Prove that $ I + J = R iff b + J $ is invertible in $ R / J$, where $ I = langle b rangle$
$begingroup$
Let $R$ be an integral domain and $I, J$ its ideals, where $I = langle b rangle$, for a $ b in R$. Prove that $ I + J = R iff b + J $ is invertible in $ R / J$
If it is a duplicate, I apologize, but I've already looked up for it.
I was able to show this:
$ I + J = R iff (forall r in R) r = i + j $, for some $ i in I, j in J iff (forall r in R) r = ba + j $, for some $ a in R, j in J$. Am I missing something or am I in a completely wrong direction? Thank you.
abstract-algebra proof-verification ring-theory ideals
asked Sep 20 '18 at 11:45
Nemanja BericNemanja Beric
37018
$endgroup$
add a comment |
$begingroup$
Let $R$ be an integral domain and $I, J$ its ideals, where $I = langle b rangle$, for a $ b in R$. Prove that $ I + J = R iff b + J $ is invertible in $ R / J$
If it is a duplicate, I apologize, but I've already looked up for it.
I was able to show this:
$ I + J = R iff (forall r in R) r = i + j $, for some $ i in I, j in J iff (forall r in R) r = ba + j $, for some $ a in R, j in J$. Am I missing something or am I in a completely wrong direction? Thank you.
abstract-algebra proof-verification ring-theory ideals
asked Sep 20 '18 at 11:45
Nemanja BericNemanja Beric
37018
$endgroup$
add a comment |
$begingroup$
Let $R$ be an integral domain and $I, J$ its ideals, where $I = langle b rangle$, for a $ b in R$. Prove that $ I + J = R iff b + J $ is invertible in $ R / J$
If it is a duplicate, I apologize, but I've already looked up for it.
I was able to show this:
$ I + J = R iff (forall r in R) r = i + j $, for some $ i in I, j in J iff (forall r in R) r = ba + j $, for some $ a in R, j in J$. Am I missing something or am I in a completely wrong direction? Thank you.
abstract-algebra proof-verification ring-theory ideals
asked Sep 20 '18 at 11:45
Nemanja BericNemanja Beric
37018
$endgroup$
Let $R$ be an integral domain and $I, J$ its ideals, where $I = langle b rangle$, for a $ b in R$. Prove that $ I + J = R iff b + J $ is invertible in $ R / J$
If it is a duplicate, I apologize, but I've already looked up for it.
I was able to show this:
$ I + J = R iff (forall r in R) r = i + j $, for some $ i in I, j in J iff (forall r in R) r = ba + j $, for some $ a in R, j in J$. Am I missing something or am I in a completely wrong direction? Thank you.
abstract-algebra proof-verification ring-theory ideals
abstract-algebra proof-verification ring-theory ideals
asked Sep 20 '18 at 11:45
Nemanja BericNemanja Beric
37018
asked Sep 20 '18 at 11:45
Nemanja BericNemanja Beric
37018
edited Dec 19 '18 at 16:21
user593746
asked Sep 20 '18 at 11:45
Nemanja BericNemanja Beric
37018
asked Sep 20 '18 at 11:45
Nemanja BericNemanja Beric
37018
asked Sep 20 '18 at 11:45
Nemanja BericNemanja Beric
37018
37018
add a comment |
add a comment |
1 Answer
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$begingroup$
You were almost there. Note that $R$ is an integral domain, so it contains the unity. Thus, $I+J=R$ if and only if $1in I+J$. That is, $I+J=R$ iff $ab+j=1$ for some $ain R$ and $jin J$. (That is, you only have to take $r$ to be $1$ in your attempt.) What next?
$endgroup$
$begingroup$
Well, if ab + j = 1, for some a, that means that b is invertible in R / J, because R / J = r + J = {r + j : j in J}, so for j = 0 (which is in J), ab + j= 1 + 0, i.e. b is invertible. Am I right?
$endgroup$
– Nemanja Beric
Sep 20 '18 at 12:01
1
$begingroup$
@NemanjaBeric I would write it in the following way: $ab+j=1$ implies $ab+J=1+J$. Thus, $(a+J)(b+J)=1+J$. That is, $a+Jin R/J$ is the inverse of $b+Jin R/J$.
$endgroup$
– user593746
Sep 20 '18 at 12:13
$begingroup$
Right, right. I must not omit the "+ J"
$endgroup$
– Nemanja Beric
Sep 20 '18 at 12:14
add a comment |
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1 Answer
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$begingroup$
You were almost there. Note that $R$ is an integral domain, so it contains the unity. Thus, $I+J=R$ if and only if $1in I+J$. That is, $I+J=R$ iff $ab+j=1$ for some $ain R$ and $jin J$. (That is, you only have to take $r$ to be $1$ in your attempt.) What next?
$endgroup$
$begingroup$
Well, if ab + j = 1, for some a, that means that b is invertible in R / J, because R / J = r + J = {r + j : j in J}, so for j = 0 (which is in J), ab + j= 1 + 0, i.e. b is invertible. Am I right?
$endgroup$
– Nemanja Beric
Sep 20 '18 at 12:01
1
$begingroup$
@NemanjaBeric I would write it in the following way: $ab+j=1$ implies $ab+J=1+J$. Thus, $(a+J)(b+J)=1+J$. That is, $a+Jin R/J$ is the inverse of $b+Jin R/J$.
$endgroup$
– user593746
Sep 20 '18 at 12:13
$begingroup$
Right, right. I must not omit the "+ J"
$endgroup$
– Nemanja Beric
Sep 20 '18 at 12:14
add a comment |
$begingroup$
You were almost there. Note that $R$ is an integral domain, so it contains the unity. Thus, $I+J=R$ if and only if $1in I+J$. That is, $I+J=R$ iff $ab+j=1$ for some $ain R$ and $jin J$. (That is, you only have to take $r$ to be $1$ in your attempt.) What next?
$endgroup$
$begingroup$
Well, if ab + j = 1, for some a, that means that b is invertible in R / J, because R / J = r + J = {r + j : j in J}, so for j = 0 (which is in J), ab + j= 1 + 0, i.e. b is invertible. Am I right?
$endgroup$
– Nemanja Beric
Sep 20 '18 at 12:01
1
$begingroup$
@NemanjaBeric I would write it in the following way: $ab+j=1$ implies $ab+J=1+J$. Thus, $(a+J)(b+J)=1+J$. That is, $a+Jin R/J$ is the inverse of $b+Jin R/J$.
$endgroup$
– user593746
Sep 20 '18 at 12:13
$begingroup$
Right, right. I must not omit the "+ J"
$endgroup$
– Nemanja Beric
Sep 20 '18 at 12:14
add a comment |
$begingroup$
You were almost there. Note that $R$ is an integral domain, so it contains the unity. Thus, $I+J=R$ if and only if $1in I+J$. That is, $I+J=R$ iff $ab+j=1$ for some $ain R$ and $jin J$. (That is, you only have to take $r$ to be $1$ in your attempt.) What next?
$endgroup$
You were almost there. Note that $R$ is an integral domain, so it contains the unity. Thus, $I+J=R$ if and only if $1in I+J$. That is, $I+J=R$ iff $ab+j=1$ for some $ain R$ and $jin J$. (That is, you only have to take $r$ to be $1$ in your attempt.) What next?
answered Sep 20 '18 at 11:51
user593746
$begingroup$
Well, if ab + j = 1, for some a, that means that b is invertible in R / J, because R / J = r + J = {r + j : j in J}, so for j = 0 (which is in J), ab + j= 1 + 0, i.e. b is invertible. Am I right?
$endgroup$
– Nemanja Beric
Sep 20 '18 at 12:01
1
$begingroup$
@NemanjaBeric I would write it in the following way: $ab+j=1$ implies $ab+J=1+J$. Thus, $(a+J)(b+J)=1+J$. That is, $a+Jin R/J$ is the inverse of $b+Jin R/J$.
$endgroup$
– user593746
Sep 20 '18 at 12:13
$begingroup$
Right, right. I must not omit the "+ J"
$endgroup$
– Nemanja Beric
Sep 20 '18 at 12:14
add a comment |
$begingroup$
Well, if ab + j = 1, for some a, that means that b is invertible in R / J, because R / J = r + J = {r + j : j in J}, so for j = 0 (which is in J), ab + j= 1 + 0, i.e. b is invertible. Am I right?
$endgroup$
– Nemanja Beric
Sep 20 '18 at 12:01
1
$begingroup$
@NemanjaBeric I would write it in the following way: $ab+j=1$ implies $ab+J=1+J$. Thus, $(a+J)(b+J)=1+J$. That is, $a+Jin R/J$ is the inverse of $b+Jin R/J$.
$endgroup$
– user593746
Sep 20 '18 at 12:13
$begingroup$
Right, right. I must not omit the "+ J"
$endgroup$
– Nemanja Beric
Sep 20 '18 at 12:14
$begingroup$
Well, if ab + j = 1, for some a, that means that b is invertible in R / J, because R / J = r + J = {r + j : j in J}, so for j = 0 (which is in J), ab + j= 1 + 0, i.e. b is invertible. Am I right?
$endgroup$
– Nemanja Beric
Sep 20 '18 at 12:01
$begingroup$
Well, if ab + j = 1, for some a, that means that b is invertible in R / J, because R / J = r + J = {r + j : j in J}, so for j = 0 (which is in J), ab + j= 1 + 0, i.e. b is invertible. Am I right?
$endgroup$
– Nemanja Beric
Sep 20 '18 at 12:01
1
1
$begingroup$
@NemanjaBeric I would write it in the following way: $ab+j=1$ implies $ab+J=1+J$. Thus, $(a+J)(b+J)=1+J$. That is, $a+Jin R/J$ is the inverse of $b+Jin R/J$.
$endgroup$
– user593746
Sep 20 '18 at 12:13
$begingroup$
@NemanjaBeric I would write it in the following way: $ab+j=1$ implies $ab+J=1+J$. Thus, $(a+J)(b+J)=1+J$. That is, $a+Jin R/J$ is the inverse of $b+Jin R/J$.
$endgroup$
– user593746
Sep 20 '18 at 12:13
$begingroup$
Right, right. I must not omit the "+ J"
$endgroup$
– Nemanja Beric
Sep 20 '18 at 12:14
$begingroup$
Right, right. I must not omit the "+ J"
$endgroup$
– Nemanja Beric
Sep 20 '18 at 12:14
add a comment |
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