Mixing
A condition equivalent to $R$ being a direct summand
0
$begingroup$
If $Rsubset S$ are rings, then why is saying that $R$ is a summand of $S$ as an $R$-module the same as saying that there is an $R$-module homomorphism $Sto R$ that fixes all elements of $R$?
The first condition means that $Ssimeq Roplus R'$ as $R$-modules for some $R$-module $R'$. The only natural map is $Roplus R'to R, (r,r')mapsto r$. But what confuses me is that $R$ isn't a submodule of $S$ now (though it is isomorphic to the submodule $Roplus {0})$.The above map indeed fixes all elements of $Roplus {0}$. But how to deal with the fact that $Rsubset S$ is not true?
Also, how to prove the other direction?
abstract-algebra modules direct-sum
asked Jan 17 at 3:33
user437309user437309
695313
$endgroup$
|
show 2 more comments
0
$begingroup$
If $Rsubset S$ are rings, then why is saying that $R$ is a summand of $S$ as an $R$-module the same as saying that there is an $R$-module homomorphism $Sto R$ that fixes all elements of $R$?
The first condition means that $Ssimeq Roplus R'$ as $R$-modules for some $R$-module $R'$. The only natural map is $Roplus R'to R, (r,r')mapsto r$. But what confuses me is that $R$ isn't a submodule of $S$ now (though it is isomorphic to the submodule $Roplus {0})$.The above map indeed fixes all elements of $Roplus {0}$. But how to deal with the fact that $Rsubset S$ is not true?
Also, how to prove the other direction?
abstract-algebra modules direct-sum
asked Jan 17 at 3:33
user437309user437309
695313
$endgroup$
$begingroup$
Are you familiar with the notion of “internal direct sum”? You seem to be viewing the direct sum only as an “external” direct sum; but one says that $R$ is a direct summand of $S$ when $Rsubseteq S$ if there is a submodule $R’$ of $S$ such that $Rcap R’={0}$ and $R+R’ = S$. In that case, it is not hard to verify that each element $sin S$ can be written uniquely as $r+r’$ with $rin R$ and $r’in R’$, so that $Scong Roplus R’$ (the “external” direct sum made up of ordered pairs with coordinate-wise operations)
$endgroup$
– Arturo Magidin
Jan 17 at 3:37
$begingroup$
For the converse, verify that the kernel of the homomorphism is a complement to $R$ (i.e., it intersects $R$ trivially, and every element of $S$ can be written as a sum of an element of $R$ and an element of the kernel)
$endgroup$
– Arturo Magidin
Jan 17 at 3:39
$begingroup$
@ArturoMagidin So if I view it as an internal direct sum, is the $R$-module homomorphism in question the inclusion map? In particular, are you talking about the inclusion map in the second comment?
$endgroup$
– user437309
Jan 17 at 3:43
$begingroup$
What do you mean “the homomorphism in question”? The only one mentioned goes from $S$ to $R$, not form $R$ to $S$, so how could it be an inclusion?
$endgroup$
– Arturo Magidin
Jan 17 at 3:50
$begingroup$
@ArturoMagidin Sorry, I mistakenly thought that it was from $R$ to $S$. But then I still need to use external direct sums to define the homomorphism? What's the advantage of thinking about internal direct sums then?
$endgroup$
– user437309
Jan 17 at 3:55
|
show 2 more comments
0
0
0
$begingroup$
If $Rsubset S$ are rings, then why is saying that $R$ is a summand of $S$ as an $R$-module the same as saying that there is an $R$-module homomorphism $Sto R$ that fixes all elements of $R$?
The first condition means that $Ssimeq Roplus R'$ as $R$-modules for some $R$-module $R'$. The only natural map is $Roplus R'to R, (r,r')mapsto r$. But what confuses me is that $R$ isn't a submodule of $S$ now (though it is isomorphic to the submodule $Roplus {0})$.The above map indeed fixes all elements of $Roplus {0}$. But how to deal with the fact that $Rsubset S$ is not true?
Also, how to prove the other direction?
abstract-algebra modules direct-sum
asked Jan 17 at 3:33
user437309user437309
695313
$endgroup$
If $Rsubset S$ are rings, then why is saying that $R$ is a summand of $S$ as an $R$-module the same as saying that there is an $R$-module homomorphism $Sto R$ that fixes all elements of $R$?
The first condition means that $Ssimeq Roplus R'$ as $R$-modules for some $R$-module $R'$. The only natural map is $Roplus R'to R, (r,r')mapsto r$. But what confuses me is that $R$ isn't a submodule of $S$ now (though it is isomorphic to the submodule $Roplus {0})$.The above map indeed fixes all elements of $Roplus {0}$. But how to deal with the fact that $Rsubset S$ is not true?
Also, how to prove the other direction?
abstract-algebra modules direct-sum
abstract-algebra modules direct-sum
asked Jan 17 at 3:33
user437309user437309
695313
asked Jan 17 at 3:33
user437309user437309
695313
asked Jan 17 at 3:33
user437309user437309
695313
asked Jan 17 at 3:33
user437309user437309
695313
asked Jan 17 at 3:33
user437309user437309
695313
695313
$begingroup$
Are you familiar with the notion of “internal direct sum”? You seem to be viewing the direct sum only as an “external” direct sum; but one says that $R$ is a direct summand of $S$ when $Rsubseteq S$ if there is a submodule $R’$ of $S$ such that $Rcap R’={0}$ and $R+R’ = S$. In that case, it is not hard to verify that each element $sin S$ can be written uniquely as $r+r’$ with $rin R$ and $r’in R’$, so that $Scong Roplus R’$ (the “external” direct sum made up of ordered pairs with coordinate-wise operations)
$endgroup$
– Arturo Magidin
Jan 17 at 3:37
$begingroup$
For the converse, verify that the kernel of the homomorphism is a complement to $R$ (i.e., it intersects $R$ trivially, and every element of $S$ can be written as a sum of an element of $R$ and an element of the kernel)
$endgroup$
– Arturo Magidin
Jan 17 at 3:39
$begingroup$
@ArturoMagidin So if I view it as an internal direct sum, is the $R$-module homomorphism in question the inclusion map? In particular, are you talking about the inclusion map in the second comment?
$endgroup$
– user437309
Jan 17 at 3:43
$begingroup$
What do you mean “the homomorphism in question”? The only one mentioned goes from $S$ to $R$, not form $R$ to $S$, so how could it be an inclusion?
$endgroup$
– Arturo Magidin
Jan 17 at 3:50
$begingroup$
@ArturoMagidin Sorry, I mistakenly thought that it was from $R$ to $S$. But then I still need to use external direct sums to define the homomorphism? What's the advantage of thinking about internal direct sums then?
$endgroup$
– user437309
Jan 17 at 3:55
|
show 2 more comments
$begingroup$
Are you familiar with the notion of “internal direct sum”? You seem to be viewing the direct sum only as an “external” direct sum; but one says that $R$ is a direct summand of $S$ when $Rsubseteq S$ if there is a submodule $R’$ of $S$ such that $Rcap R’={0}$ and $R+R’ = S$. In that case, it is not hard to verify that each element $sin S$ can be written uniquely as $r+r’$ with $rin R$ and $r’in R’$, so that $Scong Roplus R’$ (the “external” direct sum made up of ordered pairs with coordinate-wise operations)
$endgroup$
– Arturo Magidin
Jan 17 at 3:37
$begingroup$
For the converse, verify that the kernel of the homomorphism is a complement to $R$ (i.e., it intersects $R$ trivially, and every element of $S$ can be written as a sum of an element of $R$ and an element of the kernel)
$endgroup$
– Arturo Magidin
Jan 17 at 3:39
$begingroup$
@ArturoMagidin So if I view it as an internal direct sum, is the $R$-module homomorphism in question the inclusion map? In particular, are you talking about the inclusion map in the second comment?
$endgroup$
– user437309
Jan 17 at 3:43
$begingroup$
What do you mean “the homomorphism in question”? The only one mentioned goes from $S$ to $R$, not form $R$ to $S$, so how could it be an inclusion?
$endgroup$
– Arturo Magidin
Jan 17 at 3:50
$begingroup$
@ArturoMagidin Sorry, I mistakenly thought that it was from $R$ to $S$. But then I still need to use external direct sums to define the homomorphism? What's the advantage of thinking about internal direct sums then?
$endgroup$
– user437309
Jan 17 at 3:55
$begingroup$
Are you familiar with the notion of “internal direct sum”? You seem to be viewing the direct sum only as an “external” direct sum; but one says that $R$ is a direct summand of $S$ when $Rsubseteq S$ if there is a submodule $R’$ of $S$ such that $Rcap R’={0}$ and $R+R’ = S$. In that case, it is not hard to verify that each element $sin S$ can be written uniquely as $r+r’$ with $rin R$ and $r’in R’$, so that $Scong Roplus R’$ (the “external” direct sum made up of ordered pairs with coordinate-wise operations)
$endgroup$
– Arturo Magidin
Jan 17 at 3:37
$begingroup$
Are you familiar with the notion of “internal direct sum”? You seem to be viewing the direct sum only as an “external” direct sum; but one says that $R$ is a direct summand of $S$ when $Rsubseteq S$ if there is a submodule $R’$ of $S$ such that $Rcap R’={0}$ and $R+R’ = S$. In that case, it is not hard to verify that each element $sin S$ can be written uniquely as $r+r’$ with $rin R$ and $r’in R’$, so that $Scong Roplus R’$ (the “external” direct sum made up of ordered pairs with coordinate-wise operations)
$endgroup$
– Arturo Magidin
Jan 17 at 3:37
$begingroup$
For the converse, verify that the kernel of the homomorphism is a complement to $R$ (i.e., it intersects $R$ trivially, and every element of $S$ can be written as a sum of an element of $R$ and an element of the kernel)
$endgroup$
– Arturo Magidin
Jan 17 at 3:39
$begingroup$
For the converse, verify that the kernel of the homomorphism is a complement to $R$ (i.e., it intersects $R$ trivially, and every element of $S$ can be written as a sum of an element of $R$ and an element of the kernel)
$endgroup$
– Arturo Magidin
Jan 17 at 3:39
$begingroup$
@ArturoMagidin So if I view it as an internal direct sum, is the $R$-module homomorphism in question the inclusion map? In particular, are you talking about the inclusion map in the second comment?
$endgroup$
– user437309
Jan 17 at 3:43
$begingroup$
@ArturoMagidin So if I view it as an internal direct sum, is the $R$-module homomorphism in question the inclusion map? In particular, are you talking about the inclusion map in the second comment?
$endgroup$
– user437309
Jan 17 at 3:43
$begingroup$
What do you mean “the homomorphism in question”? The only one mentioned goes from $S$ to $R$, not form $R$ to $S$, so how could it be an inclusion?
$endgroup$
– Arturo Magidin
Jan 17 at 3:50
$begingroup$
What do you mean “the homomorphism in question”? The only one mentioned goes from $S$ to $R$, not form $R$ to $S$, so how could it be an inclusion?
$endgroup$
– Arturo Magidin
Jan 17 at 3:50
$begingroup$
@ArturoMagidin Sorry, I mistakenly thought that it was from $R$ to $S$. But then I still need to use external direct sums to define the homomorphism? What's the advantage of thinking about internal direct sums then?
$endgroup$
– user437309
Jan 17 at 3:55
$begingroup$
@ArturoMagidin Sorry, I mistakenly thought that it was from $R$ to $S$. But then I still need to use external direct sums to define the homomorphism? What's the advantage of thinking about internal direct sums then?
$endgroup$
– user437309
Jan 17 at 3:55
|
show 2 more comments
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$begingroup$
Are you familiar with the notion of “internal direct sum”? You seem to be viewing the direct sum only as an “external” direct sum; but one says that $R$ is a direct summand of $S$ when $Rsubseteq S$ if there is a submodule $R’$ of $S$ such that $Rcap R’={0}$ and $R+R’ = S$. In that case, it is not hard to verify that each element $sin S$ can be written uniquely as $r+r’$ with $rin R$ and $r’in R’$, so that $Scong Roplus R’$ (the “external” direct sum made up of ordered pairs with coordinate-wise operations)
$endgroup$
– Arturo Magidin
Jan 17 at 3:37
$begingroup$
For the converse, verify that the kernel of the homomorphism is a complement to $R$ (i.e., it intersects $R$ trivially, and every element of $S$ can be written as a sum of an element of $R$ and an element of the kernel)
$endgroup$
– Arturo Magidin
Jan 17 at 3:39
$begingroup$
@ArturoMagidin So if I view it as an internal direct sum, is the $R$-module homomorphism in question the inclusion map? In particular, are you talking about the inclusion map in the second comment?
$endgroup$
– user437309
Jan 17 at 3:43
$begingroup$
What do you mean “the homomorphism in question”? The only one mentioned goes from $S$ to $R$, not form $R$ to $S$, so how could it be an inclusion?
$endgroup$
– Arturo Magidin
Jan 17 at 3:50
$begingroup$
@ArturoMagidin Sorry, I mistakenly thought that it was from $R$ to $S$. But then I still need to use external direct sums to define the homomorphism? What's the advantage of thinking about internal direct sums then?
$endgroup$
– user437309
Jan 17 at 3:55