When is a Quotient Ring equal to the Zero Ring?
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I am reading Atiyah and MacDonald's 'Introduction to Commutative Algebra' and wished to check a point in one of their proofs. They state that if $A$ is a non-zero ring the following are equivalent:
(i.) $A$ is a field
(ii.) The only ideals in $A$ are $0$ and $(1)$
(iii.) Every homomorphism of $A$ to a non-zero ring $B$ is injective.
To prove that (iii.) implies (i.) they say 'Let $x$ be an element of $A$ which is not a unit. Then $(x)$ does not equal $(1)$, hence $B=A/(x)$ is not the zero ring'. To fill in the gaps here, is that because the principal ideal generated by 1 would return the original ring $A$ because you are just multiplying $1$ through by every element of the ring one-by-one, and then that gives you $B=A/A$, which is the zero ring? If anyone wants to fill in the gaps a bit more explicitly I would appreciate it.
Apologies about the basic question, as I am new to elementary abstract algebra.
abstract-algebra ring-theory commutative-algebra ideals
edited Dec 17 '18 at 20:20
rschwieb
107k12102251
asked Dec 17 '18 at 20:12
TomTom
316111
$endgroup$
add a comment |
$begingroup$
I am reading Atiyah and MacDonald's 'Introduction to Commutative Algebra' and wished to check a point in one of their proofs. They state that if $A$ is a non-zero ring the following are equivalent:
(i.) $A$ is a field
(ii.) The only ideals in $A$ are $0$ and $(1)$
(iii.) Every homomorphism of $A$ to a non-zero ring $B$ is injective.
To prove that (iii.) implies (i.) they say 'Let $x$ be an element of $A$ which is not a unit. Then $(x)$ does not equal $(1)$, hence $B=A/(x)$ is not the zero ring'. To fill in the gaps here, is that because the principal ideal generated by 1 would return the original ring $A$ because you are just multiplying $1$ through by every element of the ring one-by-one, and then that gives you $B=A/A$, which is the zero ring? If anyone wants to fill in the gaps a bit more explicitly I would appreciate it.
Apologies about the basic question, as I am new to elementary abstract algebra.
abstract-algebra ring-theory commutative-algebra ideals
edited Dec 17 '18 at 20:20
rschwieb
107k12102251
asked Dec 17 '18 at 20:12
TomTom
316111
$endgroup$
2
$begingroup$
I think you pretty much have it. Perhaps you can argue as follows: if $x$ is not invertible, then consider the ideal $langle xrangle neq R$ generated by $x$ and let $f$ be the corresponding homomorphism with kernel as $langle xrangle$. Then you have a non-injective homomorphism, which contradicts the assumption.
$endgroup$
– Anurag A
Dec 17 '18 at 20:17
add a comment |
$begingroup$
I am reading Atiyah and MacDonald's 'Introduction to Commutative Algebra' and wished to check a point in one of their proofs. They state that if $A$ is a non-zero ring the following are equivalent:
(i.) $A$ is a field
(ii.) The only ideals in $A$ are $0$ and $(1)$
(iii.) Every homomorphism of $A$ to a non-zero ring $B$ is injective.
To prove that (iii.) implies (i.) they say 'Let $x$ be an element of $A$ which is not a unit. Then $(x)$ does not equal $(1)$, hence $B=A/(x)$ is not the zero ring'. To fill in the gaps here, is that because the principal ideal generated by 1 would return the original ring $A$ because you are just multiplying $1$ through by every element of the ring one-by-one, and then that gives you $B=A/A$, which is the zero ring? If anyone wants to fill in the gaps a bit more explicitly I would appreciate it.
Apologies about the basic question, as I am new to elementary abstract algebra.
abstract-algebra ring-theory commutative-algebra ideals
edited Dec 17 '18 at 20:20
rschwieb
107k12102251
asked Dec 17 '18 at 20:12
TomTom
316111
$endgroup$
I am reading Atiyah and MacDonald's 'Introduction to Commutative Algebra' and wished to check a point in one of their proofs. They state that if $A$ is a non-zero ring the following are equivalent:
(i.) $A$ is a field
(ii.) The only ideals in $A$ are $0$ and $(1)$
(iii.) Every homomorphism of $A$ to a non-zero ring $B$ is injective.
To prove that (iii.) implies (i.) they say 'Let $x$ be an element of $A$ which is not a unit. Then $(x)$ does not equal $(1)$, hence $B=A/(x)$ is not the zero ring'. To fill in the gaps here, is that because the principal ideal generated by 1 would return the original ring $A$ because you are just multiplying $1$ through by every element of the ring one-by-one, and then that gives you $B=A/A$, which is the zero ring? If anyone wants to fill in the gaps a bit more explicitly I would appreciate it.
Apologies about the basic question, as I am new to elementary abstract algebra.
abstract-algebra ring-theory commutative-algebra ideals
abstract-algebra ring-theory commutative-algebra ideals
edited Dec 17 '18 at 20:20
rschwieb
107k12102251
asked Dec 17 '18 at 20:12
TomTom
316111
edited Dec 17 '18 at 20:20
rschwieb
107k12102251
asked Dec 17 '18 at 20:12
TomTom
316111
edited Dec 17 '18 at 20:20
rschwieb
107k12102251
edited Dec 17 '18 at 20:20
rschwieb
107k12102251
edited Dec 17 '18 at 20:20
rschwieb
107k12102251
107k12102251
asked Dec 17 '18 at 20:12
TomTom
316111
asked Dec 17 '18 at 20:12
TomTom
316111
asked Dec 17 '18 at 20:12
TomTom
316111
316111
2
$begingroup$
I think you pretty much have it. Perhaps you can argue as follows: if $x$ is not invertible, then consider the ideal $langle xrangle neq R$ generated by $x$ and let $f$ be the corresponding homomorphism with kernel as $langle xrangle$. Then you have a non-injective homomorphism, which contradicts the assumption.
$endgroup$
– Anurag A
Dec 17 '18 at 20:17
add a comment |
2
$begingroup$
I think you pretty much have it. Perhaps you can argue as follows: if $x$ is not invertible, then consider the ideal $langle xrangle neq R$ generated by $x$ and let $f$ be the corresponding homomorphism with kernel as $langle xrangle$. Then you have a non-injective homomorphism, which contradicts the assumption.
$endgroup$
– Anurag A
Dec 17 '18 at 20:17
2
2
$begingroup$
I think you pretty much have it. Perhaps you can argue as follows: if $x$ is not invertible, then consider the ideal $langle xrangle neq R$ generated by $x$ and let $f$ be the corresponding homomorphism with kernel as $langle xrangle$. Then you have a non-injective homomorphism, which contradicts the assumption.
$endgroup$
– Anurag A
Dec 17 '18 at 20:17
$begingroup$
I think you pretty much have it. Perhaps you can argue as follows: if $x$ is not invertible, then consider the ideal $langle xrangle neq R$ generated by $x$ and let $f$ be the corresponding homomorphism with kernel as $langle xrangle$. Then you have a non-injective homomorphism, which contradicts the assumption.
$endgroup$
– Anurag A
Dec 17 '18 at 20:17
add a comment |
2 Answers
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$iii)implies i):$ Every ideal $(x)$ is the kernel of the quotient homomorphism to $A/(x)$. The homomorphism being injective implies the only proper ideal is $(x)=(0)$. So for each proper ideal $(x)$, $x=0$. This means each $xneq0$ is invertible ($(x)=A$) which means the ring is a field.
answered Dec 17 '18 at 20:36
Chris CusterChris Custer
14.2k3827
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The step (iii.) $Rightarrow$ (i.) is done by contradiction. Assume for the sake of contradiction that (iii.) holds but $A$ is not a field. Then there is a non-zero element $xin A$ that is not a unit. Since it is not a unit, the ideal $I=(x)$ does not contain $1$, so $I$ is a proper ideal of $A$. Now consider the canonical homomorphism $fcolon Ato A/I$, $rmapsto r+I$. Note that $f$ is not injective since $f(x)=0$ and $xneq 0$. However, since $I$ is a proper ideal, the quotient $A/I$ is not the zero ring. (Note that $A/I$ is the zero ring if and only if $I=A$.) Hence, we got a contradiction to (iii.). We conclude that our assumption was false and that (iii.) implies (i.).
answered Dec 17 '18 at 20:26
ChristophChristoph
12.5k1642
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$begingroup$
$iii)implies i):$ Every ideal $(x)$ is the kernel of the quotient homomorphism to $A/(x)$. The homomorphism being injective implies the only proper ideal is $(x)=(0)$. So for each proper ideal $(x)$, $x=0$. This means each $xneq0$ is invertible ($(x)=A$) which means the ring is a field.
answered Dec 17 '18 at 20:36
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
$iii)implies i):$ Every ideal $(x)$ is the kernel of the quotient homomorphism to $A/(x)$. The homomorphism being injective implies the only proper ideal is $(x)=(0)$. So for each proper ideal $(x)$, $x=0$. This means each $xneq0$ is invertible ($(x)=A$) which means the ring is a field.
answered Dec 17 '18 at 20:36
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
$iii)implies i):$ Every ideal $(x)$ is the kernel of the quotient homomorphism to $A/(x)$. The homomorphism being injective implies the only proper ideal is $(x)=(0)$. So for each proper ideal $(x)$, $x=0$. This means each $xneq0$ is invertible ($(x)=A$) which means the ring is a field.
answered Dec 17 '18 at 20:36
Chris CusterChris Custer
14.2k3827
$endgroup$
$iii)implies i):$ Every ideal $(x)$ is the kernel of the quotient homomorphism to $A/(x)$. The homomorphism being injective implies the only proper ideal is $(x)=(0)$. So for each proper ideal $(x)$, $x=0$. This means each $xneq0$ is invertible ($(x)=A$) which means the ring is a field.
answered Dec 17 '18 at 20:36
Chris CusterChris Custer
14.2k3827
answered Dec 17 '18 at 20:36
Chris CusterChris Custer
14.2k3827
answered Dec 17 '18 at 20:36
Chris CusterChris Custer
14.2k3827
answered Dec 17 '18 at 20:36
Chris CusterChris Custer
14.2k3827
14.2k3827
add a comment |
add a comment |
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The step (iii.) $Rightarrow$ (i.) is done by contradiction. Assume for the sake of contradiction that (iii.) holds but $A$ is not a field. Then there is a non-zero element $xin A$ that is not a unit. Since it is not a unit, the ideal $I=(x)$ does not contain $1$, so $I$ is a proper ideal of $A$. Now consider the canonical homomorphism $fcolon Ato A/I$, $rmapsto r+I$. Note that $f$ is not injective since $f(x)=0$ and $xneq 0$. However, since $I$ is a proper ideal, the quotient $A/I$ is not the zero ring. (Note that $A/I$ is the zero ring if and only if $I=A$.) Hence, we got a contradiction to (iii.). We conclude that our assumption was false and that (iii.) implies (i.).
answered Dec 17 '18 at 20:26
ChristophChristoph
12.5k1642
$endgroup$
add a comment |
$begingroup$
The step (iii.) $Rightarrow$ (i.) is done by contradiction. Assume for the sake of contradiction that (iii.) holds but $A$ is not a field. Then there is a non-zero element $xin A$ that is not a unit. Since it is not a unit, the ideal $I=(x)$ does not contain $1$, so $I$ is a proper ideal of $A$. Now consider the canonical homomorphism $fcolon Ato A/I$, $rmapsto r+I$. Note that $f$ is not injective since $f(x)=0$ and $xneq 0$. However, since $I$ is a proper ideal, the quotient $A/I$ is not the zero ring. (Note that $A/I$ is the zero ring if and only if $I=A$.) Hence, we got a contradiction to (iii.). We conclude that our assumption was false and that (iii.) implies (i.).
answered Dec 17 '18 at 20:26
ChristophChristoph
12.5k1642
$endgroup$
add a comment |
$begingroup$
The step (iii.) $Rightarrow$ (i.) is done by contradiction. Assume for the sake of contradiction that (iii.) holds but $A$ is not a field. Then there is a non-zero element $xin A$ that is not a unit. Since it is not a unit, the ideal $I=(x)$ does not contain $1$, so $I$ is a proper ideal of $A$. Now consider the canonical homomorphism $fcolon Ato A/I$, $rmapsto r+I$. Note that $f$ is not injective since $f(x)=0$ and $xneq 0$. However, since $I$ is a proper ideal, the quotient $A/I$ is not the zero ring. (Note that $A/I$ is the zero ring if and only if $I=A$.) Hence, we got a contradiction to (iii.). We conclude that our assumption was false and that (iii.) implies (i.).
answered Dec 17 '18 at 20:26
ChristophChristoph
12.5k1642
$endgroup$
The step (iii.) $Rightarrow$ (i.) is done by contradiction. Assume for the sake of contradiction that (iii.) holds but $A$ is not a field. Then there is a non-zero element $xin A$ that is not a unit. Since it is not a unit, the ideal $I=(x)$ does not contain $1$, so $I$ is a proper ideal of $A$. Now consider the canonical homomorphism $fcolon Ato A/I$, $rmapsto r+I$. Note that $f$ is not injective since $f(x)=0$ and $xneq 0$. However, since $I$ is a proper ideal, the quotient $A/I$ is not the zero ring. (Note that $A/I$ is the zero ring if and only if $I=A$.) Hence, we got a contradiction to (iii.). We conclude that our assumption was false and that (iii.) implies (i.).
answered Dec 17 '18 at 20:26
ChristophChristoph
12.5k1642
answered Dec 17 '18 at 20:26
ChristophChristoph
12.5k1642
answered Dec 17 '18 at 20:26
ChristophChristoph
12.5k1642
answered Dec 17 '18 at 20:26
ChristophChristoph
12.5k1642
12.5k1642
add a comment |
add a comment |
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$begingroup$
I think you pretty much have it. Perhaps you can argue as follows: if $x$ is not invertible, then consider the ideal $langle xrangle neq R$ generated by $x$ and let $f$ be the corresponding homomorphism with kernel as $langle xrangle$. Then you have a non-injective homomorphism, which contradicts the assumption.
$endgroup$
– Anurag A
Dec 17 '18 at 20:17