The nilradical and Jacobson radical of $mathbb{Z}_m$











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I want to proove (finding explicit formulas) that the Jacobson radical of $mathbb{Z}_m$ equals its nilradical: i.e. $ mathcal{N}(mathbb{Z}_m)=mathcal{J}(mathbb{Z}_m) $. I prooved a formula (following Atiyah - Mac Donald, Introduction to commutative algebra - pag. 6 - Examples 1) for the ring $mathbb{Z}_m$, where $m>1$, $ m=p_1^{s_1}cdots p_t^{s_t} $ and $ I=lbrace 1,ldots,trbrace $; is
: $ mathcal{J}(mathbb{Z}_m)=bigcap_{ mathfrak{m}vartriangleleftcdot mathbb{Z}_m}mathfrak{m}=bigcap_{iin I}dfrac{p_imathbb{Z}}{mmathbb{Z}}=bigcap_{iin I}([p_i]_{mmathbb{Z}})=bigl(text{lcd} ([p_1]_{mmathbb{Z}},ldots,[p_t]_{mmathbb{Z}})bigl)=left(dfrac{[p_1]_{mmathbb{Z}}cdots[p_t]_{mmathbb{Z}}}{text{GCD} ([p_1]_{mmathbb{Z}},ldots,[p_t]_{mmathbb{Z}})}right)= ([p_1cdots p_t]_{mmathbb{Z}})=dfrac{p_1cdots p_tmathbb{Z}}{mmathbb{Z}}=dfrac{sqrt{mmathbb{Z}}}{mmathbb{Z}}=mathcal{N}(mathbb{Z}/mmathbb{Z})=mathcal{N}(mathbb{Z}_m) $



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    I want to proove (finding explicit formulas) that the Jacobson radical of $mathbb{Z}_m$ equals its nilradical: i.e. $ mathcal{N}(mathbb{Z}_m)=mathcal{J}(mathbb{Z}_m) $. I prooved a formula (following Atiyah - Mac Donald, Introduction to commutative algebra - pag. 6 - Examples 1) for the ring $mathbb{Z}_m$, where $m>1$, $ m=p_1^{s_1}cdots p_t^{s_t} $ and $ I=lbrace 1,ldots,trbrace $; is
    : $ mathcal{J}(mathbb{Z}_m)=bigcap_{ mathfrak{m}vartriangleleftcdot mathbb{Z}_m}mathfrak{m}=bigcap_{iin I}dfrac{p_imathbb{Z}}{mmathbb{Z}}=bigcap_{iin I}([p_i]_{mmathbb{Z}})=bigl(text{lcd} ([p_1]_{mmathbb{Z}},ldots,[p_t]_{mmathbb{Z}})bigl)=left(dfrac{[p_1]_{mmathbb{Z}}cdots[p_t]_{mmathbb{Z}}}{text{GCD} ([p_1]_{mmathbb{Z}},ldots,[p_t]_{mmathbb{Z}})}right)= ([p_1cdots p_t]_{mmathbb{Z}})=dfrac{p_1cdots p_tmathbb{Z}}{mmathbb{Z}}=dfrac{sqrt{mmathbb{Z}}}{mmathbb{Z}}=mathcal{N}(mathbb{Z}/mmathbb{Z})=mathcal{N}(mathbb{Z}_m) $



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      I want to proove (finding explicit formulas) that the Jacobson radical of $mathbb{Z}_m$ equals its nilradical: i.e. $ mathcal{N}(mathbb{Z}_m)=mathcal{J}(mathbb{Z}_m) $. I prooved a formula (following Atiyah - Mac Donald, Introduction to commutative algebra - pag. 6 - Examples 1) for the ring $mathbb{Z}_m$, where $m>1$, $ m=p_1^{s_1}cdots p_t^{s_t} $ and $ I=lbrace 1,ldots,trbrace $; is
      : $ mathcal{J}(mathbb{Z}_m)=bigcap_{ mathfrak{m}vartriangleleftcdot mathbb{Z}_m}mathfrak{m}=bigcap_{iin I}dfrac{p_imathbb{Z}}{mmathbb{Z}}=bigcap_{iin I}([p_i]_{mmathbb{Z}})=bigl(text{lcd} ([p_1]_{mmathbb{Z}},ldots,[p_t]_{mmathbb{Z}})bigl)=left(dfrac{[p_1]_{mmathbb{Z}}cdots[p_t]_{mmathbb{Z}}}{text{GCD} ([p_1]_{mmathbb{Z}},ldots,[p_t]_{mmathbb{Z}})}right)= ([p_1cdots p_t]_{mmathbb{Z}})=dfrac{p_1cdots p_tmathbb{Z}}{mmathbb{Z}}=dfrac{sqrt{mmathbb{Z}}}{mmathbb{Z}}=mathcal{N}(mathbb{Z}/mmathbb{Z})=mathcal{N}(mathbb{Z}_m) $



      Is it correct?










      share|cite|improve this question















      I want to proove (finding explicit formulas) that the Jacobson radical of $mathbb{Z}_m$ equals its nilradical: i.e. $ mathcal{N}(mathbb{Z}_m)=mathcal{J}(mathbb{Z}_m) $. I prooved a formula (following Atiyah - Mac Donald, Introduction to commutative algebra - pag. 6 - Examples 1) for the ring $mathbb{Z}_m$, where $m>1$, $ m=p_1^{s_1}cdots p_t^{s_t} $ and $ I=lbrace 1,ldots,trbrace $; is
      : $ mathcal{J}(mathbb{Z}_m)=bigcap_{ mathfrak{m}vartriangleleftcdot mathbb{Z}_m}mathfrak{m}=bigcap_{iin I}dfrac{p_imathbb{Z}}{mmathbb{Z}}=bigcap_{iin I}([p_i]_{mmathbb{Z}})=bigl(text{lcd} ([p_1]_{mmathbb{Z}},ldots,[p_t]_{mmathbb{Z}})bigl)=left(dfrac{[p_1]_{mmathbb{Z}}cdots[p_t]_{mmathbb{Z}}}{text{GCD} ([p_1]_{mmathbb{Z}},ldots,[p_t]_{mmathbb{Z}})}right)= ([p_1cdots p_t]_{mmathbb{Z}})=dfrac{p_1cdots p_tmathbb{Z}}{mmathbb{Z}}=dfrac{sqrt{mmathbb{Z}}}{mmathbb{Z}}=mathcal{N}(mathbb{Z}/mmathbb{Z})=mathcal{N}(mathbb{Z}_m) $



      Is it correct?







      ring-theory commutative-algebra






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      edited Nov 18 at 14:12

























      asked Nov 18 at 12:10









      Giuseppe Joseph Peppe Arnone

      33




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          Your formula is correct and in fact we have that $mathcal{J}(mathbb{Z}_{m})=mathcal{N}(mathbb{Z}_{m})$. Indeed, if $P$ is a non-zero prime ideal in $mathbb{Z}_{m}$ then $P=nmathbb{Z}/mmathbb{Z}$ for some $ninmathbb{Z}$ by the correspondence theorem (3rd isomorphism theorem) for rings. But note that $nmathbb{Z}$ is a prime ideal in $mathbb{Z}$ since $mathbb{Z}_{m}/P$ is an integral domain and
          $$mathbb{Z}_{m}/P=(mathbb{Z}/mmathbb{Z})/(nmathbb{Z}/mmathbb{Z})simeqmathbb{Z}/nmathbb{Z}$$
          by the correspondence theorem. But this implies that $mathbb{Z}/nmathbb{Z}$ is a field because every non-zero prime ideal in $mathbb{Z}$ is maximal and therefore $P$ is a maximal ideal in $mathbb{Z}_{m}$ as well. Thus
          $$mathcal{J}(mathbb{Z}_{m})=bigcaplimits_{mathfrak{m}text{ is maximal}}mathfrak{m}=bigcaplimits_{mathfrak{p}text{ is prime}}mathfrak{p}=mathcal{N}(mathbb{Z}_{m}).$$






          share|cite|improve this answer





















          • Bug: $,P = (4) = (2)$ is prime in $Bbb Z_6$ but $,Bbb Z_6/4 notcong Bbb Z/4 $
            – Bill Dubuque
            Nov 18 at 15:05












          • There really isn't a bug @BillDubuque since $nmid m$ by the correspondence theorem and $4notmid 6$, though I could have been more explicit.
            – YumekuiMath
            Nov 20 at 15:24










          • Yes, the bug can easily be fixed, but without mention of that in the proof it is still a bug, as the example I gave shoiws.
            – Bill Dubuque
            Nov 20 at 15:32













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          Your formula is correct and in fact we have that $mathcal{J}(mathbb{Z}_{m})=mathcal{N}(mathbb{Z}_{m})$. Indeed, if $P$ is a non-zero prime ideal in $mathbb{Z}_{m}$ then $P=nmathbb{Z}/mmathbb{Z}$ for some $ninmathbb{Z}$ by the correspondence theorem (3rd isomorphism theorem) for rings. But note that $nmathbb{Z}$ is a prime ideal in $mathbb{Z}$ since $mathbb{Z}_{m}/P$ is an integral domain and
          $$mathbb{Z}_{m}/P=(mathbb{Z}/mmathbb{Z})/(nmathbb{Z}/mmathbb{Z})simeqmathbb{Z}/nmathbb{Z}$$
          by the correspondence theorem. But this implies that $mathbb{Z}/nmathbb{Z}$ is a field because every non-zero prime ideal in $mathbb{Z}$ is maximal and therefore $P$ is a maximal ideal in $mathbb{Z}_{m}$ as well. Thus
          $$mathcal{J}(mathbb{Z}_{m})=bigcaplimits_{mathfrak{m}text{ is maximal}}mathfrak{m}=bigcaplimits_{mathfrak{p}text{ is prime}}mathfrak{p}=mathcal{N}(mathbb{Z}_{m}).$$






          share|cite|improve this answer





















          • Bug: $,P = (4) = (2)$ is prime in $Bbb Z_6$ but $,Bbb Z_6/4 notcong Bbb Z/4 $
            – Bill Dubuque
            Nov 18 at 15:05












          • There really isn't a bug @BillDubuque since $nmid m$ by the correspondence theorem and $4notmid 6$, though I could have been more explicit.
            – YumekuiMath
            Nov 20 at 15:24










          • Yes, the bug can easily be fixed, but without mention of that in the proof it is still a bug, as the example I gave shoiws.
            – Bill Dubuque
            Nov 20 at 15:32

















          up vote
          0
          down vote



          accepted










          Your formula is correct and in fact we have that $mathcal{J}(mathbb{Z}_{m})=mathcal{N}(mathbb{Z}_{m})$. Indeed, if $P$ is a non-zero prime ideal in $mathbb{Z}_{m}$ then $P=nmathbb{Z}/mmathbb{Z}$ for some $ninmathbb{Z}$ by the correspondence theorem (3rd isomorphism theorem) for rings. But note that $nmathbb{Z}$ is a prime ideal in $mathbb{Z}$ since $mathbb{Z}_{m}/P$ is an integral domain and
          $$mathbb{Z}_{m}/P=(mathbb{Z}/mmathbb{Z})/(nmathbb{Z}/mmathbb{Z})simeqmathbb{Z}/nmathbb{Z}$$
          by the correspondence theorem. But this implies that $mathbb{Z}/nmathbb{Z}$ is a field because every non-zero prime ideal in $mathbb{Z}$ is maximal and therefore $P$ is a maximal ideal in $mathbb{Z}_{m}$ as well. Thus
          $$mathcal{J}(mathbb{Z}_{m})=bigcaplimits_{mathfrak{m}text{ is maximal}}mathfrak{m}=bigcaplimits_{mathfrak{p}text{ is prime}}mathfrak{p}=mathcal{N}(mathbb{Z}_{m}).$$






          share|cite|improve this answer





















          • Bug: $,P = (4) = (2)$ is prime in $Bbb Z_6$ but $,Bbb Z_6/4 notcong Bbb Z/4 $
            – Bill Dubuque
            Nov 18 at 15:05












          • There really isn't a bug @BillDubuque since $nmid m$ by the correspondence theorem and $4notmid 6$, though I could have been more explicit.
            – YumekuiMath
            Nov 20 at 15:24










          • Yes, the bug can easily be fixed, but without mention of that in the proof it is still a bug, as the example I gave shoiws.
            – Bill Dubuque
            Nov 20 at 15:32















          up vote
          0
          down vote



          accepted







          up vote
          0
          down vote



          accepted






          Your formula is correct and in fact we have that $mathcal{J}(mathbb{Z}_{m})=mathcal{N}(mathbb{Z}_{m})$. Indeed, if $P$ is a non-zero prime ideal in $mathbb{Z}_{m}$ then $P=nmathbb{Z}/mmathbb{Z}$ for some $ninmathbb{Z}$ by the correspondence theorem (3rd isomorphism theorem) for rings. But note that $nmathbb{Z}$ is a prime ideal in $mathbb{Z}$ since $mathbb{Z}_{m}/P$ is an integral domain and
          $$mathbb{Z}_{m}/P=(mathbb{Z}/mmathbb{Z})/(nmathbb{Z}/mmathbb{Z})simeqmathbb{Z}/nmathbb{Z}$$
          by the correspondence theorem. But this implies that $mathbb{Z}/nmathbb{Z}$ is a field because every non-zero prime ideal in $mathbb{Z}$ is maximal and therefore $P$ is a maximal ideal in $mathbb{Z}_{m}$ as well. Thus
          $$mathcal{J}(mathbb{Z}_{m})=bigcaplimits_{mathfrak{m}text{ is maximal}}mathfrak{m}=bigcaplimits_{mathfrak{p}text{ is prime}}mathfrak{p}=mathcal{N}(mathbb{Z}_{m}).$$






          share|cite|improve this answer












          Your formula is correct and in fact we have that $mathcal{J}(mathbb{Z}_{m})=mathcal{N}(mathbb{Z}_{m})$. Indeed, if $P$ is a non-zero prime ideal in $mathbb{Z}_{m}$ then $P=nmathbb{Z}/mmathbb{Z}$ for some $ninmathbb{Z}$ by the correspondence theorem (3rd isomorphism theorem) for rings. But note that $nmathbb{Z}$ is a prime ideal in $mathbb{Z}$ since $mathbb{Z}_{m}/P$ is an integral domain and
          $$mathbb{Z}_{m}/P=(mathbb{Z}/mmathbb{Z})/(nmathbb{Z}/mmathbb{Z})simeqmathbb{Z}/nmathbb{Z}$$
          by the correspondence theorem. But this implies that $mathbb{Z}/nmathbb{Z}$ is a field because every non-zero prime ideal in $mathbb{Z}$ is maximal and therefore $P$ is a maximal ideal in $mathbb{Z}_{m}$ as well. Thus
          $$mathcal{J}(mathbb{Z}_{m})=bigcaplimits_{mathfrak{m}text{ is maximal}}mathfrak{m}=bigcaplimits_{mathfrak{p}text{ is prime}}mathfrak{p}=mathcal{N}(mathbb{Z}_{m}).$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 18 at 14:25









          YumekuiMath

          34114




          34114












          • Bug: $,P = (4) = (2)$ is prime in $Bbb Z_6$ but $,Bbb Z_6/4 notcong Bbb Z/4 $
            – Bill Dubuque
            Nov 18 at 15:05












          • There really isn't a bug @BillDubuque since $nmid m$ by the correspondence theorem and $4notmid 6$, though I could have been more explicit.
            – YumekuiMath
            Nov 20 at 15:24










          • Yes, the bug can easily be fixed, but without mention of that in the proof it is still a bug, as the example I gave shoiws.
            – Bill Dubuque
            Nov 20 at 15:32




















          • Bug: $,P = (4) = (2)$ is prime in $Bbb Z_6$ but $,Bbb Z_6/4 notcong Bbb Z/4 $
            – Bill Dubuque
            Nov 18 at 15:05












          • There really isn't a bug @BillDubuque since $nmid m$ by the correspondence theorem and $4notmid 6$, though I could have been more explicit.
            – YumekuiMath
            Nov 20 at 15:24










          • Yes, the bug can easily be fixed, but without mention of that in the proof it is still a bug, as the example I gave shoiws.
            – Bill Dubuque
            Nov 20 at 15:32


















          Bug: $,P = (4) = (2)$ is prime in $Bbb Z_6$ but $,Bbb Z_6/4 notcong Bbb Z/4 $
          – Bill Dubuque
          Nov 18 at 15:05






          Bug: $,P = (4) = (2)$ is prime in $Bbb Z_6$ but $,Bbb Z_6/4 notcong Bbb Z/4 $
          – Bill Dubuque
          Nov 18 at 15:05














          There really isn't a bug @BillDubuque since $nmid m$ by the correspondence theorem and $4notmid 6$, though I could have been more explicit.
          – YumekuiMath
          Nov 20 at 15:24




          There really isn't a bug @BillDubuque since $nmid m$ by the correspondence theorem and $4notmid 6$, though I could have been more explicit.
          – YumekuiMath
          Nov 20 at 15:24












          Yes, the bug can easily be fixed, but without mention of that in the proof it is still a bug, as the example I gave shoiws.
          – Bill Dubuque
          Nov 20 at 15:32






          Yes, the bug can easily be fixed, but without mention of that in the proof it is still a bug, as the example I gave shoiws.
          – Bill Dubuque
          Nov 20 at 15:32




















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