image of injective function from $Z_{pqr} to Z_{pq}×Z_{qr}×Z_{rp}$












0












$begingroup$



Fix three distinct primes $p, q, r$, prove that the map



$Z_{pqr}→Z_{pq}×Z_{qr}×Z_{pr}$ by $[x]_{pqr} → ([x]_{pq}, [x]_{qr}, [x]_{pr})$



is injective and determine its image.




My attempt:



To prove it is injective, I said let there be $x$ and $y$ such that $[x]_{pq}=[y]_{pq}$. This means $pq|(x−y)$ So, one of $p$ or $q$ must divide $(x−y)$. Similarly, either $p$ or $r$ and either $q$ or $r$ must divide $x−y$ as well. Therefore, one of $p,q,r$ must divide $x−y$.



This implies $pqr|(x−y)$, So, $[x]_{pqr}=[y]_{pqr}$ This proves that the function is injective (if I'm correct in my implications).



Now, how do I determine its image?
Maybe I can use the chinese remainder theorem and establish a bijection from $Z_p×Z_q×Z_r$ to $Z_{pqr}$ and use that somehow, but can anyone help me on this?






elementary-number-theory modular-arithmetic







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is the kernel of the map?
    $endgroup$
    – user 170039
    Dec 13 '18 at 4:27










  • $begingroup$
    I am not sure what you mean by that, I apologize.
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 4:30










  • $begingroup$
    Is the map a homomorphism?
    $endgroup$
    – user 170039
    Dec 13 '18 at 4:31










  • $begingroup$
    Yes, it is. Could you please elaborate on how that should help me?
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 4:33






  • 1




    $begingroup$
    But how does this help me determine which elements this function (homomorphism in this case) maps to? I have not been formally introduced to group theory, so I'm trying to make sense of the things you're saying as I go.
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 6:32
















0












$begingroup$



Fix three distinct primes $p, q, r$, prove that the map



$Z_{pqr}→Z_{pq}×Z_{qr}×Z_{pr}$ by $[x]_{pqr} → ([x]_{pq}, [x]_{qr}, [x]_{pr})$



is injective and determine its image.




My attempt:



To prove it is injective, I said let there be $x$ and $y$ such that $[x]_{pq}=[y]_{pq}$. This means $pq|(x−y)$ So, one of $p$ or $q$ must divide $(x−y)$. Similarly, either $p$ or $r$ and either $q$ or $r$ must divide $x−y$ as well. Therefore, one of $p,q,r$ must divide $x−y$.



This implies $pqr|(x−y)$, So, $[x]_{pqr}=[y]_{pqr}$ This proves that the function is injective (if I'm correct in my implications).



Now, how do I determine its image?
Maybe I can use the chinese remainder theorem and establish a bijection from $Z_p×Z_q×Z_r$ to $Z_{pqr}$ and use that somehow, but can anyone help me on this?






elementary-number-theory modular-arithmetic







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is the kernel of the map?
    $endgroup$
    – user 170039
    Dec 13 '18 at 4:27










  • $begingroup$
    I am not sure what you mean by that, I apologize.
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 4:30










  • $begingroup$
    Is the map a homomorphism?
    $endgroup$
    – user 170039
    Dec 13 '18 at 4:31










  • $begingroup$
    Yes, it is. Could you please elaborate on how that should help me?
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 4:33






  • 1




    $begingroup$
    But how does this help me determine which elements this function (homomorphism in this case) maps to? I have not been formally introduced to group theory, so I'm trying to make sense of the things you're saying as I go.
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 6:32














0












0








0





$begingroup$



Fix three distinct primes $p, q, r$, prove that the map



$Z_{pqr}→Z_{pq}×Z_{qr}×Z_{pr}$ by $[x]_{pqr} → ([x]_{pq}, [x]_{qr}, [x]_{pr})$



is injective and determine its image.




My attempt:



To prove it is injective, I said let there be $x$ and $y$ such that $[x]_{pq}=[y]_{pq}$. This means $pq|(x−y)$ So, one of $p$ or $q$ must divide $(x−y)$. Similarly, either $p$ or $r$ and either $q$ or $r$ must divide $x−y$ as well. Therefore, one of $p,q,r$ must divide $x−y$.



This implies $pqr|(x−y)$, So, $[x]_{pqr}=[y]_{pqr}$ This proves that the function is injective (if I'm correct in my implications).



Now, how do I determine its image?
Maybe I can use the chinese remainder theorem and establish a bijection from $Z_p×Z_q×Z_r$ to $Z_{pqr}$ and use that somehow, but can anyone help me on this?






elementary-number-theory modular-arithmetic







share|cite|improve this question











$endgroup$





Fix three distinct primes $p, q, r$, prove that the map



$Z_{pqr}→Z_{pq}×Z_{qr}×Z_{pr}$ by $[x]_{pqr} → ([x]_{pq}, [x]_{qr}, [x]_{pr})$



is injective and determine its image.




My attempt:



To prove it is injective, I said let there be $x$ and $y$ such that $[x]_{pq}=[y]_{pq}$. This means $pq|(x−y)$ So, one of $p$ or $q$ must divide $(x−y)$. Similarly, either $p$ or $r$ and either $q$ or $r$ must divide $x−y$ as well. Therefore, one of $p,q,r$ must divide $x−y$.



This implies $pqr|(x−y)$, So, $[x]_{pqr}=[y]_{pqr}$ This proves that the function is injective (if I'm correct in my implications).



Now, how do I determine its image?
Maybe I can use the chinese remainder theorem and establish a bijection from $Z_p×Z_q×Z_r$ to $Z_{pqr}$ and use that somehow, but can anyone help me on this?






elementary-number-theory modular-arithmetic




elementary-number-theory modular-arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 4:26









user 170039

10.5k42466




10.5k42466










asked Dec 13 '18 at 3:02









childishsadbinochildishsadbino

1148




1148












  • $begingroup$
    What is the kernel of the map?
    $endgroup$
    – user 170039
    Dec 13 '18 at 4:27










  • $begingroup$
    I am not sure what you mean by that, I apologize.
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 4:30










  • $begingroup$
    Is the map a homomorphism?
    $endgroup$
    – user 170039
    Dec 13 '18 at 4:31










  • $begingroup$
    Yes, it is. Could you please elaborate on how that should help me?
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 4:33






  • 1




    $begingroup$
    But how does this help me determine which elements this function (homomorphism in this case) maps to? I have not been formally introduced to group theory, so I'm trying to make sense of the things you're saying as I go.
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 6:32


















  • $begingroup$
    What is the kernel of the map?
    $endgroup$
    – user 170039
    Dec 13 '18 at 4:27










  • $begingroup$
    I am not sure what you mean by that, I apologize.
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 4:30










  • $begingroup$
    Is the map a homomorphism?
    $endgroup$
    – user 170039
    Dec 13 '18 at 4:31










  • $begingroup$
    Yes, it is. Could you please elaborate on how that should help me?
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 4:33






  • 1




    $begingroup$
    But how does this help me determine which elements this function (homomorphism in this case) maps to? I have not been formally introduced to group theory, so I'm trying to make sense of the things you're saying as I go.
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 6:32
















$begingroup$
What is the kernel of the map?
$endgroup$
– user 170039
Dec 13 '18 at 4:27




$begingroup$
What is the kernel of the map?
$endgroup$
– user 170039
Dec 13 '18 at 4:27












$begingroup$
I am not sure what you mean by that, I apologize.
$endgroup$
– childishsadbino
Dec 13 '18 at 4:30




$begingroup$
I am not sure what you mean by that, I apologize.
$endgroup$
– childishsadbino
Dec 13 '18 at 4:30












$begingroup$
Is the map a homomorphism?
$endgroup$
– user 170039
Dec 13 '18 at 4:31




$begingroup$
Is the map a homomorphism?
$endgroup$
– user 170039
Dec 13 '18 at 4:31












$begingroup$
Yes, it is. Could you please elaborate on how that should help me?
$endgroup$
– childishsadbino
Dec 13 '18 at 4:33




$begingroup$
Yes, it is. Could you please elaborate on how that should help me?
$endgroup$
– childishsadbino
Dec 13 '18 at 4:33




1




1




$begingroup$
But how does this help me determine which elements this function (homomorphism in this case) maps to? I have not been formally introduced to group theory, so I'm trying to make sense of the things you're saying as I go.
$endgroup$
– childishsadbino
Dec 13 '18 at 6:32




$begingroup$
But how does this help me determine which elements this function (homomorphism in this case) maps to? I have not been formally introduced to group theory, so I'm trying to make sense of the things you're saying as I go.
$endgroup$
– childishsadbino
Dec 13 '18 at 6:32










2 Answers
2






active

oldest

votes


















1












$begingroup$

As the other answer indicates, your proof is not quite right. You are not trying to show that a single prime divides $x-y$, but that all three do.



As for characterizing the image, suppose we have some triple $(a,b,c) in Z_{pq}times Z_{qr} times Z_{pr}$. How can we check that it actually came from an element $xin Z_{pqr}$? One approach is to compare, for example, the image of $a$ in $Z_p$ with the image of $c$ in $Z_p$. If there is such an $x$, these two things have to be equal.



Do the same thing for $Z_q$ and $Z_r$ you'll have three conditions on $(a,b,c)$. To actually construct $x$ using these conditions, you can use the Chinese Remainder Theorem.



More explicitly, consider the map $psi(a,b,c) = (a-b, b-c, c-a) in Z_q times Z_r times Z_p$. Prove that everything in the image of your map $phi$ is in the kernel of $psi$ (manipulate the definitions), and prove that everything in the kernel of $psi$ is in the image of $phi$ (requires the Chinese Remainder Theorem).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How do the images of $a$ and $c$ in $Z_p$ being equal imply that $x$ must be in $Z_{pqr}$?
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 18:00












  • $begingroup$
    It doesn't, you need to use the same conditions for $Z_q$ and $Z_r$. And the point isn't to show anything about $x$, which isn't given--the point is to construct $x$ in the first place! Here is an additional hint: use the Chinese Remainder Theorem to choose some integer $x$ that has the right remainders modulo $p$, $q$, and $r$.
    $endgroup$
    – Slade
    Dec 13 '18 at 18:56












  • $begingroup$
    But what is this "right remainder" you mention?
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 18:58










  • $begingroup$
    The point is to use the condition $a equiv c pmod{p}$ and construct an $x$ that is congruent to both.
    $endgroup$
    – Slade
    Dec 13 '18 at 18:59












  • $begingroup$
    You say "if there is such an x, these two things have to be equal". Why is that?
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 19:00



















0












$begingroup$

Your proof of the first part is completely wrong. "One of $p,q,r$ divides ..." does not imply "All of $p,q,r$ divide ...". I suggest you fix that problem first before trying the second part. Also, it is easy and has nothing to do with homomorphisms and kernels. Hint: if $x mid n$ then $n = ax$ for some integer $a$, and if $y mid n$ such that $gcd(x,y) = 1$, then $n equiv ax pmod{y}$ and $x$ is invertible mod $y$, so ...



I'm not sure what the question means by saying "determine its image", since it seems the simplest description of its image is literally ${ ([x]_{pq},[x]_{qr},[x]_{rp}) : x in mathbb{Z} }$.



If you wish to invert the mapping, note that $x bmod p = ( x bmod pq ) bmod p = ( x bmod rp ) bmod p$, so each triple $(a,b,c)$ in the image must satisfy $p mid c-a$ and $q mid a-b$ and $r mid b-c$. What else can we say?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    well, $Z_{pqr}$ has pqr number of elements and the set it is mapping to has $(pqr)^2$ elements, so the question is asking to specify which elements it hits.
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 17:22










  • $begingroup$
    @childishsadbino: I know that, and I did literally that in my answer. I don't see a simpler way to "specify which elements it hits" than simply saying it is the image of the given function. I would say that that question is too vague to be properly answerable.
    $endgroup$
    – user21820
    Dec 13 '18 at 18:26










  • $begingroup$
    @user21820 See my answer. There is a nice characterization of the image intrinsic to the target: it consists of the triples $(a,b,c)$ that map to zero under the map $(a,b,c) mapsto (a-b, b-c, c-a) in Z_q times Z_r times Z_p$.
    $endgroup$
    – Slade
    Dec 13 '18 at 19:09













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

As the other answer indicates, your proof is not quite right. You are not trying to show that a single prime divides $x-y$, but that all three do.



As for characterizing the image, suppose we have some triple $(a,b,c) in Z_{pq}times Z_{qr} times Z_{pr}$. How can we check that it actually came from an element $xin Z_{pqr}$? One approach is to compare, for example, the image of $a$ in $Z_p$ with the image of $c$ in $Z_p$. If there is such an $x$, these two things have to be equal.



Do the same thing for $Z_q$ and $Z_r$ you'll have three conditions on $(a,b,c)$. To actually construct $x$ using these conditions, you can use the Chinese Remainder Theorem.



More explicitly, consider the map $psi(a,b,c) = (a-b, b-c, c-a) in Z_q times Z_r times Z_p$. Prove that everything in the image of your map $phi$ is in the kernel of $psi$ (manipulate the definitions), and prove that everything in the kernel of $psi$ is in the image of $phi$ (requires the Chinese Remainder Theorem).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How do the images of $a$ and $c$ in $Z_p$ being equal imply that $x$ must be in $Z_{pqr}$?
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 18:00












  • $begingroup$
    It doesn't, you need to use the same conditions for $Z_q$ and $Z_r$. And the point isn't to show anything about $x$, which isn't given--the point is to construct $x$ in the first place! Here is an additional hint: use the Chinese Remainder Theorem to choose some integer $x$ that has the right remainders modulo $p$, $q$, and $r$.
    $endgroup$
    – Slade
    Dec 13 '18 at 18:56












  • $begingroup$
    But what is this "right remainder" you mention?
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 18:58










  • $begingroup$
    The point is to use the condition $a equiv c pmod{p}$ and construct an $x$ that is congruent to both.
    $endgroup$
    – Slade
    Dec 13 '18 at 18:59












  • $begingroup$
    You say "if there is such an x, these two things have to be equal". Why is that?
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 19:00
















1












$begingroup$

As the other answer indicates, your proof is not quite right. You are not trying to show that a single prime divides $x-y$, but that all three do.



As for characterizing the image, suppose we have some triple $(a,b,c) in Z_{pq}times Z_{qr} times Z_{pr}$. How can we check that it actually came from an element $xin Z_{pqr}$? One approach is to compare, for example, the image of $a$ in $Z_p$ with the image of $c$ in $Z_p$. If there is such an $x$, these two things have to be equal.



Do the same thing for $Z_q$ and $Z_r$ you'll have three conditions on $(a,b,c)$. To actually construct $x$ using these conditions, you can use the Chinese Remainder Theorem.



More explicitly, consider the map $psi(a,b,c) = (a-b, b-c, c-a) in Z_q times Z_r times Z_p$. Prove that everything in the image of your map $phi$ is in the kernel of $psi$ (manipulate the definitions), and prove that everything in the kernel of $psi$ is in the image of $phi$ (requires the Chinese Remainder Theorem).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How do the images of $a$ and $c$ in $Z_p$ being equal imply that $x$ must be in $Z_{pqr}$?
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 18:00












  • $begingroup$
    It doesn't, you need to use the same conditions for $Z_q$ and $Z_r$. And the point isn't to show anything about $x$, which isn't given--the point is to construct $x$ in the first place! Here is an additional hint: use the Chinese Remainder Theorem to choose some integer $x$ that has the right remainders modulo $p$, $q$, and $r$.
    $endgroup$
    – Slade
    Dec 13 '18 at 18:56












  • $begingroup$
    But what is this "right remainder" you mention?
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 18:58










  • $begingroup$
    The point is to use the condition $a equiv c pmod{p}$ and construct an $x$ that is congruent to both.
    $endgroup$
    – Slade
    Dec 13 '18 at 18:59












  • $begingroup$
    You say "if there is such an x, these two things have to be equal". Why is that?
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 19:00














1












1








1





$begingroup$

As the other answer indicates, your proof is not quite right. You are not trying to show that a single prime divides $x-y$, but that all three do.



As for characterizing the image, suppose we have some triple $(a,b,c) in Z_{pq}times Z_{qr} times Z_{pr}$. How can we check that it actually came from an element $xin Z_{pqr}$? One approach is to compare, for example, the image of $a$ in $Z_p$ with the image of $c$ in $Z_p$. If there is such an $x$, these two things have to be equal.



Do the same thing for $Z_q$ and $Z_r$ you'll have three conditions on $(a,b,c)$. To actually construct $x$ using these conditions, you can use the Chinese Remainder Theorem.



More explicitly, consider the map $psi(a,b,c) = (a-b, b-c, c-a) in Z_q times Z_r times Z_p$. Prove that everything in the image of your map $phi$ is in the kernel of $psi$ (manipulate the definitions), and prove that everything in the kernel of $psi$ is in the image of $phi$ (requires the Chinese Remainder Theorem).






share|cite|improve this answer











$endgroup$



As the other answer indicates, your proof is not quite right. You are not trying to show that a single prime divides $x-y$, but that all three do.



As for characterizing the image, suppose we have some triple $(a,b,c) in Z_{pq}times Z_{qr} times Z_{pr}$. How can we check that it actually came from an element $xin Z_{pqr}$? One approach is to compare, for example, the image of $a$ in $Z_p$ with the image of $c$ in $Z_p$. If there is such an $x$, these two things have to be equal.



Do the same thing for $Z_q$ and $Z_r$ you'll have three conditions on $(a,b,c)$. To actually construct $x$ using these conditions, you can use the Chinese Remainder Theorem.



More explicitly, consider the map $psi(a,b,c) = (a-b, b-c, c-a) in Z_q times Z_r times Z_p$. Prove that everything in the image of your map $phi$ is in the kernel of $psi$ (manipulate the definitions), and prove that everything in the kernel of $psi$ is in the image of $phi$ (requires the Chinese Remainder Theorem).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 13 '18 at 19:21

























answered Dec 13 '18 at 10:15









SladeSlade

25.2k12665




25.2k12665












  • $begingroup$
    How do the images of $a$ and $c$ in $Z_p$ being equal imply that $x$ must be in $Z_{pqr}$?
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 18:00












  • $begingroup$
    It doesn't, you need to use the same conditions for $Z_q$ and $Z_r$. And the point isn't to show anything about $x$, which isn't given--the point is to construct $x$ in the first place! Here is an additional hint: use the Chinese Remainder Theorem to choose some integer $x$ that has the right remainders modulo $p$, $q$, and $r$.
    $endgroup$
    – Slade
    Dec 13 '18 at 18:56












  • $begingroup$
    But what is this "right remainder" you mention?
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 18:58










  • $begingroup$
    The point is to use the condition $a equiv c pmod{p}$ and construct an $x$ that is congruent to both.
    $endgroup$
    – Slade
    Dec 13 '18 at 18:59












  • $begingroup$
    You say "if there is such an x, these two things have to be equal". Why is that?
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 19:00


















  • $begingroup$
    How do the images of $a$ and $c$ in $Z_p$ being equal imply that $x$ must be in $Z_{pqr}$?
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 18:00












  • $begingroup$
    It doesn't, you need to use the same conditions for $Z_q$ and $Z_r$. And the point isn't to show anything about $x$, which isn't given--the point is to construct $x$ in the first place! Here is an additional hint: use the Chinese Remainder Theorem to choose some integer $x$ that has the right remainders modulo $p$, $q$, and $r$.
    $endgroup$
    – Slade
    Dec 13 '18 at 18:56












  • $begingroup$
    But what is this "right remainder" you mention?
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 18:58










  • $begingroup$
    The point is to use the condition $a equiv c pmod{p}$ and construct an $x$ that is congruent to both.
    $endgroup$
    – Slade
    Dec 13 '18 at 18:59












  • $begingroup$
    You say "if there is such an x, these two things have to be equal". Why is that?
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 19:00
















$begingroup$
How do the images of $a$ and $c$ in $Z_p$ being equal imply that $x$ must be in $Z_{pqr}$?
$endgroup$
– childishsadbino
Dec 13 '18 at 18:00






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How do the images of $a$ and $c$ in $Z_p$ being equal imply that $x$ must be in $Z_{pqr}$?
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– childishsadbino
Dec 13 '18 at 18:00














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It doesn't, you need to use the same conditions for $Z_q$ and $Z_r$. And the point isn't to show anything about $x$, which isn't given--the point is to construct $x$ in the first place! Here is an additional hint: use the Chinese Remainder Theorem to choose some integer $x$ that has the right remainders modulo $p$, $q$, and $r$.
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– Slade
Dec 13 '18 at 18:56






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It doesn't, you need to use the same conditions for $Z_q$ and $Z_r$. And the point isn't to show anything about $x$, which isn't given--the point is to construct $x$ in the first place! Here is an additional hint: use the Chinese Remainder Theorem to choose some integer $x$ that has the right remainders modulo $p$, $q$, and $r$.
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– Slade
Dec 13 '18 at 18:56














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But what is this "right remainder" you mention?
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– childishsadbino
Dec 13 '18 at 18:58




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But what is this "right remainder" you mention?
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– childishsadbino
Dec 13 '18 at 18:58












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The point is to use the condition $a equiv c pmod{p}$ and construct an $x$ that is congruent to both.
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– Slade
Dec 13 '18 at 18:59






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The point is to use the condition $a equiv c pmod{p}$ and construct an $x$ that is congruent to both.
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– Slade
Dec 13 '18 at 18:59














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You say "if there is such an x, these two things have to be equal". Why is that?
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– childishsadbino
Dec 13 '18 at 19:00




$begingroup$
You say "if there is such an x, these two things have to be equal". Why is that?
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– childishsadbino
Dec 13 '18 at 19:00











0












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Your proof of the first part is completely wrong. "One of $p,q,r$ divides ..." does not imply "All of $p,q,r$ divide ...". I suggest you fix that problem first before trying the second part. Also, it is easy and has nothing to do with homomorphisms and kernels. Hint: if $x mid n$ then $n = ax$ for some integer $a$, and if $y mid n$ such that $gcd(x,y) = 1$, then $n equiv ax pmod{y}$ and $x$ is invertible mod $y$, so ...



I'm not sure what the question means by saying "determine its image", since it seems the simplest description of its image is literally ${ ([x]_{pq},[x]_{qr},[x]_{rp}) : x in mathbb{Z} }$.



If you wish to invert the mapping, note that $x bmod p = ( x bmod pq ) bmod p = ( x bmod rp ) bmod p$, so each triple $(a,b,c)$ in the image must satisfy $p mid c-a$ and $q mid a-b$ and $r mid b-c$. What else can we say?






share|cite|improve this answer











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  • $begingroup$
    well, $Z_{pqr}$ has pqr number of elements and the set it is mapping to has $(pqr)^2$ elements, so the question is asking to specify which elements it hits.
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 17:22










  • $begingroup$
    @childishsadbino: I know that, and I did literally that in my answer. I don't see a simpler way to "specify which elements it hits" than simply saying it is the image of the given function. I would say that that question is too vague to be properly answerable.
    $endgroup$
    – user21820
    Dec 13 '18 at 18:26










  • $begingroup$
    @user21820 See my answer. There is a nice characterization of the image intrinsic to the target: it consists of the triples $(a,b,c)$ that map to zero under the map $(a,b,c) mapsto (a-b, b-c, c-a) in Z_q times Z_r times Z_p$.
    $endgroup$
    – Slade
    Dec 13 '18 at 19:09


















0












$begingroup$

Your proof of the first part is completely wrong. "One of $p,q,r$ divides ..." does not imply "All of $p,q,r$ divide ...". I suggest you fix that problem first before trying the second part. Also, it is easy and has nothing to do with homomorphisms and kernels. Hint: if $x mid n$ then $n = ax$ for some integer $a$, and if $y mid n$ such that $gcd(x,y) = 1$, then $n equiv ax pmod{y}$ and $x$ is invertible mod $y$, so ...



I'm not sure what the question means by saying "determine its image", since it seems the simplest description of its image is literally ${ ([x]_{pq},[x]_{qr},[x]_{rp}) : x in mathbb{Z} }$.



If you wish to invert the mapping, note that $x bmod p = ( x bmod pq ) bmod p = ( x bmod rp ) bmod p$, so each triple $(a,b,c)$ in the image must satisfy $p mid c-a$ and $q mid a-b$ and $r mid b-c$. What else can we say?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    well, $Z_{pqr}$ has pqr number of elements and the set it is mapping to has $(pqr)^2$ elements, so the question is asking to specify which elements it hits.
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 17:22










  • $begingroup$
    @childishsadbino: I know that, and I did literally that in my answer. I don't see a simpler way to "specify which elements it hits" than simply saying it is the image of the given function. I would say that that question is too vague to be properly answerable.
    $endgroup$
    – user21820
    Dec 13 '18 at 18:26










  • $begingroup$
    @user21820 See my answer. There is a nice characterization of the image intrinsic to the target: it consists of the triples $(a,b,c)$ that map to zero under the map $(a,b,c) mapsto (a-b, b-c, c-a) in Z_q times Z_r times Z_p$.
    $endgroup$
    – Slade
    Dec 13 '18 at 19:09
















0












0








0





$begingroup$

Your proof of the first part is completely wrong. "One of $p,q,r$ divides ..." does not imply "All of $p,q,r$ divide ...". I suggest you fix that problem first before trying the second part. Also, it is easy and has nothing to do with homomorphisms and kernels. Hint: if $x mid n$ then $n = ax$ for some integer $a$, and if $y mid n$ such that $gcd(x,y) = 1$, then $n equiv ax pmod{y}$ and $x$ is invertible mod $y$, so ...



I'm not sure what the question means by saying "determine its image", since it seems the simplest description of its image is literally ${ ([x]_{pq},[x]_{qr},[x]_{rp}) : x in mathbb{Z} }$.



If you wish to invert the mapping, note that $x bmod p = ( x bmod pq ) bmod p = ( x bmod rp ) bmod p$, so each triple $(a,b,c)$ in the image must satisfy $p mid c-a$ and $q mid a-b$ and $r mid b-c$. What else can we say?






share|cite|improve this answer











$endgroup$



Your proof of the first part is completely wrong. "One of $p,q,r$ divides ..." does not imply "All of $p,q,r$ divide ...". I suggest you fix that problem first before trying the second part. Also, it is easy and has nothing to do with homomorphisms and kernels. Hint: if $x mid n$ then $n = ax$ for some integer $a$, and if $y mid n$ such that $gcd(x,y) = 1$, then $n equiv ax pmod{y}$ and $x$ is invertible mod $y$, so ...



I'm not sure what the question means by saying "determine its image", since it seems the simplest description of its image is literally ${ ([x]_{pq},[x]_{qr},[x]_{rp}) : x in mathbb{Z} }$.



If you wish to invert the mapping, note that $x bmod p = ( x bmod pq ) bmod p = ( x bmod rp ) bmod p$, so each triple $(a,b,c)$ in the image must satisfy $p mid c-a$ and $q mid a-b$ and $r mid b-c$. What else can we say?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 13 '18 at 19:32

























answered Dec 13 '18 at 9:33









user21820user21820

39.2k543154




39.2k543154












  • $begingroup$
    well, $Z_{pqr}$ has pqr number of elements and the set it is mapping to has $(pqr)^2$ elements, so the question is asking to specify which elements it hits.
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 17:22










  • $begingroup$
    @childishsadbino: I know that, and I did literally that in my answer. I don't see a simpler way to "specify which elements it hits" than simply saying it is the image of the given function. I would say that that question is too vague to be properly answerable.
    $endgroup$
    – user21820
    Dec 13 '18 at 18:26










  • $begingroup$
    @user21820 See my answer. There is a nice characterization of the image intrinsic to the target: it consists of the triples $(a,b,c)$ that map to zero under the map $(a,b,c) mapsto (a-b, b-c, c-a) in Z_q times Z_r times Z_p$.
    $endgroup$
    – Slade
    Dec 13 '18 at 19:09




















  • $begingroup$
    well, $Z_{pqr}$ has pqr number of elements and the set it is mapping to has $(pqr)^2$ elements, so the question is asking to specify which elements it hits.
    $endgroup$
    – childishsadbino
    Dec 13 '18 at 17:22










  • $begingroup$
    @childishsadbino: I know that, and I did literally that in my answer. I don't see a simpler way to "specify which elements it hits" than simply saying it is the image of the given function. I would say that that question is too vague to be properly answerable.
    $endgroup$
    – user21820
    Dec 13 '18 at 18:26










  • $begingroup$
    @user21820 See my answer. There is a nice characterization of the image intrinsic to the target: it consists of the triples $(a,b,c)$ that map to zero under the map $(a,b,c) mapsto (a-b, b-c, c-a) in Z_q times Z_r times Z_p$.
    $endgroup$
    – Slade
    Dec 13 '18 at 19:09


















$begingroup$
well, $Z_{pqr}$ has pqr number of elements and the set it is mapping to has $(pqr)^2$ elements, so the question is asking to specify which elements it hits.
$endgroup$
– childishsadbino
Dec 13 '18 at 17:22




$begingroup$
well, $Z_{pqr}$ has pqr number of elements and the set it is mapping to has $(pqr)^2$ elements, so the question is asking to specify which elements it hits.
$endgroup$
– childishsadbino
Dec 13 '18 at 17:22












$begingroup$
@childishsadbino: I know that, and I did literally that in my answer. I don't see a simpler way to "specify which elements it hits" than simply saying it is the image of the given function. I would say that that question is too vague to be properly answerable.
$endgroup$
– user21820
Dec 13 '18 at 18:26




$begingroup$
@childishsadbino: I know that, and I did literally that in my answer. I don't see a simpler way to "specify which elements it hits" than simply saying it is the image of the given function. I would say that that question is too vague to be properly answerable.
$endgroup$
– user21820
Dec 13 '18 at 18:26












$begingroup$
@user21820 See my answer. There is a nice characterization of the image intrinsic to the target: it consists of the triples $(a,b,c)$ that map to zero under the map $(a,b,c) mapsto (a-b, b-c, c-a) in Z_q times Z_r times Z_p$.
$endgroup$
– Slade
Dec 13 '18 at 19:09






$begingroup$
@user21820 See my answer. There is a nice characterization of the image intrinsic to the target: it consists of the triples $(a,b,c)$ that map to zero under the map $(a,b,c) mapsto (a-b, b-c, c-a) in Z_q times Z_r times Z_p$.
$endgroup$
– Slade
Dec 13 '18 at 19:09




















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