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







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    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.
    – Anurag A
    6 hours ago

















up vote
3
down vote

favorite












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







share|cite|improve this question




















  • 2




    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.
    – Anurag A
    6 hours ago















up vote
3
down vote

favorite









up vote
3
down vote

favorite











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







share|cite|improve this question















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






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share|cite|improve this question













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edited 6 hours ago









rschwieb

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asked 6 hours ago









Tom

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  • 2




    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.
    – Anurag A
    6 hours ago
















  • 2




    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.
    – Anurag A
    6 hours ago










2




2




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.
– Anurag A
6 hours ago






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.
– Anurag A
6 hours ago












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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.






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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.).






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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.






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        accepted










        $iii)implies i):$ 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.






        share|cite|improve this answer























          up vote
          4
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          accepted







          up vote
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          down vote



          accepted






          $iii)implies i):$ 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.






          share|cite|improve this answer












          $iii)implies i):$ 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.







          share|cite|improve this answer












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          share|cite|improve this answer










          answered 5 hours ago









          Chris Custer

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          10.3k3724






















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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.).






              share|cite|improve this answer

























                up vote
                4
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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.).






                share|cite|improve this answer























                  up vote
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                  up vote
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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.).






                  share|cite|improve this answer












                  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.).







                  share|cite|improve this answer












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                  share|cite|improve this answer










                  answered 6 hours ago









                  Christoph

                  11.7k1541




                  11.7k1541






























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