Showing that $(I+J)/(Icap J)cong (I+J)/Itimes (I+J)/J$.
$begingroup$
Show that $(I+J)/(Icap J)cong (I+J)/Itimes (I+J)/J$, where $I,J$ are ideals in a commutative ring $R$ with identity.
What I did was first show that $phi: I+Jto (I+J)/Itimes (I+J)/J$ defined by $phi(r+s)=(r+s+I,r+s+J)$ is a ring homomorphism, and then I applied the first isomorphism theorem.
I only showed that $phi$ is well-defined and operation-preserving before applying the first isomorphism theorem. I also found that $ker(phi)=Icap J$.
Is this sufficient enough to show it is an isomorphism? My teacher said I need to show it is surjective too. Why?
abstract-algebra ring-theory ring-isomorphism
asked Dec 10 '18 at 3:12
numericalorangenumericalorange
1,735311
$endgroup$
add a comment |
$begingroup$
Show that $(I+J)/(Icap J)cong (I+J)/Itimes (I+J)/J$, where $I,J$ are ideals in a commutative ring $R$ with identity.
What I did was first show that $phi: I+Jto (I+J)/Itimes (I+J)/J$ defined by $phi(r+s)=(r+s+I,r+s+J)$ is a ring homomorphism, and then I applied the first isomorphism theorem.
I only showed that $phi$ is well-defined and operation-preserving before applying the first isomorphism theorem. I also found that $ker(phi)=Icap J$.
Is this sufficient enough to show it is an isomorphism? My teacher said I need to show it is surjective too. Why?
abstract-algebra ring-theory ring-isomorphism
asked Dec 10 '18 at 3:12
numericalorangenumericalorange
1,735311
$endgroup$
1
$begingroup$
The first isomorphism theorem says that if $phi: R longrightarrow S$ is a ring homomorphism with kernel $I$,. then $R/I cong text{Im}(phi)$. So for your proof to work you need to show that the image of the map is your ring on the right side, in other words you need to show that it is onto.
$endgroup$
– Anurag A
Dec 10 '18 at 3:16
1
$begingroup$
I think you try to use the fundamental theorem of homomorphism: $R/mathrm {Ker}, phi cong mathrm {Im}, varphi$, so you need to show that $mathrm {Im }, varphi= (I+J)/I times (I+J)/J$, i.e. the mapping is surjective.
$endgroup$
– xbh
Dec 10 '18 at 3:18
$begingroup$
Thanks for the input. I get it now! I will make sure to remember to show surjectivity.
$endgroup$
– numericalorange
Dec 10 '18 at 3:19
add a comment |
$begingroup$
Show that $(I+J)/(Icap J)cong (I+J)/Itimes (I+J)/J$, where $I,J$ are ideals in a commutative ring $R$ with identity.
What I did was first show that $phi: I+Jto (I+J)/Itimes (I+J)/J$ defined by $phi(r+s)=(r+s+I,r+s+J)$ is a ring homomorphism, and then I applied the first isomorphism theorem.
I only showed that $phi$ is well-defined and operation-preserving before applying the first isomorphism theorem. I also found that $ker(phi)=Icap J$.
Is this sufficient enough to show it is an isomorphism? My teacher said I need to show it is surjective too. Why?
abstract-algebra ring-theory ring-isomorphism
asked Dec 10 '18 at 3:12
numericalorangenumericalorange
1,735311
$endgroup$
Show that $(I+J)/(Icap J)cong (I+J)/Itimes (I+J)/J$, where $I,J$ are ideals in a commutative ring $R$ with identity.
What I did was first show that $phi: I+Jto (I+J)/Itimes (I+J)/J$ defined by $phi(r+s)=(r+s+I,r+s+J)$ is a ring homomorphism, and then I applied the first isomorphism theorem.
I only showed that $phi$ is well-defined and operation-preserving before applying the first isomorphism theorem. I also found that $ker(phi)=Icap J$.
Is this sufficient enough to show it is an isomorphism? My teacher said I need to show it is surjective too. Why?
abstract-algebra ring-theory ring-isomorphism
abstract-algebra ring-theory ring-isomorphism
asked Dec 10 '18 at 3:12
numericalorangenumericalorange
1,735311
asked Dec 10 '18 at 3:12
numericalorangenumericalorange
1,735311
asked Dec 10 '18 at 3:12
numericalorangenumericalorange
1,735311
asked Dec 10 '18 at 3:12
numericalorangenumericalorange
1,735311
asked Dec 10 '18 at 3:12
numericalorangenumericalorange
1,735311
1,735311
1
$begingroup$
The first isomorphism theorem says that if $phi: R longrightarrow S$ is a ring homomorphism with kernel $I$,. then $R/I cong text{Im}(phi)$. So for your proof to work you need to show that the image of the map is your ring on the right side, in other words you need to show that it is onto.
$endgroup$
– Anurag A
Dec 10 '18 at 3:16
1
$begingroup$
I think you try to use the fundamental theorem of homomorphism: $R/mathrm {Ker}, phi cong mathrm {Im}, varphi$, so you need to show that $mathrm {Im }, varphi= (I+J)/I times (I+J)/J$, i.e. the mapping is surjective.
$endgroup$
– xbh
Dec 10 '18 at 3:18
$begingroup$
Thanks for the input. I get it now! I will make sure to remember to show surjectivity.
$endgroup$
– numericalorange
Dec 10 '18 at 3:19
add a comment |
1
$begingroup$
The first isomorphism theorem says that if $phi: R longrightarrow S$ is a ring homomorphism with kernel $I$,. then $R/I cong text{Im}(phi)$. So for your proof to work you need to show that the image of the map is your ring on the right side, in other words you need to show that it is onto.
$endgroup$
– Anurag A
Dec 10 '18 at 3:16
1
$begingroup$
I think you try to use the fundamental theorem of homomorphism: $R/mathrm {Ker}, phi cong mathrm {Im}, varphi$, so you need to show that $mathrm {Im }, varphi= (I+J)/I times (I+J)/J$, i.e. the mapping is surjective.
$endgroup$
– xbh
Dec 10 '18 at 3:18
$begingroup$
Thanks for the input. I get it now! I will make sure to remember to show surjectivity.
$endgroup$
– numericalorange
Dec 10 '18 at 3:19
1
1
$begingroup$
The first isomorphism theorem says that if $phi: R longrightarrow S$ is a ring homomorphism with kernel $I$,. then $R/I cong text{Im}(phi)$. So for your proof to work you need to show that the image of the map is your ring on the right side, in other words you need to show that it is onto.
$endgroup$
– Anurag A
Dec 10 '18 at 3:16
$begingroup$
The first isomorphism theorem says that if $phi: R longrightarrow S$ is a ring homomorphism with kernel $I$,. then $R/I cong text{Im}(phi)$. So for your proof to work you need to show that the image of the map is your ring on the right side, in other words you need to show that it is onto.
$endgroup$
– Anurag A
Dec 10 '18 at 3:16
1
1
$begingroup$
I think you try to use the fundamental theorem of homomorphism: $R/mathrm {Ker}, phi cong mathrm {Im}, varphi$, so you need to show that $mathrm {Im }, varphi= (I+J)/I times (I+J)/J$, i.e. the mapping is surjective.
$endgroup$
– xbh
Dec 10 '18 at 3:18
$begingroup$
I think you try to use the fundamental theorem of homomorphism: $R/mathrm {Ker}, phi cong mathrm {Im}, varphi$, so you need to show that $mathrm {Im }, varphi= (I+J)/I times (I+J)/J$, i.e. the mapping is surjective.
$endgroup$
– xbh
Dec 10 '18 at 3:18
$begingroup$
Thanks for the input. I get it now! I will make sure to remember to show surjectivity.
$endgroup$
– numericalorange
Dec 10 '18 at 3:19
$begingroup$
Thanks for the input. I get it now! I will make sure to remember to show surjectivity.
$endgroup$
– numericalorange
Dec 10 '18 at 3:19
add a comment |
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$begingroup$
The first isomorphism theorem says that if $phi: R longrightarrow S$ is a ring homomorphism with kernel $I$,. then $R/I cong text{Im}(phi)$. So for your proof to work you need to show that the image of the map is your ring on the right side, in other words you need to show that it is onto.
$endgroup$
– Anurag A
Dec 10 '18 at 3:16
1
$begingroup$
I think you try to use the fundamental theorem of homomorphism: $R/mathrm {Ker}, phi cong mathrm {Im}, varphi$, so you need to show that $mathrm {Im }, varphi= (I+J)/I times (I+J)/J$, i.e. the mapping is surjective.
$endgroup$
– xbh
Dec 10 '18 at 3:18
$begingroup$
Thanks for the input. I get it now! I will make sure to remember to show surjectivity.
$endgroup$
– numericalorange
Dec 10 '18 at 3:19