Commutative ring with unity












2












$begingroup$


I'm given the following question :



$mathbb{Z} × mathbb{Z}= {(a,b) : a, b in mathbb{Z}}$, where $mathbb{Z}$ is the set of integers.
Prove that it is a commutative ring with unity.



I am able to prove that the set is closed under addition. But to prove this for multiplication, I arrive at : $(a,b)cdot(c,d)$
How should I carry out this multiplication?






abstract-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    First you should think about how multiplication and addition can be defined.
    $endgroup$
    – Paul K
    Dec 7 '18 at 8:37










  • $begingroup$
    answers.yahoo.com/question/…
    $endgroup$
    – 1ENİGMA1
    Dec 7 '18 at 8:45










  • $begingroup$
    Its component-wise.
    $endgroup$
    – Wuestenfux
    Dec 7 '18 at 8:47










  • $begingroup$
    You do have an appropriate software to write it accurately, though; it is just done with pure text: $Bbb ZtimesBbb Z$ gives $Bbb Ztimes Bbb Z$. For more information, you can check out this page.
    $endgroup$
    – Arthur
    Dec 7 '18 at 8:49










  • $begingroup$
    @Wuestenfux Since OP isn't you, how know it's componentwise? Or did you find a source for the problem?
    $endgroup$
    – coffeemath
    Dec 7 '18 at 10:10
















2












$begingroup$


I'm given the following question :



$mathbb{Z} × mathbb{Z}= {(a,b) : a, b in mathbb{Z}}$, where $mathbb{Z}$ is the set of integers.
Prove that it is a commutative ring with unity.



I am able to prove that the set is closed under addition. But to prove this for multiplication, I arrive at : $(a,b)cdot(c,d)$
How should I carry out this multiplication?






abstract-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    First you should think about how multiplication and addition can be defined.
    $endgroup$
    – Paul K
    Dec 7 '18 at 8:37










  • $begingroup$
    answers.yahoo.com/question/…
    $endgroup$
    – 1ENİGMA1
    Dec 7 '18 at 8:45










  • $begingroup$
    Its component-wise.
    $endgroup$
    – Wuestenfux
    Dec 7 '18 at 8:47










  • $begingroup$
    You do have an appropriate software to write it accurately, though; it is just done with pure text: $Bbb ZtimesBbb Z$ gives $Bbb Ztimes Bbb Z$. For more information, you can check out this page.
    $endgroup$
    – Arthur
    Dec 7 '18 at 8:49










  • $begingroup$
    @Wuestenfux Since OP isn't you, how know it's componentwise? Or did you find a source for the problem?
    $endgroup$
    – coffeemath
    Dec 7 '18 at 10:10














2












2








2


0



$begingroup$


I'm given the following question :



$mathbb{Z} × mathbb{Z}= {(a,b) : a, b in mathbb{Z}}$, where $mathbb{Z}$ is the set of integers.
Prove that it is a commutative ring with unity.



I am able to prove that the set is closed under addition. But to prove this for multiplication, I arrive at : $(a,b)cdot(c,d)$
How should I carry out this multiplication?






abstract-algebra







share|cite|improve this question











$endgroup$




I'm given the following question :



$mathbb{Z} × mathbb{Z}= {(a,b) : a, b in mathbb{Z}}$, where $mathbb{Z}$ is the set of integers.
Prove that it is a commutative ring with unity.



I am able to prove that the set is closed under addition. But to prove this for multiplication, I arrive at : $(a,b)cdot(c,d)$
How should I carry out this multiplication?






abstract-algebra




abstract-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 8:55









egreg

181k1485203




181k1485203










asked Dec 7 '18 at 8:35









JasmineJasmine

313




313












  • $begingroup$
    First you should think about how multiplication and addition can be defined.
    $endgroup$
    – Paul K
    Dec 7 '18 at 8:37










  • $begingroup$
    answers.yahoo.com/question/…
    $endgroup$
    – 1ENİGMA1
    Dec 7 '18 at 8:45










  • $begingroup$
    Its component-wise.
    $endgroup$
    – Wuestenfux
    Dec 7 '18 at 8:47










  • $begingroup$
    You do have an appropriate software to write it accurately, though; it is just done with pure text: $Bbb ZtimesBbb Z$ gives $Bbb Ztimes Bbb Z$. For more information, you can check out this page.
    $endgroup$
    – Arthur
    Dec 7 '18 at 8:49










  • $begingroup$
    @Wuestenfux Since OP isn't you, how know it's componentwise? Or did you find a source for the problem?
    $endgroup$
    – coffeemath
    Dec 7 '18 at 10:10


















  • $begingroup$
    First you should think about how multiplication and addition can be defined.
    $endgroup$
    – Paul K
    Dec 7 '18 at 8:37










  • $begingroup$
    answers.yahoo.com/question/…
    $endgroup$
    – 1ENİGMA1
    Dec 7 '18 at 8:45










  • $begingroup$
    Its component-wise.
    $endgroup$
    – Wuestenfux
    Dec 7 '18 at 8:47










  • $begingroup$
    You do have an appropriate software to write it accurately, though; it is just done with pure text: $Bbb ZtimesBbb Z$ gives $Bbb Ztimes Bbb Z$. For more information, you can check out this page.
    $endgroup$
    – Arthur
    Dec 7 '18 at 8:49










  • $begingroup$
    @Wuestenfux Since OP isn't you, how know it's componentwise? Or did you find a source for the problem?
    $endgroup$
    – coffeemath
    Dec 7 '18 at 10:10
















$begingroup$
First you should think about how multiplication and addition can be defined.
$endgroup$
– Paul K
Dec 7 '18 at 8:37




$begingroup$
First you should think about how multiplication and addition can be defined.
$endgroup$
– Paul K
Dec 7 '18 at 8:37












$begingroup$
answers.yahoo.com/question/…
$endgroup$
– 1ENİGMA1
Dec 7 '18 at 8:45




$begingroup$
answers.yahoo.com/question/…
$endgroup$
– 1ENİGMA1
Dec 7 '18 at 8:45












$begingroup$
Its component-wise.
$endgroup$
– Wuestenfux
Dec 7 '18 at 8:47




$begingroup$
Its component-wise.
$endgroup$
– Wuestenfux
Dec 7 '18 at 8:47












$begingroup$
You do have an appropriate software to write it accurately, though; it is just done with pure text: $Bbb ZtimesBbb Z$ gives $Bbb Ztimes Bbb Z$. For more information, you can check out this page.
$endgroup$
– Arthur
Dec 7 '18 at 8:49




$begingroup$
You do have an appropriate software to write it accurately, though; it is just done with pure text: $Bbb ZtimesBbb Z$ gives $Bbb Ztimes Bbb Z$. For more information, you can check out this page.
$endgroup$
– Arthur
Dec 7 '18 at 8:49












$begingroup$
@Wuestenfux Since OP isn't you, how know it's componentwise? Or did you find a source for the problem?
$endgroup$
– coffeemath
Dec 7 '18 at 10:10




$begingroup$
@Wuestenfux Since OP isn't you, how know it's componentwise? Or did you find a source for the problem?
$endgroup$
– coffeemath
Dec 7 '18 at 10:10










1 Answer
1






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2












$begingroup$

Before saying a set is a ring, you have to specify the operations.



There are infinitely many ways to turn $mathbb{Z}timesmathbb{Z}$ into a commutative ring with unity, as well as infinitely many ways to turn it into a noncommutative ring with unity.



Here are three out of the infinitely many possibilities. In all cases, the addition is $(a,b)+(c,d)=(a+c,b+d)$.




  1. $(a,b)(c,d)=(ac,bd)$

  2. $(a,b)(c,d)=(ac,ad+bc)$

  3. $(a,b)(c,d)=(ac-bd,ad+bc)$






share|cite|improve this answer









$endgroup$













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    1 Answer
    1






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    active

    oldest

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    active

    oldest

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    2












    $begingroup$

    Before saying a set is a ring, you have to specify the operations.



    There are infinitely many ways to turn $mathbb{Z}timesmathbb{Z}$ into a commutative ring with unity, as well as infinitely many ways to turn it into a noncommutative ring with unity.



    Here are three out of the infinitely many possibilities. In all cases, the addition is $(a,b)+(c,d)=(a+c,b+d)$.




    1. $(a,b)(c,d)=(ac,bd)$

    2. $(a,b)(c,d)=(ac,ad+bc)$

    3. $(a,b)(c,d)=(ac-bd,ad+bc)$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Before saying a set is a ring, you have to specify the operations.



      There are infinitely many ways to turn $mathbb{Z}timesmathbb{Z}$ into a commutative ring with unity, as well as infinitely many ways to turn it into a noncommutative ring with unity.



      Here are three out of the infinitely many possibilities. In all cases, the addition is $(a,b)+(c,d)=(a+c,b+d)$.




      1. $(a,b)(c,d)=(ac,bd)$

      2. $(a,b)(c,d)=(ac,ad+bc)$

      3. $(a,b)(c,d)=(ac-bd,ad+bc)$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Before saying a set is a ring, you have to specify the operations.



        There are infinitely many ways to turn $mathbb{Z}timesmathbb{Z}$ into a commutative ring with unity, as well as infinitely many ways to turn it into a noncommutative ring with unity.



        Here are three out of the infinitely many possibilities. In all cases, the addition is $(a,b)+(c,d)=(a+c,b+d)$.




        1. $(a,b)(c,d)=(ac,bd)$

        2. $(a,b)(c,d)=(ac,ad+bc)$

        3. $(a,b)(c,d)=(ac-bd,ad+bc)$






        share|cite|improve this answer









        $endgroup$



        Before saying a set is a ring, you have to specify the operations.



        There are infinitely many ways to turn $mathbb{Z}timesmathbb{Z}$ into a commutative ring with unity, as well as infinitely many ways to turn it into a noncommutative ring with unity.



        Here are three out of the infinitely many possibilities. In all cases, the addition is $(a,b)+(c,d)=(a+c,b+d)$.




        1. $(a,b)(c,d)=(ac,bd)$

        2. $(a,b)(c,d)=(ac,ad+bc)$

        3. $(a,b)(c,d)=(ac-bd,ad+bc)$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 7 '18 at 8:59









        egregegreg

        181k1485203




        181k1485203






























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