If $fcolonmathbb{Z}to F$ is an onto morphism and $F$ is a field, then $mathbb{Z}_pcong F$ where $p$ is a...











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Let $F$ a field ed $fcolonmathbb{Z}to F$ an onto morphism. We know that $ker f$ is an ideal of $mathbb{Z}$, then $ker f=(n)=nmathbb{Z}$ for same $ninmathbb{Z}$. For the first isomorphism theorem we have that $$mathbb{Z}_n:=mathbb{Z}/nmathbb{Z}=mathbb{Z}/ker fcong F.$$



Now, if $n=0$, the canonical projection is $picolonmathbb{Z}to mathbb{Z}$, but the only non-zero morphism from $mathbb{Z}$ to $mathbb{Z}$ is $id_{mathbb{Z}}$ which is, in particular, injective. Since $ker f=nmathbb{Z}$, then $tilde{f}colonmathbb{Z}to F$ is injective, but for the first isomorphism theorem $f=tilde{f}circpi$, then $f$ is injective, moreover, for hypotesis, $f$ is onto, then $f$ is an isomorphism. But this is absurd, because $mathbb{Z}$ is not a field. Therefore $n$ can not be $0$.



On the other hand $mathbb{Z}_ncong F$, then $mathbb{Z}_n$ is a field, then $n$ must be prime.



Correct?



Thanks!






abstract-algebra proof-verification proof-explanation







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  • 1




    I don't understand your claim. $mathbb{Z}$ is not isomorphic to a field because it is not a field: $2$ has no multiplicative inverse. And there is an onto homomorphism from $mathbb{Z}$ to a field, take the quotient map $mathbb{Z}tomathbb{Z}/pmathbb{Z}$ for prime $p$. But obviously not every field is a homomorphic image of $mathbb{Z}$ (e.g. reals).
    – freakish
    Nov 20 at 9:11












  • Sorry but I have not explained well, now I correct my question.
    – Jack J.
    Nov 20 at 9:18










  • The kernel cannot be zero, since this would imply that $2$, for exemple, has multiplicative inverse (since its image in F does). Now, by the isomorphism theorem you have $Z/nZ cong F$. Thus, $Z/nZ$ is a field. Now prove that $n$ must be prime.
    – math.h
    Nov 20 at 10:17












  • @math.h I would like to know if the way I have shown that $n$ can not be zero is correct
    – Jack J.
    Nov 20 at 10:23






  • 1




    Yes, it is. But I think you are "talking too much" (no offense, really) for just a simple thing. The morphism $Zrightarrow F$ cannot be injective (otherwise the morphism would be bijective), since $F$ is a field. You seem to know that $Z$ can't be isomorphic to a field, so you gotta accept this argument. As you said, the kernel should be of the form: $nZ$. As it is not zero (the kernel), then $n$ cannot be zero.
    – math.h
    Nov 20 at 10:35

















up vote
0
down vote

favorite












Let $F$ a field ed $fcolonmathbb{Z}to F$ an onto morphism. We know that $ker f$ is an ideal of $mathbb{Z}$, then $ker f=(n)=nmathbb{Z}$ for same $ninmathbb{Z}$. For the first isomorphism theorem we have that $$mathbb{Z}_n:=mathbb{Z}/nmathbb{Z}=mathbb{Z}/ker fcong F.$$



Now, if $n=0$, the canonical projection is $picolonmathbb{Z}to mathbb{Z}$, but the only non-zero morphism from $mathbb{Z}$ to $mathbb{Z}$ is $id_{mathbb{Z}}$ which is, in particular, injective. Since $ker f=nmathbb{Z}$, then $tilde{f}colonmathbb{Z}to F$ is injective, but for the first isomorphism theorem $f=tilde{f}circpi$, then $f$ is injective, moreover, for hypotesis, $f$ is onto, then $f$ is an isomorphism. But this is absurd, because $mathbb{Z}$ is not a field. Therefore $n$ can not be $0$.



On the other hand $mathbb{Z}_ncong F$, then $mathbb{Z}_n$ is a field, then $n$ must be prime.



Correct?



Thanks!






abstract-algebra proof-verification proof-explanation







share|cite|improve this question




















  • 1




    I don't understand your claim. $mathbb{Z}$ is not isomorphic to a field because it is not a field: $2$ has no multiplicative inverse. And there is an onto homomorphism from $mathbb{Z}$ to a field, take the quotient map $mathbb{Z}tomathbb{Z}/pmathbb{Z}$ for prime $p$. But obviously not every field is a homomorphic image of $mathbb{Z}$ (e.g. reals).
    – freakish
    Nov 20 at 9:11












  • Sorry but I have not explained well, now I correct my question.
    – Jack J.
    Nov 20 at 9:18










  • The kernel cannot be zero, since this would imply that $2$, for exemple, has multiplicative inverse (since its image in F does). Now, by the isomorphism theorem you have $Z/nZ cong F$. Thus, $Z/nZ$ is a field. Now prove that $n$ must be prime.
    – math.h
    Nov 20 at 10:17












  • @math.h I would like to know if the way I have shown that $n$ can not be zero is correct
    – Jack J.
    Nov 20 at 10:23






  • 1




    Yes, it is. But I think you are "talking too much" (no offense, really) for just a simple thing. The morphism $Zrightarrow F$ cannot be injective (otherwise the morphism would be bijective), since $F$ is a field. You seem to know that $Z$ can't be isomorphic to a field, so you gotta accept this argument. As you said, the kernel should be of the form: $nZ$. As it is not zero (the kernel), then $n$ cannot be zero.
    – math.h
    Nov 20 at 10:35















up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $F$ a field ed $fcolonmathbb{Z}to F$ an onto morphism. We know that $ker f$ is an ideal of $mathbb{Z}$, then $ker f=(n)=nmathbb{Z}$ for same $ninmathbb{Z}$. For the first isomorphism theorem we have that $$mathbb{Z}_n:=mathbb{Z}/nmathbb{Z}=mathbb{Z}/ker fcong F.$$



Now, if $n=0$, the canonical projection is $picolonmathbb{Z}to mathbb{Z}$, but the only non-zero morphism from $mathbb{Z}$ to $mathbb{Z}$ is $id_{mathbb{Z}}$ which is, in particular, injective. Since $ker f=nmathbb{Z}$, then $tilde{f}colonmathbb{Z}to F$ is injective, but for the first isomorphism theorem $f=tilde{f}circpi$, then $f$ is injective, moreover, for hypotesis, $f$ is onto, then $f$ is an isomorphism. But this is absurd, because $mathbb{Z}$ is not a field. Therefore $n$ can not be $0$.



On the other hand $mathbb{Z}_ncong F$, then $mathbb{Z}_n$ is a field, then $n$ must be prime.



Correct?



Thanks!






abstract-algebra proof-verification proof-explanation







share|cite|improve this question















Let $F$ a field ed $fcolonmathbb{Z}to F$ an onto morphism. We know that $ker f$ is an ideal of $mathbb{Z}$, then $ker f=(n)=nmathbb{Z}$ for same $ninmathbb{Z}$. For the first isomorphism theorem we have that $$mathbb{Z}_n:=mathbb{Z}/nmathbb{Z}=mathbb{Z}/ker fcong F.$$



Now, if $n=0$, the canonical projection is $picolonmathbb{Z}to mathbb{Z}$, but the only non-zero morphism from $mathbb{Z}$ to $mathbb{Z}$ is $id_{mathbb{Z}}$ which is, in particular, injective. Since $ker f=nmathbb{Z}$, then $tilde{f}colonmathbb{Z}to F$ is injective, but for the first isomorphism theorem $f=tilde{f}circpi$, then $f$ is injective, moreover, for hypotesis, $f$ is onto, then $f$ is an isomorphism. But this is absurd, because $mathbb{Z}$ is not a field. Therefore $n$ can not be $0$.



On the other hand $mathbb{Z}_ncong F$, then $mathbb{Z}_n$ is a field, then $n$ must be prime.



Correct?



Thanks!






abstract-algebra proof-verification proof-explanation




abstract-algebra proof-verification proof-explanation






share|cite|improve this question















share|cite|improve this question













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edited Nov 20 at 9:21

























asked Nov 20 at 9:04









Jack J.

6031319




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  • 1




    I don't understand your claim. $mathbb{Z}$ is not isomorphic to a field because it is not a field: $2$ has no multiplicative inverse. And there is an onto homomorphism from $mathbb{Z}$ to a field, take the quotient map $mathbb{Z}tomathbb{Z}/pmathbb{Z}$ for prime $p$. But obviously not every field is a homomorphic image of $mathbb{Z}$ (e.g. reals).
    – freakish
    Nov 20 at 9:11












  • Sorry but I have not explained well, now I correct my question.
    – Jack J.
    Nov 20 at 9:18










  • The kernel cannot be zero, since this would imply that $2$, for exemple, has multiplicative inverse (since its image in F does). Now, by the isomorphism theorem you have $Z/nZ cong F$. Thus, $Z/nZ$ is a field. Now prove that $n$ must be prime.
    – math.h
    Nov 20 at 10:17












  • @math.h I would like to know if the way I have shown that $n$ can not be zero is correct
    – Jack J.
    Nov 20 at 10:23






  • 1




    Yes, it is. But I think you are "talking too much" (no offense, really) for just a simple thing. The morphism $Zrightarrow F$ cannot be injective (otherwise the morphism would be bijective), since $F$ is a field. You seem to know that $Z$ can't be isomorphic to a field, so you gotta accept this argument. As you said, the kernel should be of the form: $nZ$. As it is not zero (the kernel), then $n$ cannot be zero.
    – math.h
    Nov 20 at 10:35
















  • 1




    I don't understand your claim. $mathbb{Z}$ is not isomorphic to a field because it is not a field: $2$ has no multiplicative inverse. And there is an onto homomorphism from $mathbb{Z}$ to a field, take the quotient map $mathbb{Z}tomathbb{Z}/pmathbb{Z}$ for prime $p$. But obviously not every field is a homomorphic image of $mathbb{Z}$ (e.g. reals).
    – freakish
    Nov 20 at 9:11












  • Sorry but I have not explained well, now I correct my question.
    – Jack J.
    Nov 20 at 9:18










  • The kernel cannot be zero, since this would imply that $2$, for exemple, has multiplicative inverse (since its image in F does). Now, by the isomorphism theorem you have $Z/nZ cong F$. Thus, $Z/nZ$ is a field. Now prove that $n$ must be prime.
    – math.h
    Nov 20 at 10:17












  • @math.h I would like to know if the way I have shown that $n$ can not be zero is correct
    – Jack J.
    Nov 20 at 10:23






  • 1




    Yes, it is. But I think you are "talking too much" (no offense, really) for just a simple thing. The morphism $Zrightarrow F$ cannot be injective (otherwise the morphism would be bijective), since $F$ is a field. You seem to know that $Z$ can't be isomorphic to a field, so you gotta accept this argument. As you said, the kernel should be of the form: $nZ$. As it is not zero (the kernel), then $n$ cannot be zero.
    – math.h
    Nov 20 at 10:35










1




1




I don't understand your claim. $mathbb{Z}$ is not isomorphic to a field because it is not a field: $2$ has no multiplicative inverse. And there is an onto homomorphism from $mathbb{Z}$ to a field, take the quotient map $mathbb{Z}tomathbb{Z}/pmathbb{Z}$ for prime $p$. But obviously not every field is a homomorphic image of $mathbb{Z}$ (e.g. reals).
– freakish
Nov 20 at 9:11






I don't understand your claim. $mathbb{Z}$ is not isomorphic to a field because it is not a field: $2$ has no multiplicative inverse. And there is an onto homomorphism from $mathbb{Z}$ to a field, take the quotient map $mathbb{Z}tomathbb{Z}/pmathbb{Z}$ for prime $p$. But obviously not every field is a homomorphic image of $mathbb{Z}$ (e.g. reals).
– freakish
Nov 20 at 9:11














Sorry but I have not explained well, now I correct my question.
– Jack J.
Nov 20 at 9:18




Sorry but I have not explained well, now I correct my question.
– Jack J.
Nov 20 at 9:18












The kernel cannot be zero, since this would imply that $2$, for exemple, has multiplicative inverse (since its image in F does). Now, by the isomorphism theorem you have $Z/nZ cong F$. Thus, $Z/nZ$ is a field. Now prove that $n$ must be prime.
– math.h
Nov 20 at 10:17






The kernel cannot be zero, since this would imply that $2$, for exemple, has multiplicative inverse (since its image in F does). Now, by the isomorphism theorem you have $Z/nZ cong F$. Thus, $Z/nZ$ is a field. Now prove that $n$ must be prime.
– math.h
Nov 20 at 10:17














@math.h I would like to know if the way I have shown that $n$ can not be zero is correct
– Jack J.
Nov 20 at 10:23




@math.h I would like to know if the way I have shown that $n$ can not be zero is correct
– Jack J.
Nov 20 at 10:23




1




1




Yes, it is. But I think you are "talking too much" (no offense, really) for just a simple thing. The morphism $Zrightarrow F$ cannot be injective (otherwise the morphism would be bijective), since $F$ is a field. You seem to know that $Z$ can't be isomorphic to a field, so you gotta accept this argument. As you said, the kernel should be of the form: $nZ$. As it is not zero (the kernel), then $n$ cannot be zero.
– math.h
Nov 20 at 10:35






Yes, it is. But I think you are "talking too much" (no offense, really) for just a simple thing. The morphism $Zrightarrow F$ cannot be injective (otherwise the morphism would be bijective), since $F$ is a field. You seem to know that $Z$ can't be isomorphic to a field, so you gotta accept this argument. As you said, the kernel should be of the form: $nZ$. As it is not zero (the kernel), then $n$ cannot be zero.
– math.h
Nov 20 at 10:35












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We can assume $ker(f)neq (0)$, since this would imply that $Bbb Z$ is a field.



Since $Bbb Z$ is a PID, we can assume $ker(f)=(n)$, where $ninBbb Z^+$. You are ofcourse right that $Bbb Z_n$ is a field $iff n$ prime. One way to show this:



Suppose $n=ab$ so that $ker(f)=(ab)$, then $f(ab)=f(a)f(b)=0$ in $F$. Since $F$ is a field, and so an integral domain we need to have $f(a)=0$ or $f(b)=0$, which forces either $a,b$ to be a unit. This means by definition that $n$ is prime.






share|cite|improve this answer





















  • @cansomeonehelpmeoutThanks for yourn answer. So my reasoning to show that $n$ can not be zero is correct?
    – Jack J.
    Nov 20 at 10:14











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up vote
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We can assume $ker(f)neq (0)$, since this would imply that $Bbb Z$ is a field.



Since $Bbb Z$ is a PID, we can assume $ker(f)=(n)$, where $ninBbb Z^+$. You are ofcourse right that $Bbb Z_n$ is a field $iff n$ prime. One way to show this:



Suppose $n=ab$ so that $ker(f)=(ab)$, then $f(ab)=f(a)f(b)=0$ in $F$. Since $F$ is a field, and so an integral domain we need to have $f(a)=0$ or $f(b)=0$, which forces either $a,b$ to be a unit. This means by definition that $n$ is prime.






share|cite|improve this answer





















  • @cansomeonehelpmeoutThanks for yourn answer. So my reasoning to show that $n$ can not be zero is correct?
    – Jack J.
    Nov 20 at 10:14















up vote
1
down vote



accepted










We can assume $ker(f)neq (0)$, since this would imply that $Bbb Z$ is a field.



Since $Bbb Z$ is a PID, we can assume $ker(f)=(n)$, where $ninBbb Z^+$. You are ofcourse right that $Bbb Z_n$ is a field $iff n$ prime. One way to show this:



Suppose $n=ab$ so that $ker(f)=(ab)$, then $f(ab)=f(a)f(b)=0$ in $F$. Since $F$ is a field, and so an integral domain we need to have $f(a)=0$ or $f(b)=0$, which forces either $a,b$ to be a unit. This means by definition that $n$ is prime.






share|cite|improve this answer





















  • @cansomeonehelpmeoutThanks for yourn answer. So my reasoning to show that $n$ can not be zero is correct?
    – Jack J.
    Nov 20 at 10:14













up vote
1
down vote



accepted







up vote
1
down vote



accepted






We can assume $ker(f)neq (0)$, since this would imply that $Bbb Z$ is a field.



Since $Bbb Z$ is a PID, we can assume $ker(f)=(n)$, where $ninBbb Z^+$. You are ofcourse right that $Bbb Z_n$ is a field $iff n$ prime. One way to show this:



Suppose $n=ab$ so that $ker(f)=(ab)$, then $f(ab)=f(a)f(b)=0$ in $F$. Since $F$ is a field, and so an integral domain we need to have $f(a)=0$ or $f(b)=0$, which forces either $a,b$ to be a unit. This means by definition that $n$ is prime.






share|cite|improve this answer












We can assume $ker(f)neq (0)$, since this would imply that $Bbb Z$ is a field.



Since $Bbb Z$ is a PID, we can assume $ker(f)=(n)$, where $ninBbb Z^+$. You are ofcourse right that $Bbb Z_n$ is a field $iff n$ prime. One way to show this:



Suppose $n=ab$ so that $ker(f)=(ab)$, then $f(ab)=f(a)f(b)=0$ in $F$. Since $F$ is a field, and so an integral domain we need to have $f(a)=0$ or $f(b)=0$, which forces either $a,b$ to be a unit. This means by definition that $n$ is prime.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 20 at 9:55









cansomeonehelpmeout

6,5933834




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  • @cansomeonehelpmeoutThanks for yourn answer. So my reasoning to show that $n$ can not be zero is correct?
    – Jack J.
    Nov 20 at 10:14


















  • @cansomeonehelpmeoutThanks for yourn answer. So my reasoning to show that $n$ can not be zero is correct?
    – Jack J.
    Nov 20 at 10:14
















@cansomeonehelpmeoutThanks for yourn answer. So my reasoning to show that $n$ can not be zero is correct?
– Jack J.
Nov 20 at 10:14




@cansomeonehelpmeoutThanks for yourn answer. So my reasoning to show that $n$ can not be zero is correct?
– Jack J.
Nov 20 at 10:14


















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