Ring of $mathbb{Z}_2$-valued functions











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Let $R ={f:{1,2,3,4,cdots ,10}to mathbb{Z}_2 }$ be the set of
all $mathbb{Z}_2$ -valued functions on the set ${1,2,3......10}$ of the first ten positive integer. Then $R$ is a commutative ring with point-wise addition and multiplication of functions. Which of the following is true?




1) $R$ has a unique maximal ideal.



2) Every prime ideal of $R$ is also maximal.



3 ) The number of proper ideals of $R$ is $511$.



4 ) Every element of $R$ is idempotent.




R is finite commutative ring so every prime ideal is maximlal ideal ,so I think option 2 is true .But for the other options I am helpless .










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  • what is "Z2" is it "$mathbb{Z}_2 = mathbb{Z}/2mathbb{Z} = mathbb{F}_2$" the field with 2 elements? or $mathbb{Z}^2$ a lattice? or something completely different?
    – mdave16
    Jun 19 '17 at 12:17












  • If you want $mathbb{Z}^2$ just change $mathbb{Z}_2$ to $mathbb{Z}^2$.
    – kingW3
    Jun 19 '17 at 12:20












  • Yes you are right Z2 is field with 2 elements F2
    – Gilll
    Jun 19 '17 at 12:24










  • R is finite commutative ring so every prime ideal is maximlal ideal ,so I think option 2 is true .But for the other options I am helpless
    – Gilll
    Jun 19 '17 at 12:29















up vote
2
down vote

favorite
3












Let $R ={f:{1,2,3,4,cdots ,10}to mathbb{Z}_2 }$ be the set of
all $mathbb{Z}_2$ -valued functions on the set ${1,2,3......10}$ of the first ten positive integer. Then $R$ is a commutative ring with point-wise addition and multiplication of functions. Which of the following is true?




1) $R$ has a unique maximal ideal.



2) Every prime ideal of $R$ is also maximal.



3 ) The number of proper ideals of $R$ is $511$.



4 ) Every element of $R$ is idempotent.




R is finite commutative ring so every prime ideal is maximlal ideal ,so I think option 2 is true .But for the other options I am helpless .










share|cite|improve this question
























  • what is "Z2" is it "$mathbb{Z}_2 = mathbb{Z}/2mathbb{Z} = mathbb{F}_2$" the field with 2 elements? or $mathbb{Z}^2$ a lattice? or something completely different?
    – mdave16
    Jun 19 '17 at 12:17












  • If you want $mathbb{Z}^2$ just change $mathbb{Z}_2$ to $mathbb{Z}^2$.
    – kingW3
    Jun 19 '17 at 12:20












  • Yes you are right Z2 is field with 2 elements F2
    – Gilll
    Jun 19 '17 at 12:24










  • R is finite commutative ring so every prime ideal is maximlal ideal ,so I think option 2 is true .But for the other options I am helpless
    – Gilll
    Jun 19 '17 at 12:29













up vote
2
down vote

favorite
3









up vote
2
down vote

favorite
3






3





Let $R ={f:{1,2,3,4,cdots ,10}to mathbb{Z}_2 }$ be the set of
all $mathbb{Z}_2$ -valued functions on the set ${1,2,3......10}$ of the first ten positive integer. Then $R$ is a commutative ring with point-wise addition and multiplication of functions. Which of the following is true?




1) $R$ has a unique maximal ideal.



2) Every prime ideal of $R$ is also maximal.



3 ) The number of proper ideals of $R$ is $511$.



4 ) Every element of $R$ is idempotent.




R is finite commutative ring so every prime ideal is maximlal ideal ,so I think option 2 is true .But for the other options I am helpless .










share|cite|improve this question















Let $R ={f:{1,2,3,4,cdots ,10}to mathbb{Z}_2 }$ be the set of
all $mathbb{Z}_2$ -valued functions on the set ${1,2,3......10}$ of the first ten positive integer. Then $R$ is a commutative ring with point-wise addition and multiplication of functions. Which of the following is true?




1) $R$ has a unique maximal ideal.



2) Every prime ideal of $R$ is also maximal.



3 ) The number of proper ideals of $R$ is $511$.



4 ) Every element of $R$ is idempotent.




R is finite commutative ring so every prime ideal is maximlal ideal ,so I think option 2 is true .But for the other options I am helpless .







ring-theory maximal-and-prime-ideals idempotents






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edited Jun 19 '17 at 12:27

























asked Jun 19 '17 at 12:13









Gilll

1378




1378












  • what is "Z2" is it "$mathbb{Z}_2 = mathbb{Z}/2mathbb{Z} = mathbb{F}_2$" the field with 2 elements? or $mathbb{Z}^2$ a lattice? or something completely different?
    – mdave16
    Jun 19 '17 at 12:17












  • If you want $mathbb{Z}^2$ just change $mathbb{Z}_2$ to $mathbb{Z}^2$.
    – kingW3
    Jun 19 '17 at 12:20












  • Yes you are right Z2 is field with 2 elements F2
    – Gilll
    Jun 19 '17 at 12:24










  • R is finite commutative ring so every prime ideal is maximlal ideal ,so I think option 2 is true .But for the other options I am helpless
    – Gilll
    Jun 19 '17 at 12:29


















  • what is "Z2" is it "$mathbb{Z}_2 = mathbb{Z}/2mathbb{Z} = mathbb{F}_2$" the field with 2 elements? or $mathbb{Z}^2$ a lattice? or something completely different?
    – mdave16
    Jun 19 '17 at 12:17












  • If you want $mathbb{Z}^2$ just change $mathbb{Z}_2$ to $mathbb{Z}^2$.
    – kingW3
    Jun 19 '17 at 12:20












  • Yes you are right Z2 is field with 2 elements F2
    – Gilll
    Jun 19 '17 at 12:24










  • R is finite commutative ring so every prime ideal is maximlal ideal ,so I think option 2 is true .But for the other options I am helpless
    – Gilll
    Jun 19 '17 at 12:29
















what is "Z2" is it "$mathbb{Z}_2 = mathbb{Z}/2mathbb{Z} = mathbb{F}_2$" the field with 2 elements? or $mathbb{Z}^2$ a lattice? or something completely different?
– mdave16
Jun 19 '17 at 12:17






what is "Z2" is it "$mathbb{Z}_2 = mathbb{Z}/2mathbb{Z} = mathbb{F}_2$" the field with 2 elements? or $mathbb{Z}^2$ a lattice? or something completely different?
– mdave16
Jun 19 '17 at 12:17














If you want $mathbb{Z}^2$ just change $mathbb{Z}_2$ to $mathbb{Z}^2$.
– kingW3
Jun 19 '17 at 12:20






If you want $mathbb{Z}^2$ just change $mathbb{Z}_2$ to $mathbb{Z}^2$.
– kingW3
Jun 19 '17 at 12:20














Yes you are right Z2 is field with 2 elements F2
– Gilll
Jun 19 '17 at 12:24




Yes you are right Z2 is field with 2 elements F2
– Gilll
Jun 19 '17 at 12:24












R is finite commutative ring so every prime ideal is maximlal ideal ,so I think option 2 is true .But for the other options I am helpless
– Gilll
Jun 19 '17 at 12:29




R is finite commutative ring so every prime ideal is maximlal ideal ,so I think option 2 is true .But for the other options I am helpless
– Gilll
Jun 19 '17 at 12:29










3 Answers
3






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You shouldn't have any trouble showing that your ring is isomorphic to $mathbb Z_2^{10}$ (the product ring of $10$ copies of $mathbb Z_2$.) The idea is that you send $phi :{1,2,ldots, 10}tomathbb Z_2$ to $(phi(1),phi(2),ldots,phi(10))in mathbb Z_2^{10}$.



Of course $mathbb Z_2^{10}$ is a boolean ring since it is a product of boolean rings. This takes care of $(4)$.



It is well-known and easy to prove that the prime ideals are maximal in any boolean ring. Of course, it's also well-known and easy to prove that the prime ideals are maximal in finite rings. This takes care of $(2)$.



The ideals of a finite product of rings are easy to describe: they are all possible products of the ideals in the component rings. Since you have $10$ component rings, each of which have $2$ ideals, you have $2^{10}$ possible ideals. The whole ring is the one non-proper ideal you'll have to discard. This takes care of $(3)$.



The maximal ideals of a finite product of rings are also easy to describe: they are all of the form of the ideals described above, of course. It is easy to show that they are exactly those where you have a maximal ideal in one position, and the full ring in all other positions. So that gives you exactly $10$ maximal ideals, one for each position. This takes care of $(1)$.






share|cite|improve this answer




























    up vote
    1
    down vote













    4) is true because every element of $mathbb{Z}_2={0,1}$ is idempotent, so $forall Phi in R, forall kin {1,...,10}, Phi^2(k)= Phi(k) * Phi(k) = Phi(k)$.






    share|cite|improve this answer




























      up vote
      0
      down vote













      Yes as you claim (b) is true. Call A as the set of first 10 natural numbers .Now Let us see some ideals of this ring. Let $ain A$, Define: $I_{a}={ f:A to mathbb{Z}_{2}|f(a) =0 } $. Observe this is an ideal of this ring. Moreover, it is a maximal ideal. It is obvious that (d) is correct because $f^{2}=f$. Infact, this is an example of a boolean ring.
      (a) is incorrect is pretty clear as we have a quite a few maximal ideals as we will vary a from 1 to 10.



      Now about the number of ideals. observe first that if $Bsubseteq A$ we can define $I_{B}= { f:A to mathbb{Z}_{2}|f(b) =0$ $forall$ b $in B } $.



      Clearly $I_{B}$ is an ideal. If $B=A$, then $I_{B}$ is the zero ideal.



      Now how many such proper ideal we have already.



      we claim if B is empty then $I_{B}$ is the whole ring. To see this if B is empty we have for each $a in A$ , there exists $f_{a} in I$ such that $f_{a}(a)=1$.



      Define: $g_{a}=f_{a}.h_{a}$ where $h_{a}(x)=1$ if x=a or zero otherwise. clearly $g_{a} in I$. Observe $sum_{ain A}g_{a}=1$, where denotes the identity function. Hence $I_{B}=R$.



      So, if B is non-empty we have $I_{B}$ is clearly a proper ideal of R. How many proper subset B of A are there? $2^{10}-1$ isn't it? 1023 ideals we have already have which are proper. So (c) is incorrect.






      share|cite|improve this answer





















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        3 Answers
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        3 Answers
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        up vote
        3
        down vote













        You shouldn't have any trouble showing that your ring is isomorphic to $mathbb Z_2^{10}$ (the product ring of $10$ copies of $mathbb Z_2$.) The idea is that you send $phi :{1,2,ldots, 10}tomathbb Z_2$ to $(phi(1),phi(2),ldots,phi(10))in mathbb Z_2^{10}$.



        Of course $mathbb Z_2^{10}$ is a boolean ring since it is a product of boolean rings. This takes care of $(4)$.



        It is well-known and easy to prove that the prime ideals are maximal in any boolean ring. Of course, it's also well-known and easy to prove that the prime ideals are maximal in finite rings. This takes care of $(2)$.



        The ideals of a finite product of rings are easy to describe: they are all possible products of the ideals in the component rings. Since you have $10$ component rings, each of which have $2$ ideals, you have $2^{10}$ possible ideals. The whole ring is the one non-proper ideal you'll have to discard. This takes care of $(3)$.



        The maximal ideals of a finite product of rings are also easy to describe: they are all of the form of the ideals described above, of course. It is easy to show that they are exactly those where you have a maximal ideal in one position, and the full ring in all other positions. So that gives you exactly $10$ maximal ideals, one for each position. This takes care of $(1)$.






        share|cite|improve this answer

























          up vote
          3
          down vote













          You shouldn't have any trouble showing that your ring is isomorphic to $mathbb Z_2^{10}$ (the product ring of $10$ copies of $mathbb Z_2$.) The idea is that you send $phi :{1,2,ldots, 10}tomathbb Z_2$ to $(phi(1),phi(2),ldots,phi(10))in mathbb Z_2^{10}$.



          Of course $mathbb Z_2^{10}$ is a boolean ring since it is a product of boolean rings. This takes care of $(4)$.



          It is well-known and easy to prove that the prime ideals are maximal in any boolean ring. Of course, it's also well-known and easy to prove that the prime ideals are maximal in finite rings. This takes care of $(2)$.



          The ideals of a finite product of rings are easy to describe: they are all possible products of the ideals in the component rings. Since you have $10$ component rings, each of which have $2$ ideals, you have $2^{10}$ possible ideals. The whole ring is the one non-proper ideal you'll have to discard. This takes care of $(3)$.



          The maximal ideals of a finite product of rings are also easy to describe: they are all of the form of the ideals described above, of course. It is easy to show that they are exactly those where you have a maximal ideal in one position, and the full ring in all other positions. So that gives you exactly $10$ maximal ideals, one for each position. This takes care of $(1)$.






          share|cite|improve this answer























            up vote
            3
            down vote










            up vote
            3
            down vote









            You shouldn't have any trouble showing that your ring is isomorphic to $mathbb Z_2^{10}$ (the product ring of $10$ copies of $mathbb Z_2$.) The idea is that you send $phi :{1,2,ldots, 10}tomathbb Z_2$ to $(phi(1),phi(2),ldots,phi(10))in mathbb Z_2^{10}$.



            Of course $mathbb Z_2^{10}$ is a boolean ring since it is a product of boolean rings. This takes care of $(4)$.



            It is well-known and easy to prove that the prime ideals are maximal in any boolean ring. Of course, it's also well-known and easy to prove that the prime ideals are maximal in finite rings. This takes care of $(2)$.



            The ideals of a finite product of rings are easy to describe: they are all possible products of the ideals in the component rings. Since you have $10$ component rings, each of which have $2$ ideals, you have $2^{10}$ possible ideals. The whole ring is the one non-proper ideal you'll have to discard. This takes care of $(3)$.



            The maximal ideals of a finite product of rings are also easy to describe: they are all of the form of the ideals described above, of course. It is easy to show that they are exactly those where you have a maximal ideal in one position, and the full ring in all other positions. So that gives you exactly $10$ maximal ideals, one for each position. This takes care of $(1)$.






            share|cite|improve this answer












            You shouldn't have any trouble showing that your ring is isomorphic to $mathbb Z_2^{10}$ (the product ring of $10$ copies of $mathbb Z_2$.) The idea is that you send $phi :{1,2,ldots, 10}tomathbb Z_2$ to $(phi(1),phi(2),ldots,phi(10))in mathbb Z_2^{10}$.



            Of course $mathbb Z_2^{10}$ is a boolean ring since it is a product of boolean rings. This takes care of $(4)$.



            It is well-known and easy to prove that the prime ideals are maximal in any boolean ring. Of course, it's also well-known and easy to prove that the prime ideals are maximal in finite rings. This takes care of $(2)$.



            The ideals of a finite product of rings are easy to describe: they are all possible products of the ideals in the component rings. Since you have $10$ component rings, each of which have $2$ ideals, you have $2^{10}$ possible ideals. The whole ring is the one non-proper ideal you'll have to discard. This takes care of $(3)$.



            The maximal ideals of a finite product of rings are also easy to describe: they are all of the form of the ideals described above, of course. It is easy to show that they are exactly those where you have a maximal ideal in one position, and the full ring in all other positions. So that gives you exactly $10$ maximal ideals, one for each position. This takes care of $(1)$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jun 19 '17 at 14:18









            rschwieb

            104k1299238




            104k1299238






















                up vote
                1
                down vote













                4) is true because every element of $mathbb{Z}_2={0,1}$ is idempotent, so $forall Phi in R, forall kin {1,...,10}, Phi^2(k)= Phi(k) * Phi(k) = Phi(k)$.






                share|cite|improve this answer

























                  up vote
                  1
                  down vote













                  4) is true because every element of $mathbb{Z}_2={0,1}$ is idempotent, so $forall Phi in R, forall kin {1,...,10}, Phi^2(k)= Phi(k) * Phi(k) = Phi(k)$.






                  share|cite|improve this answer























                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    4) is true because every element of $mathbb{Z}_2={0,1}$ is idempotent, so $forall Phi in R, forall kin {1,...,10}, Phi^2(k)= Phi(k) * Phi(k) = Phi(k)$.






                    share|cite|improve this answer












                    4) is true because every element of $mathbb{Z}_2={0,1}$ is idempotent, so $forall Phi in R, forall kin {1,...,10}, Phi^2(k)= Phi(k) * Phi(k) = Phi(k)$.







                    share|cite|improve this answer












                    share|cite|improve this answer



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                    answered Jun 19 '17 at 12:54









                    Evargalo

                    2,41618




                    2,41618






















                        up vote
                        0
                        down vote













                        Yes as you claim (b) is true. Call A as the set of first 10 natural numbers .Now Let us see some ideals of this ring. Let $ain A$, Define: $I_{a}={ f:A to mathbb{Z}_{2}|f(a) =0 } $. Observe this is an ideal of this ring. Moreover, it is a maximal ideal. It is obvious that (d) is correct because $f^{2}=f$. Infact, this is an example of a boolean ring.
                        (a) is incorrect is pretty clear as we have a quite a few maximal ideals as we will vary a from 1 to 10.



                        Now about the number of ideals. observe first that if $Bsubseteq A$ we can define $I_{B}= { f:A to mathbb{Z}_{2}|f(b) =0$ $forall$ b $in B } $.



                        Clearly $I_{B}$ is an ideal. If $B=A$, then $I_{B}$ is the zero ideal.



                        Now how many such proper ideal we have already.



                        we claim if B is empty then $I_{B}$ is the whole ring. To see this if B is empty we have for each $a in A$ , there exists $f_{a} in I$ such that $f_{a}(a)=1$.



                        Define: $g_{a}=f_{a}.h_{a}$ where $h_{a}(x)=1$ if x=a or zero otherwise. clearly $g_{a} in I$. Observe $sum_{ain A}g_{a}=1$, where denotes the identity function. Hence $I_{B}=R$.



                        So, if B is non-empty we have $I_{B}$ is clearly a proper ideal of R. How many proper subset B of A are there? $2^{10}-1$ isn't it? 1023 ideals we have already have which are proper. So (c) is incorrect.






                        share|cite|improve this answer

























                          up vote
                          0
                          down vote













                          Yes as you claim (b) is true. Call A as the set of first 10 natural numbers .Now Let us see some ideals of this ring. Let $ain A$, Define: $I_{a}={ f:A to mathbb{Z}_{2}|f(a) =0 } $. Observe this is an ideal of this ring. Moreover, it is a maximal ideal. It is obvious that (d) is correct because $f^{2}=f$. Infact, this is an example of a boolean ring.
                          (a) is incorrect is pretty clear as we have a quite a few maximal ideals as we will vary a from 1 to 10.



                          Now about the number of ideals. observe first that if $Bsubseteq A$ we can define $I_{B}= { f:A to mathbb{Z}_{2}|f(b) =0$ $forall$ b $in B } $.



                          Clearly $I_{B}$ is an ideal. If $B=A$, then $I_{B}$ is the zero ideal.



                          Now how many such proper ideal we have already.



                          we claim if B is empty then $I_{B}$ is the whole ring. To see this if B is empty we have for each $a in A$ , there exists $f_{a} in I$ such that $f_{a}(a)=1$.



                          Define: $g_{a}=f_{a}.h_{a}$ where $h_{a}(x)=1$ if x=a or zero otherwise. clearly $g_{a} in I$. Observe $sum_{ain A}g_{a}=1$, where denotes the identity function. Hence $I_{B}=R$.



                          So, if B is non-empty we have $I_{B}$ is clearly a proper ideal of R. How many proper subset B of A are there? $2^{10}-1$ isn't it? 1023 ideals we have already have which are proper. So (c) is incorrect.






                          share|cite|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            Yes as you claim (b) is true. Call A as the set of first 10 natural numbers .Now Let us see some ideals of this ring. Let $ain A$, Define: $I_{a}={ f:A to mathbb{Z}_{2}|f(a) =0 } $. Observe this is an ideal of this ring. Moreover, it is a maximal ideal. It is obvious that (d) is correct because $f^{2}=f$. Infact, this is an example of a boolean ring.
                            (a) is incorrect is pretty clear as we have a quite a few maximal ideals as we will vary a from 1 to 10.



                            Now about the number of ideals. observe first that if $Bsubseteq A$ we can define $I_{B}= { f:A to mathbb{Z}_{2}|f(b) =0$ $forall$ b $in B } $.



                            Clearly $I_{B}$ is an ideal. If $B=A$, then $I_{B}$ is the zero ideal.



                            Now how many such proper ideal we have already.



                            we claim if B is empty then $I_{B}$ is the whole ring. To see this if B is empty we have for each $a in A$ , there exists $f_{a} in I$ such that $f_{a}(a)=1$.



                            Define: $g_{a}=f_{a}.h_{a}$ where $h_{a}(x)=1$ if x=a or zero otherwise. clearly $g_{a} in I$. Observe $sum_{ain A}g_{a}=1$, where denotes the identity function. Hence $I_{B}=R$.



                            So, if B is non-empty we have $I_{B}$ is clearly a proper ideal of R. How many proper subset B of A are there? $2^{10}-1$ isn't it? 1023 ideals we have already have which are proper. So (c) is incorrect.






                            share|cite|improve this answer












                            Yes as you claim (b) is true. Call A as the set of first 10 natural numbers .Now Let us see some ideals of this ring. Let $ain A$, Define: $I_{a}={ f:A to mathbb{Z}_{2}|f(a) =0 } $. Observe this is an ideal of this ring. Moreover, it is a maximal ideal. It is obvious that (d) is correct because $f^{2}=f$. Infact, this is an example of a boolean ring.
                            (a) is incorrect is pretty clear as we have a quite a few maximal ideals as we will vary a from 1 to 10.



                            Now about the number of ideals. observe first that if $Bsubseteq A$ we can define $I_{B}= { f:A to mathbb{Z}_{2}|f(b) =0$ $forall$ b $in B } $.



                            Clearly $I_{B}$ is an ideal. If $B=A$, then $I_{B}$ is the zero ideal.



                            Now how many such proper ideal we have already.



                            we claim if B is empty then $I_{B}$ is the whole ring. To see this if B is empty we have for each $a in A$ , there exists $f_{a} in I$ such that $f_{a}(a)=1$.



                            Define: $g_{a}=f_{a}.h_{a}$ where $h_{a}(x)=1$ if x=a or zero otherwise. clearly $g_{a} in I$. Observe $sum_{ain A}g_{a}=1$, where denotes the identity function. Hence $I_{B}=R$.



                            So, if B is non-empty we have $I_{B}$ is clearly a proper ideal of R. How many proper subset B of A are there? $2^{10}-1$ isn't it? 1023 ideals we have already have which are proper. So (c) is incorrect.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jun 19 '17 at 13:58









                            Riju

                            2,205314




                            2,205314






























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