Intersection of ideals of $mathbb{Z}$












1












$begingroup$


Let $R=mathbb{Z}$, and let $p$ a fixed prime. We condiser $$bigcap_{kge 1} (p^k).$$
I know that this intersection is empty for the Foundamental Theorem, but How can I properly formalize the proof?



Thanks






proof-verification proof-explanation







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The intersection isn't empty but is the zero ideal. The integers in the intersection are those integers that are divisible by all powers of $p$. But any nonzero integer has a highest power of $p$ that divides it, as follows from unique prime factorization. So this intersection doesn't contain nonzero elements, and hence is $=(0)={0}$.
    $endgroup$
    – Olivier Bégassat
    Dec 4 '18 at 17:55










  • $begingroup$
    Thanks for your answer. The theorem of unique factorization is it also true for negative integer?
    $endgroup$
    – Jack J.
    Dec 4 '18 at 18:01










  • $begingroup$
    First of all the intersection is an ideal. If it is not $(0)$, then it contains a minimal positive integer. But $p^ktoinfty$ gives a contradiction.
    $endgroup$
    – dan_fulea
    Dec 4 '18 at 18:22










  • $begingroup$
    Thanks for your answer. But I would like a proof to consider the unique factorization theorem.
    $endgroup$
    – Jack J.
    Dec 4 '18 at 19:08


















1












$begingroup$


Let $R=mathbb{Z}$, and let $p$ a fixed prime. We condiser $$bigcap_{kge 1} (p^k).$$
I know that this intersection is empty for the Foundamental Theorem, but How can I properly formalize the proof?



Thanks






proof-verification proof-explanation







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The intersection isn't empty but is the zero ideal. The integers in the intersection are those integers that are divisible by all powers of $p$. But any nonzero integer has a highest power of $p$ that divides it, as follows from unique prime factorization. So this intersection doesn't contain nonzero elements, and hence is $=(0)={0}$.
    $endgroup$
    – Olivier Bégassat
    Dec 4 '18 at 17:55










  • $begingroup$
    Thanks for your answer. The theorem of unique factorization is it also true for negative integer?
    $endgroup$
    – Jack J.
    Dec 4 '18 at 18:01










  • $begingroup$
    First of all the intersection is an ideal. If it is not $(0)$, then it contains a minimal positive integer. But $p^ktoinfty$ gives a contradiction.
    $endgroup$
    – dan_fulea
    Dec 4 '18 at 18:22










  • $begingroup$
    Thanks for your answer. But I would like a proof to consider the unique factorization theorem.
    $endgroup$
    – Jack J.
    Dec 4 '18 at 19:08
















1












1








1


1



$begingroup$


Let $R=mathbb{Z}$, and let $p$ a fixed prime. We condiser $$bigcap_{kge 1} (p^k).$$
I know that this intersection is empty for the Foundamental Theorem, but How can I properly formalize the proof?



Thanks






proof-verification proof-explanation







share|cite|improve this question









$endgroup$




Let $R=mathbb{Z}$, and let $p$ a fixed prime. We condiser $$bigcap_{kge 1} (p^k).$$
I know that this intersection is empty for the Foundamental Theorem, but How can I properly formalize the proof?



Thanks






proof-verification proof-explanation




proof-verification proof-explanation






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 4 '18 at 17:52









Jack J.Jack J.

4851419




4851419








  • 1




    $begingroup$
    The intersection isn't empty but is the zero ideal. The integers in the intersection are those integers that are divisible by all powers of $p$. But any nonzero integer has a highest power of $p$ that divides it, as follows from unique prime factorization. So this intersection doesn't contain nonzero elements, and hence is $=(0)={0}$.
    $endgroup$
    – Olivier Bégassat
    Dec 4 '18 at 17:55










  • $begingroup$
    Thanks for your answer. The theorem of unique factorization is it also true for negative integer?
    $endgroup$
    – Jack J.
    Dec 4 '18 at 18:01










  • $begingroup$
    First of all the intersection is an ideal. If it is not $(0)$, then it contains a minimal positive integer. But $p^ktoinfty$ gives a contradiction.
    $endgroup$
    – dan_fulea
    Dec 4 '18 at 18:22










  • $begingroup$
    Thanks for your answer. But I would like a proof to consider the unique factorization theorem.
    $endgroup$
    – Jack J.
    Dec 4 '18 at 19:08
















  • 1




    $begingroup$
    The intersection isn't empty but is the zero ideal. The integers in the intersection are those integers that are divisible by all powers of $p$. But any nonzero integer has a highest power of $p$ that divides it, as follows from unique prime factorization. So this intersection doesn't contain nonzero elements, and hence is $=(0)={0}$.
    $endgroup$
    – Olivier Bégassat
    Dec 4 '18 at 17:55










  • $begingroup$
    Thanks for your answer. The theorem of unique factorization is it also true for negative integer?
    $endgroup$
    – Jack J.
    Dec 4 '18 at 18:01










  • $begingroup$
    First of all the intersection is an ideal. If it is not $(0)$, then it contains a minimal positive integer. But $p^ktoinfty$ gives a contradiction.
    $endgroup$
    – dan_fulea
    Dec 4 '18 at 18:22










  • $begingroup$
    Thanks for your answer. But I would like a proof to consider the unique factorization theorem.
    $endgroup$
    – Jack J.
    Dec 4 '18 at 19:08










1




1




$begingroup$
The intersection isn't empty but is the zero ideal. The integers in the intersection are those integers that are divisible by all powers of $p$. But any nonzero integer has a highest power of $p$ that divides it, as follows from unique prime factorization. So this intersection doesn't contain nonzero elements, and hence is $=(0)={0}$.
$endgroup$
– Olivier Bégassat
Dec 4 '18 at 17:55




$begingroup$
The intersection isn't empty but is the zero ideal. The integers in the intersection are those integers that are divisible by all powers of $p$. But any nonzero integer has a highest power of $p$ that divides it, as follows from unique prime factorization. So this intersection doesn't contain nonzero elements, and hence is $=(0)={0}$.
$endgroup$
– Olivier Bégassat
Dec 4 '18 at 17:55












$begingroup$
Thanks for your answer. The theorem of unique factorization is it also true for negative integer?
$endgroup$
– Jack J.
Dec 4 '18 at 18:01




$begingroup$
Thanks for your answer. The theorem of unique factorization is it also true for negative integer?
$endgroup$
– Jack J.
Dec 4 '18 at 18:01












$begingroup$
First of all the intersection is an ideal. If it is not $(0)$, then it contains a minimal positive integer. But $p^ktoinfty$ gives a contradiction.
$endgroup$
– dan_fulea
Dec 4 '18 at 18:22




$begingroup$
First of all the intersection is an ideal. If it is not $(0)$, then it contains a minimal positive integer. But $p^ktoinfty$ gives a contradiction.
$endgroup$
– dan_fulea
Dec 4 '18 at 18:22












$begingroup$
Thanks for your answer. But I would like a proof to consider the unique factorization theorem.
$endgroup$
– Jack J.
Dec 4 '18 at 19:08






$begingroup$
Thanks for your answer. But I would like a proof to consider the unique factorization theorem.
$endgroup$
– Jack J.
Dec 4 '18 at 19:08












1 Answer
1






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oldest

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$begingroup$

Follows what I would consider a more-or-less formal proof of the assertion than $bigcap_{k ge 1} (p^k) = {0}$, where $p in Bbb P$ is a prime:



If



$displaystyle bigcap_{kge 1} (p^k) ne {0}, tag 1$



then



$exists 0 ne z in Bbb Z, ; z in displaystyle bigcap_{k ge 1} (p^k), tag 2$



and since $bigcap_{k ge 1} (p^k)$ is an ideal,



$z in displaystyle bigcap_{k ge 1} (p^k) Longleftrightarrow -z in displaystyle bigcap_{k ge 1} (p^k); tag 3$



thus we may assume



$z > 0; tag 4$



now by (2) we have



$forall k ge 1, ; z in (p^k); tag 5$



then



$forall k ge 1 ; exists 0 < m_k in Bbb Z, z = m_k p^k; tag 6$



we observe that if $z ne 0$ then this forces $z > 1$; however, then this contradicts the fundamental theorem of arithmetic, which asserts that the prime factorization of every positive integer is unique. We thus conclude that (1) is false, i.e. that



$displaystyle bigcap_{kge 1} (p^k) = {0}. tag 7$






share|cite|improve this answer









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    2












    $begingroup$

    Follows what I would consider a more-or-less formal proof of the assertion than $bigcap_{k ge 1} (p^k) = {0}$, where $p in Bbb P$ is a prime:



    If



    $displaystyle bigcap_{kge 1} (p^k) ne {0}, tag 1$



    then



    $exists 0 ne z in Bbb Z, ; z in displaystyle bigcap_{k ge 1} (p^k), tag 2$



    and since $bigcap_{k ge 1} (p^k)$ is an ideal,



    $z in displaystyle bigcap_{k ge 1} (p^k) Longleftrightarrow -z in displaystyle bigcap_{k ge 1} (p^k); tag 3$



    thus we may assume



    $z > 0; tag 4$



    now by (2) we have



    $forall k ge 1, ; z in (p^k); tag 5$



    then



    $forall k ge 1 ; exists 0 < m_k in Bbb Z, z = m_k p^k; tag 6$



    we observe that if $z ne 0$ then this forces $z > 1$; however, then this contradicts the fundamental theorem of arithmetic, which asserts that the prime factorization of every positive integer is unique. We thus conclude that (1) is false, i.e. that



    $displaystyle bigcap_{kge 1} (p^k) = {0}. tag 7$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Follows what I would consider a more-or-less formal proof of the assertion than $bigcap_{k ge 1} (p^k) = {0}$, where $p in Bbb P$ is a prime:



      If



      $displaystyle bigcap_{kge 1} (p^k) ne {0}, tag 1$



      then



      $exists 0 ne z in Bbb Z, ; z in displaystyle bigcap_{k ge 1} (p^k), tag 2$



      and since $bigcap_{k ge 1} (p^k)$ is an ideal,



      $z in displaystyle bigcap_{k ge 1} (p^k) Longleftrightarrow -z in displaystyle bigcap_{k ge 1} (p^k); tag 3$



      thus we may assume



      $z > 0; tag 4$



      now by (2) we have



      $forall k ge 1, ; z in (p^k); tag 5$



      then



      $forall k ge 1 ; exists 0 < m_k in Bbb Z, z = m_k p^k; tag 6$



      we observe that if $z ne 0$ then this forces $z > 1$; however, then this contradicts the fundamental theorem of arithmetic, which asserts that the prime factorization of every positive integer is unique. We thus conclude that (1) is false, i.e. that



      $displaystyle bigcap_{kge 1} (p^k) = {0}. tag 7$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Follows what I would consider a more-or-less formal proof of the assertion than $bigcap_{k ge 1} (p^k) = {0}$, where $p in Bbb P$ is a prime:



        If



        $displaystyle bigcap_{kge 1} (p^k) ne {0}, tag 1$



        then



        $exists 0 ne z in Bbb Z, ; z in displaystyle bigcap_{k ge 1} (p^k), tag 2$



        and since $bigcap_{k ge 1} (p^k)$ is an ideal,



        $z in displaystyle bigcap_{k ge 1} (p^k) Longleftrightarrow -z in displaystyle bigcap_{k ge 1} (p^k); tag 3$



        thus we may assume



        $z > 0; tag 4$



        now by (2) we have



        $forall k ge 1, ; z in (p^k); tag 5$



        then



        $forall k ge 1 ; exists 0 < m_k in Bbb Z, z = m_k p^k; tag 6$



        we observe that if $z ne 0$ then this forces $z > 1$; however, then this contradicts the fundamental theorem of arithmetic, which asserts that the prime factorization of every positive integer is unique. We thus conclude that (1) is false, i.e. that



        $displaystyle bigcap_{kge 1} (p^k) = {0}. tag 7$






        share|cite|improve this answer









        $endgroup$



        Follows what I would consider a more-or-less formal proof of the assertion than $bigcap_{k ge 1} (p^k) = {0}$, where $p in Bbb P$ is a prime:



        If



        $displaystyle bigcap_{kge 1} (p^k) ne {0}, tag 1$



        then



        $exists 0 ne z in Bbb Z, ; z in displaystyle bigcap_{k ge 1} (p^k), tag 2$



        and since $bigcap_{k ge 1} (p^k)$ is an ideal,



        $z in displaystyle bigcap_{k ge 1} (p^k) Longleftrightarrow -z in displaystyle bigcap_{k ge 1} (p^k); tag 3$



        thus we may assume



        $z > 0; tag 4$



        now by (2) we have



        $forall k ge 1, ; z in (p^k); tag 5$



        then



        $forall k ge 1 ; exists 0 < m_k in Bbb Z, z = m_k p^k; tag 6$



        we observe that if $z ne 0$ then this forces $z > 1$; however, then this contradicts the fundamental theorem of arithmetic, which asserts that the prime factorization of every positive integer is unique. We thus conclude that (1) is false, i.e. that



        $displaystyle bigcap_{kge 1} (p^k) = {0}. tag 7$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 4 '18 at 19:17









        Robert LewisRobert Lewis

        44.9k22964




        44.9k22964






























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