Statement about semisimple quotient ring.
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I was doing some reading and I came across the following statement:
"Let $R$ be a ring and let $I$ be a two-sided ideal of $R$ such that $R/I$ is semisimple. Then $I$ is a finite intersection of maximal ideals $M subset R$ such that $R/M$ is simple Artinian. Therefore, we may as well assume that $R/I$ is simple Artinian."
I have difficulties understanding this statement. Does anyone care to explain?
abstract-algebra ring-theory
asked Nov 12 at 16:34
J.Bosser
197110
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1
down vote
favorite
I was doing some reading and I came across the following statement:
"Let $R$ be a ring and let $I$ be a two-sided ideal of $R$ such that $R/I$ is semisimple. Then $I$ is a finite intersection of maximal ideals $M subset R$ such that $R/M$ is simple Artinian. Therefore, we may as well assume that $R/I$ is simple Artinian."
I have difficulties understanding this statement. Does anyone care to explain?
abstract-algebra ring-theory
asked Nov 12 at 16:34
J.Bosser
197110
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I was doing some reading and I came across the following statement:
"Let $R$ be a ring and let $I$ be a two-sided ideal of $R$ such that $R/I$ is semisimple. Then $I$ is a finite intersection of maximal ideals $M subset R$ such that $R/M$ is simple Artinian. Therefore, we may as well assume that $R/I$ is simple Artinian."
I have difficulties understanding this statement. Does anyone care to explain?
abstract-algebra ring-theory
asked Nov 12 at 16:34
J.Bosser
197110
I was doing some reading and I came across the following statement:
"Let $R$ be a ring and let $I$ be a two-sided ideal of $R$ such that $R/I$ is semisimple. Then $I$ is a finite intersection of maximal ideals $M subset R$ such that $R/M$ is simple Artinian. Therefore, we may as well assume that $R/I$ is simple Artinian."
I have difficulties understanding this statement. Does anyone care to explain?
abstract-algebra ring-theory
abstract-algebra ring-theory
asked Nov 12 at 16:34
J.Bosser
197110
asked Nov 12 at 16:34
J.Bosser
197110
asked Nov 12 at 16:34
J.Bosser
197110
asked Nov 12 at 16:34
J.Bosser
197110
asked Nov 12 at 16:34
J.Bosser
197110
197110
add a comment |
add a comment |
1 Answer
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A semisimple ring $prod_{i=1}^n R_i$ (each $R_i$ simple Artinian) has exactly $n$ maximal ideals, one for each coordinate. They are all of the form $prod_{i=1}^n I_i$ where $I_j={0}$ for one $j$ and $I_i=R_i$ for $ineq j$.
Now, the maximal ideals of $R/I$ correspond exactly to the maximal ideals of $R$ containing $I$. Therefore there are $n$ maximal ideals of $R$ containing $I$, and their intersection is $I$.
I do not know precisely what the context is, so I cannot comment on why it is acceptable to handle the case where $R/I$ is simple Artinian. You would have to provide more background information.
answered Nov 12 at 17:06
rschwieb
103k1299238
This statement is the start of the solution of Exercise 3.20, as formulated on page 44 of the book "Exercises in Classical Ring Theory", by T.Y. Lam.
– J.Bosser
Nov 12 at 17:24
Exercise 3.20: Let $R$ be a ring and $I=bigcap_{x in X}I_x$, where $I_x$ is an ideal of $R$ for each $x in X$. Prove that if $X$ is finite and every $R/I_x$ is semisimple, then $R/I$ is semisimple.
– J.Bosser
Nov 12 at 17:25
@J.Bosser OK, then in that case, you see that if $I$ is a finite intersection of ideals such that quotients by those ideals are semisimple. then the lemma above lets you rewrite the problem equivalently assuming the collection consists purely of ideals which create a simple artinian quotient. You'll still only have finitely many. It just lets you use the "apparently stronger" (although actually just equivalent) statement that you have a finite collection of ideals intersecting to $I$ which are maximal.
– rschwieb
Nov 12 at 17:41
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
A semisimple ring $prod_{i=1}^n R_i$ (each $R_i$ simple Artinian) has exactly $n$ maximal ideals, one for each coordinate. They are all of the form $prod_{i=1}^n I_i$ where $I_j={0}$ for one $j$ and $I_i=R_i$ for $ineq j$.
Now, the maximal ideals of $R/I$ correspond exactly to the maximal ideals of $R$ containing $I$. Therefore there are $n$ maximal ideals of $R$ containing $I$, and their intersection is $I$.
I do not know precisely what the context is, so I cannot comment on why it is acceptable to handle the case where $R/I$ is simple Artinian. You would have to provide more background information.
answered Nov 12 at 17:06
rschwieb
103k1299238
This statement is the start of the solution of Exercise 3.20, as formulated on page 44 of the book "Exercises in Classical Ring Theory", by T.Y. Lam.
– J.Bosser
Nov 12 at 17:24
Exercise 3.20: Let $R$ be a ring and $I=bigcap_{x in X}I_x$, where $I_x$ is an ideal of $R$ for each $x in X$. Prove that if $X$ is finite and every $R/I_x$ is semisimple, then $R/I$ is semisimple.
– J.Bosser
Nov 12 at 17:25
@J.Bosser OK, then in that case, you see that if $I$ is a finite intersection of ideals such that quotients by those ideals are semisimple. then the lemma above lets you rewrite the problem equivalently assuming the collection consists purely of ideals which create a simple artinian quotient. You'll still only have finitely many. It just lets you use the "apparently stronger" (although actually just equivalent) statement that you have a finite collection of ideals intersecting to $I$ which are maximal.
– rschwieb
Nov 12 at 17:41
add a comment |
up vote
1
down vote
A semisimple ring $prod_{i=1}^n R_i$ (each $R_i$ simple Artinian) has exactly $n$ maximal ideals, one for each coordinate. They are all of the form $prod_{i=1}^n I_i$ where $I_j={0}$ for one $j$ and $I_i=R_i$ for $ineq j$.
Now, the maximal ideals of $R/I$ correspond exactly to the maximal ideals of $R$ containing $I$. Therefore there are $n$ maximal ideals of $R$ containing $I$, and their intersection is $I$.
I do not know precisely what the context is, so I cannot comment on why it is acceptable to handle the case where $R/I$ is simple Artinian. You would have to provide more background information.
answered Nov 12 at 17:06
rschwieb
103k1299238
This statement is the start of the solution of Exercise 3.20, as formulated on page 44 of the book "Exercises in Classical Ring Theory", by T.Y. Lam.
– J.Bosser
Nov 12 at 17:24
Exercise 3.20: Let $R$ be a ring and $I=bigcap_{x in X}I_x$, where $I_x$ is an ideal of $R$ for each $x in X$. Prove that if $X$ is finite and every $R/I_x$ is semisimple, then $R/I$ is semisimple.
– J.Bosser
Nov 12 at 17:25
@J.Bosser OK, then in that case, you see that if $I$ is a finite intersection of ideals such that quotients by those ideals are semisimple. then the lemma above lets you rewrite the problem equivalently assuming the collection consists purely of ideals which create a simple artinian quotient. You'll still only have finitely many. It just lets you use the "apparently stronger" (although actually just equivalent) statement that you have a finite collection of ideals intersecting to $I$ which are maximal.
– rschwieb
Nov 12 at 17:41
add a comment |
up vote
1
down vote
up vote
1
down vote
A semisimple ring $prod_{i=1}^n R_i$ (each $R_i$ simple Artinian) has exactly $n$ maximal ideals, one for each coordinate. They are all of the form $prod_{i=1}^n I_i$ where $I_j={0}$ for one $j$ and $I_i=R_i$ for $ineq j$.
Now, the maximal ideals of $R/I$ correspond exactly to the maximal ideals of $R$ containing $I$. Therefore there are $n$ maximal ideals of $R$ containing $I$, and their intersection is $I$.
I do not know precisely what the context is, so I cannot comment on why it is acceptable to handle the case where $R/I$ is simple Artinian. You would have to provide more background information.
answered Nov 12 at 17:06
rschwieb
103k1299238
A semisimple ring $prod_{i=1}^n R_i$ (each $R_i$ simple Artinian) has exactly $n$ maximal ideals, one for each coordinate. They are all of the form $prod_{i=1}^n I_i$ where $I_j={0}$ for one $j$ and $I_i=R_i$ for $ineq j$.
Now, the maximal ideals of $R/I$ correspond exactly to the maximal ideals of $R$ containing $I$. Therefore there are $n$ maximal ideals of $R$ containing $I$, and their intersection is $I$.
I do not know precisely what the context is, so I cannot comment on why it is acceptable to handle the case where $R/I$ is simple Artinian. You would have to provide more background information.
answered Nov 12 at 17:06
rschwieb
103k1299238
answered Nov 12 at 17:06
rschwieb
103k1299238
answered Nov 12 at 17:06
rschwieb
103k1299238
answered Nov 12 at 17:06
rschwieb
103k1299238
103k1299238
This statement is the start of the solution of Exercise 3.20, as formulated on page 44 of the book "Exercises in Classical Ring Theory", by T.Y. Lam.
– J.Bosser
Nov 12 at 17:24
Exercise 3.20: Let $R$ be a ring and $I=bigcap_{x in X}I_x$, where $I_x$ is an ideal of $R$ for each $x in X$. Prove that if $X$ is finite and every $R/I_x$ is semisimple, then $R/I$ is semisimple.
– J.Bosser
Nov 12 at 17:25
@J.Bosser OK, then in that case, you see that if $I$ is a finite intersection of ideals such that quotients by those ideals are semisimple. then the lemma above lets you rewrite the problem equivalently assuming the collection consists purely of ideals which create a simple artinian quotient. You'll still only have finitely many. It just lets you use the "apparently stronger" (although actually just equivalent) statement that you have a finite collection of ideals intersecting to $I$ which are maximal.
– rschwieb
Nov 12 at 17:41
add a comment |
This statement is the start of the solution of Exercise 3.20, as formulated on page 44 of the book "Exercises in Classical Ring Theory", by T.Y. Lam.
– J.Bosser
Nov 12 at 17:24
Exercise 3.20: Let $R$ be a ring and $I=bigcap_{x in X}I_x$, where $I_x$ is an ideal of $R$ for each $x in X$. Prove that if $X$ is finite and every $R/I_x$ is semisimple, then $R/I$ is semisimple.
– J.Bosser
Nov 12 at 17:25
@J.Bosser OK, then in that case, you see that if $I$ is a finite intersection of ideals such that quotients by those ideals are semisimple. then the lemma above lets you rewrite the problem equivalently assuming the collection consists purely of ideals which create a simple artinian quotient. You'll still only have finitely many. It just lets you use the "apparently stronger" (although actually just equivalent) statement that you have a finite collection of ideals intersecting to $I$ which are maximal.
– rschwieb
Nov 12 at 17:41
This statement is the start of the solution of Exercise 3.20, as formulated on page 44 of the book "Exercises in Classical Ring Theory", by T.Y. Lam.
– J.Bosser
Nov 12 at 17:24
This statement is the start of the solution of Exercise 3.20, as formulated on page 44 of the book "Exercises in Classical Ring Theory", by T.Y. Lam.
– J.Bosser
Nov 12 at 17:24
Exercise 3.20: Let $R$ be a ring and $I=bigcap_{x in X}I_x$, where $I_x$ is an ideal of $R$ for each $x in X$. Prove that if $X$ is finite and every $R/I_x$ is semisimple, then $R/I$ is semisimple.
– J.Bosser
Nov 12 at 17:25
Exercise 3.20: Let $R$ be a ring and $I=bigcap_{x in X}I_x$, where $I_x$ is an ideal of $R$ for each $x in X$. Prove that if $X$ is finite and every $R/I_x$ is semisimple, then $R/I$ is semisimple.
– J.Bosser
Nov 12 at 17:25
@J.Bosser OK, then in that case, you see that if $I$ is a finite intersection of ideals such that quotients by those ideals are semisimple. then the lemma above lets you rewrite the problem equivalently assuming the collection consists purely of ideals which create a simple artinian quotient. You'll still only have finitely many. It just lets you use the "apparently stronger" (although actually just equivalent) statement that you have a finite collection of ideals intersecting to $I$ which are maximal.
– rschwieb
Nov 12 at 17:41
@J.Bosser OK, then in that case, you see that if $I$ is a finite intersection of ideals such that quotients by those ideals are semisimple. then the lemma above lets you rewrite the problem equivalently assuming the collection consists purely of ideals which create a simple artinian quotient. You'll still only have finitely many. It just lets you use the "apparently stronger" (although actually just equivalent) statement that you have a finite collection of ideals intersecting to $I$ which are maximal.
– rschwieb
Nov 12 at 17:41
add a comment |
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