jacobson radical nilpotence, associative algebras.
$begingroup$
In $Pierce, Associative Algebras$, it is mentioned that the Jacobson radical of an associative algebra $A$ is nilpotent, meaning $J(A)^k=0$ for a natural number $k$. However, my professor said that this isn't meant in the sense that $a^k=0$ for all elements $ain J(A)$, rather that the Jacobson radical is $0$, when written as a linear combination of $k$ elements. What does this mean, and, is there an example for this?
I am now trying to write a code in GAP to calculate the quiver for non-commutative Algebras. However, I am stuck at this point, as I don't know how this linear combination is supposed to look.
abstract-algebra representation-theory gap
asked Dec 16 '18 at 10:22
UnexpectedExpectationUnexpectedExpectation
518
$endgroup$
add a comment |
$begingroup$
In $Pierce, Associative Algebras$, it is mentioned that the Jacobson radical of an associative algebra $A$ is nilpotent, meaning $J(A)^k=0$ for a natural number $k$. However, my professor said that this isn't meant in the sense that $a^k=0$ for all elements $ain J(A)$, rather that the Jacobson radical is $0$, when written as a linear combination of $k$ elements. What does this mean, and, is there an example for this?
I am now trying to write a code in GAP to calculate the quiver for non-commutative Algebras. However, I am stuck at this point, as I don't know how this linear combination is supposed to look.
abstract-algebra representation-theory gap
asked Dec 16 '18 at 10:22
UnexpectedExpectationUnexpectedExpectation
518
$endgroup$
1
$begingroup$
There seems to be a missing assumption for $A$, since the Jacobson radical of a ring is not nilpotent in general (take $A=mathbb{R}[[x]]$, for example). Is it possible that Pierce assumes that $A$ is Artinian?
$endgroup$
– Pierre-Guy Plamondon
Dec 16 '18 at 12:37
$begingroup$
@Pierre-GuyPlamondon Oh, sorry. Indeed, $A$ is assumed to be right-Artinian.
$endgroup$
– UnexpectedExpectation
Dec 16 '18 at 13:05
add a comment |
$begingroup$
In $Pierce, Associative Algebras$, it is mentioned that the Jacobson radical of an associative algebra $A$ is nilpotent, meaning $J(A)^k=0$ for a natural number $k$. However, my professor said that this isn't meant in the sense that $a^k=0$ for all elements $ain J(A)$, rather that the Jacobson radical is $0$, when written as a linear combination of $k$ elements. What does this mean, and, is there an example for this?
I am now trying to write a code in GAP to calculate the quiver for non-commutative Algebras. However, I am stuck at this point, as I don't know how this linear combination is supposed to look.
abstract-algebra representation-theory gap
asked Dec 16 '18 at 10:22
UnexpectedExpectationUnexpectedExpectation
518
$endgroup$
In $Pierce, Associative Algebras$, it is mentioned that the Jacobson radical of an associative algebra $A$ is nilpotent, meaning $J(A)^k=0$ for a natural number $k$. However, my professor said that this isn't meant in the sense that $a^k=0$ for all elements $ain J(A)$, rather that the Jacobson radical is $0$, when written as a linear combination of $k$ elements. What does this mean, and, is there an example for this?
I am now trying to write a code in GAP to calculate the quiver for non-commutative Algebras. However, I am stuck at this point, as I don't know how this linear combination is supposed to look.
abstract-algebra representation-theory gap
abstract-algebra representation-theory gap
asked Dec 16 '18 at 10:22
UnexpectedExpectationUnexpectedExpectation
518
asked Dec 16 '18 at 10:22
UnexpectedExpectationUnexpectedExpectation
518
asked Dec 16 '18 at 10:22
UnexpectedExpectationUnexpectedExpectation
518
asked Dec 16 '18 at 10:22
UnexpectedExpectationUnexpectedExpectation
518
asked Dec 16 '18 at 10:22
UnexpectedExpectationUnexpectedExpectation
518
518
1
$begingroup$
There seems to be a missing assumption for $A$, since the Jacobson radical of a ring is not nilpotent in general (take $A=mathbb{R}[[x]]$, for example). Is it possible that Pierce assumes that $A$ is Artinian?
$endgroup$
– Pierre-Guy Plamondon
Dec 16 '18 at 12:37
$begingroup$
@Pierre-GuyPlamondon Oh, sorry. Indeed, $A$ is assumed to be right-Artinian.
$endgroup$
– UnexpectedExpectation
Dec 16 '18 at 13:05
add a comment |
1
$begingroup$
There seems to be a missing assumption for $A$, since the Jacobson radical of a ring is not nilpotent in general (take $A=mathbb{R}[[x]]$, for example). Is it possible that Pierce assumes that $A$ is Artinian?
$endgroup$
– Pierre-Guy Plamondon
Dec 16 '18 at 12:37
$begingroup$
@Pierre-GuyPlamondon Oh, sorry. Indeed, $A$ is assumed to be right-Artinian.
$endgroup$
– UnexpectedExpectation
Dec 16 '18 at 13:05
1
1
$begingroup$
There seems to be a missing assumption for $A$, since the Jacobson radical of a ring is not nilpotent in general (take $A=mathbb{R}[[x]]$, for example). Is it possible that Pierce assumes that $A$ is Artinian?
$endgroup$
– Pierre-Guy Plamondon
Dec 16 '18 at 12:37
$begingroup$
There seems to be a missing assumption for $A$, since the Jacobson radical of a ring is not nilpotent in general (take $A=mathbb{R}[[x]]$, for example). Is it possible that Pierce assumes that $A$ is Artinian?
$endgroup$
– Pierre-Guy Plamondon
Dec 16 '18 at 12:37
$begingroup$
@Pierre-GuyPlamondon Oh, sorry. Indeed, $A$ is assumed to be right-Artinian.
$endgroup$
– UnexpectedExpectation
Dec 16 '18 at 13:05
$begingroup$
@Pierre-GuyPlamondon Oh, sorry. Indeed, $A$ is assumed to be right-Artinian.
$endgroup$
– UnexpectedExpectation
Dec 16 '18 at 13:05
add a comment |
1 Answer
1
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oldest
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$begingroup$
An ideal $I$ is said to be a nilpotent ideal if the ideal product $I^k={0}$ for some natural number $k$.
Now in general, the ideal product $I^k$ would consist of all finite sums of products of $k$ elements in $I$. In other words, each element looks like $sum_{j=1}^n prod_{m=1}^k i_{jm}$ where all $i_{jm}in I$ and $n$ can be any natural number (but $k$ is the one we specified.)
But in our case, if $I^k={0}$, any product of $k$ elements of $I$ is zero, and hence any finite sum of them is zero too. So, this is equivalent to every product of length $k$ of elements of $I$ being $0$. So, you can see that this is strictly stronger than every element simply being nilpotent.
An ideal for which every element is nilpotent is called a nil ideal. The Jacobson radical also need not be nil, either. Here are some rings whose Jacobson radicals aren't nil and here are some whose Jacobson radical isn't nilpotent, but is nil. There are even rings whose Jacobson radicals are nilpotent but are not Artinian.
the Jacobson radical is ${0}$, when written as a linear combination of $k$ elements. What does this mean, and, is there an example for this?
This seems to be what I'm describing above, although perhaps a little garbled. The things that are being combined come from $J(R)$ and we are not really talking about linear combinations, but sums of products.
I think an easy to understand example would be something like $k[x,y]/(x,y)^2$. The Jacobson radical is easily seen to be $(x,y)$, and not only are $x^2=y^2=0$, but also $xy=0$.
answered Dec 16 '18 at 14:22
rschwiebrschwieb
106k12102249
$endgroup$
$begingroup$
So, if I have determined the generators of the Jacobson radical, I can determine $J(A)^2$ by taking the products of each pair of generators?
$endgroup$
– UnexpectedExpectation
Dec 16 '18 at 14:47
$begingroup$
@UnexpectedExpectation if you have a generating set for $J(R)$, the products will be of the form $g_irg_js$, after you distribute. It doesn’t really simplify anything. The products of pairs of generators is only a small subset of those...
$endgroup$
– rschwieb
Dec 16 '18 at 18:11
add a comment |
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$begingroup$
An ideal $I$ is said to be a nilpotent ideal if the ideal product $I^k={0}$ for some natural number $k$.
Now in general, the ideal product $I^k$ would consist of all finite sums of products of $k$ elements in $I$. In other words, each element looks like $sum_{j=1}^n prod_{m=1}^k i_{jm}$ where all $i_{jm}in I$ and $n$ can be any natural number (but $k$ is the one we specified.)
But in our case, if $I^k={0}$, any product of $k$ elements of $I$ is zero, and hence any finite sum of them is zero too. So, this is equivalent to every product of length $k$ of elements of $I$ being $0$. So, you can see that this is strictly stronger than every element simply being nilpotent.
An ideal for which every element is nilpotent is called a nil ideal. The Jacobson radical also need not be nil, either. Here are some rings whose Jacobson radicals aren't nil and here are some whose Jacobson radical isn't nilpotent, but is nil. There are even rings whose Jacobson radicals are nilpotent but are not Artinian.
the Jacobson radical is ${0}$, when written as a linear combination of $k$ elements. What does this mean, and, is there an example for this?
This seems to be what I'm describing above, although perhaps a little garbled. The things that are being combined come from $J(R)$ and we are not really talking about linear combinations, but sums of products.
I think an easy to understand example would be something like $k[x,y]/(x,y)^2$. The Jacobson radical is easily seen to be $(x,y)$, and not only are $x^2=y^2=0$, but also $xy=0$.
answered Dec 16 '18 at 14:22
rschwiebrschwieb
106k12102249
$endgroup$
$begingroup$
So, if I have determined the generators of the Jacobson radical, I can determine $J(A)^2$ by taking the products of each pair of generators?
$endgroup$
– UnexpectedExpectation
Dec 16 '18 at 14:47
$begingroup$
@UnexpectedExpectation if you have a generating set for $J(R)$, the products will be of the form $g_irg_js$, after you distribute. It doesn’t really simplify anything. The products of pairs of generators is only a small subset of those...
$endgroup$
– rschwieb
Dec 16 '18 at 18:11
add a comment |
$begingroup$
An ideal $I$ is said to be a nilpotent ideal if the ideal product $I^k={0}$ for some natural number $k$.
Now in general, the ideal product $I^k$ would consist of all finite sums of products of $k$ elements in $I$. In other words, each element looks like $sum_{j=1}^n prod_{m=1}^k i_{jm}$ where all $i_{jm}in I$ and $n$ can be any natural number (but $k$ is the one we specified.)
But in our case, if $I^k={0}$, any product of $k$ elements of $I$ is zero, and hence any finite sum of them is zero too. So, this is equivalent to every product of length $k$ of elements of $I$ being $0$. So, you can see that this is strictly stronger than every element simply being nilpotent.
An ideal for which every element is nilpotent is called a nil ideal. The Jacobson radical also need not be nil, either. Here are some rings whose Jacobson radicals aren't nil and here are some whose Jacobson radical isn't nilpotent, but is nil. There are even rings whose Jacobson radicals are nilpotent but are not Artinian.
the Jacobson radical is ${0}$, when written as a linear combination of $k$ elements. What does this mean, and, is there an example for this?
This seems to be what I'm describing above, although perhaps a little garbled. The things that are being combined come from $J(R)$ and we are not really talking about linear combinations, but sums of products.
I think an easy to understand example would be something like $k[x,y]/(x,y)^2$. The Jacobson radical is easily seen to be $(x,y)$, and not only are $x^2=y^2=0$, but also $xy=0$.
answered Dec 16 '18 at 14:22
rschwiebrschwieb
106k12102249
$endgroup$
$begingroup$
So, if I have determined the generators of the Jacobson radical, I can determine $J(A)^2$ by taking the products of each pair of generators?
$endgroup$
– UnexpectedExpectation
Dec 16 '18 at 14:47
$begingroup$
@UnexpectedExpectation if you have a generating set for $J(R)$, the products will be of the form $g_irg_js$, after you distribute. It doesn’t really simplify anything. The products of pairs of generators is only a small subset of those...
$endgroup$
– rschwieb
Dec 16 '18 at 18:11
add a comment |
$begingroup$
An ideal $I$ is said to be a nilpotent ideal if the ideal product $I^k={0}$ for some natural number $k$.
Now in general, the ideal product $I^k$ would consist of all finite sums of products of $k$ elements in $I$. In other words, each element looks like $sum_{j=1}^n prod_{m=1}^k i_{jm}$ where all $i_{jm}in I$ and $n$ can be any natural number (but $k$ is the one we specified.)
But in our case, if $I^k={0}$, any product of $k$ elements of $I$ is zero, and hence any finite sum of them is zero too. So, this is equivalent to every product of length $k$ of elements of $I$ being $0$. So, you can see that this is strictly stronger than every element simply being nilpotent.
An ideal for which every element is nilpotent is called a nil ideal. The Jacobson radical also need not be nil, either. Here are some rings whose Jacobson radicals aren't nil and here are some whose Jacobson radical isn't nilpotent, but is nil. There are even rings whose Jacobson radicals are nilpotent but are not Artinian.
the Jacobson radical is ${0}$, when written as a linear combination of $k$ elements. What does this mean, and, is there an example for this?
This seems to be what I'm describing above, although perhaps a little garbled. The things that are being combined come from $J(R)$ and we are not really talking about linear combinations, but sums of products.
I think an easy to understand example would be something like $k[x,y]/(x,y)^2$. The Jacobson radical is easily seen to be $(x,y)$, and not only are $x^2=y^2=0$, but also $xy=0$.
answered Dec 16 '18 at 14:22
rschwiebrschwieb
106k12102249
$endgroup$
An ideal $I$ is said to be a nilpotent ideal if the ideal product $I^k={0}$ for some natural number $k$.
Now in general, the ideal product $I^k$ would consist of all finite sums of products of $k$ elements in $I$. In other words, each element looks like $sum_{j=1}^n prod_{m=1}^k i_{jm}$ where all $i_{jm}in I$ and $n$ can be any natural number (but $k$ is the one we specified.)
But in our case, if $I^k={0}$, any product of $k$ elements of $I$ is zero, and hence any finite sum of them is zero too. So, this is equivalent to every product of length $k$ of elements of $I$ being $0$. So, you can see that this is strictly stronger than every element simply being nilpotent.
An ideal for which every element is nilpotent is called a nil ideal. The Jacobson radical also need not be nil, either. Here are some rings whose Jacobson radicals aren't nil and here are some whose Jacobson radical isn't nilpotent, but is nil. There are even rings whose Jacobson radicals are nilpotent but are not Artinian.
the Jacobson radical is ${0}$, when written as a linear combination of $k$ elements. What does this mean, and, is there an example for this?
This seems to be what I'm describing above, although perhaps a little garbled. The things that are being combined come from $J(R)$ and we are not really talking about linear combinations, but sums of products.
I think an easy to understand example would be something like $k[x,y]/(x,y)^2$. The Jacobson radical is easily seen to be $(x,y)$, and not only are $x^2=y^2=0$, but also $xy=0$.
answered Dec 16 '18 at 14:22
rschwiebrschwieb
106k12102249
edited Dec 16 '18 at 14:42
answered Dec 16 '18 at 14:22
rschwiebrschwieb
106k12102249
answered Dec 16 '18 at 14:22
rschwiebrschwieb
106k12102249
answered Dec 16 '18 at 14:22
rschwiebrschwieb
106k12102249
106k12102249
$begingroup$
So, if I have determined the generators of the Jacobson radical, I can determine $J(A)^2$ by taking the products of each pair of generators?
$endgroup$
– UnexpectedExpectation
Dec 16 '18 at 14:47
$begingroup$
@UnexpectedExpectation if you have a generating set for $J(R)$, the products will be of the form $g_irg_js$, after you distribute. It doesn’t really simplify anything. The products of pairs of generators is only a small subset of those...
$endgroup$
– rschwieb
Dec 16 '18 at 18:11
add a comment |
$begingroup$
So, if I have determined the generators of the Jacobson radical, I can determine $J(A)^2$ by taking the products of each pair of generators?
$endgroup$
– UnexpectedExpectation
Dec 16 '18 at 14:47
$begingroup$
@UnexpectedExpectation if you have a generating set for $J(R)$, the products will be of the form $g_irg_js$, after you distribute. It doesn’t really simplify anything. The products of pairs of generators is only a small subset of those...
$endgroup$
– rschwieb
Dec 16 '18 at 18:11
$begingroup$
So, if I have determined the generators of the Jacobson radical, I can determine $J(A)^2$ by taking the products of each pair of generators?
$endgroup$
– UnexpectedExpectation
Dec 16 '18 at 14:47
$begingroup$
So, if I have determined the generators of the Jacobson radical, I can determine $J(A)^2$ by taking the products of each pair of generators?
$endgroup$
– UnexpectedExpectation
Dec 16 '18 at 14:47
$begingroup$
@UnexpectedExpectation if you have a generating set for $J(R)$, the products will be of the form $g_irg_js$, after you distribute. It doesn’t really simplify anything. The products of pairs of generators is only a small subset of those...
$endgroup$
– rschwieb
Dec 16 '18 at 18:11
$begingroup$
@UnexpectedExpectation if you have a generating set for $J(R)$, the products will be of the form $g_irg_js$, after you distribute. It doesn’t really simplify anything. The products of pairs of generators is only a small subset of those...
$endgroup$
– rschwieb
Dec 16 '18 at 18:11
add a comment |
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There seems to be a missing assumption for $A$, since the Jacobson radical of a ring is not nilpotent in general (take $A=mathbb{R}[[x]]$, for example). Is it possible that Pierce assumes that $A$ is Artinian?
$endgroup$
– Pierre-Guy Plamondon
Dec 16 '18 at 12:37
$begingroup$
@Pierre-GuyPlamondon Oh, sorry. Indeed, $A$ is assumed to be right-Artinian.
$endgroup$
– UnexpectedExpectation
Dec 16 '18 at 13:05