i in I$ or $n+m-ige m$ in which case $b^{n+m-i}$. In both cases $a^ib^{n+m-i} in I$.
answered Jan 2 at 15:17
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
The fact is that in the binomial expansion you have a sum of products $a^ib^{n+m-i}$ with $0 le i le n+m$. So either $ige n$ in which case $a^i in I$ or $n+m-ige m$ in which case $b^{n+m-i}$. In both cases $a^ib^{n+m-i} in I$.
answered Jan 2 at 15:17
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 2 at 15:17
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 2 at 15:17
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 2 at 15:17
mathcounterexamples.netmathcounterexamples.net
27k22158
27k22158
$begingroup$
I'm reminded yet again that one shouldn't be lazy in ones thinking. Thank you.
$endgroup$
– ToricTorus
Jan 2 at 15:29
add a comment |
$begingroup$
I'm reminded yet again that one shouldn't be lazy in ones thinking. Thank you.
$endgroup$
– ToricTorus
Jan 2 at 15:29
$begingroup$
I'm reminded yet again that one shouldn't be lazy in ones thinking. Thank you.
$endgroup$
– ToricTorus
Jan 2 at 15:29
$begingroup$
I'm reminded yet again that one shouldn't be lazy in ones thinking. Thank you.
$endgroup$
– ToricTorus
Jan 2 at 15:29
add a comment |
$begingroup$
Suppose, for instance, that $m=3$ and that $n=2$. Then$$(a+b)^{m+n}=(a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5.$$Then $a^5$, $a^4b$, $a^3b^2$ and $a^2b^3$ all belong to $I$, since $a^2$ does. And $ab^4$ and $b^5$ also belong to $I$, since $b^3$ does. Therefore, $(a+b)^5in I$.
answered Jan 2 at 15:14
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
Suppose, for instance, that $m=3$ and that $n=2$. Then$$(a+b)^{m+n}=(a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5.$$Then $a^5$, $a^4b$, $a^3b^2$ and $a^2b^3$ all belong to $I$, since $a^2$ does. And $ab^4$ and $b^5$ also belong to $I$, since $b^3$ does. Therefore, $(a+b)^5in I$.
answered Jan 2 at 15:14
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
Suppose, for instance, that $m=3$ and that $n=2$. Then$$(a+b)^{m+n}=(a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5.$$Then $a^5$, $a^4b$, $a^3b^2$ and $a^2b^3$ all belong to $I$, since $a^2$ does. And $ab^4$ and $b^5$ also belong to $I$, since $b^3$ does. Therefore, $(a+b)^5in I$.
answered Jan 2 at 15:14
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
Suppose, for instance, that $m=3$ and that $n=2$. Then$$(a+b)^{m+n}=(a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5.$$Then $a^5$, $a^4b$, $a^3b^2$ and $a^2b^3$ all belong to $I$, since $a^2$ does. And $ab^4$ and $b^5$ also belong to $I$, since $b^3$ does. Therefore, $(a+b)^5in I$.
answered Jan 2 at 15:14
José Carlos SantosJosé Carlos Santos
171k23132240
answered Jan 2 at 15:14
José Carlos SantosJosé Carlos Santos
171k23132240
answered Jan 2 at 15:14
José Carlos SantosJosé Carlos Santos
171k23132240
answered Jan 2 at 15:14
José Carlos SantosJosé Carlos Santos
171k23132240
171k23132240
add a comment |
add a comment |
