Commutative ring with unity
$begingroup$
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
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|
show 2 more comments
$begingroup$
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
$endgroup$
$begingroup$
First you should think about how multiplication and addition can be defined.
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– Paul K
Dec 7 '18 at 8:37
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answers.yahoo.com/question/…
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– 1ENİGMA1
Dec 7 '18 at 8:45
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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.
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– Arthur
Dec 7 '18 at 8:49
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@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
|
show 2 more comments
$begingroup$
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
$endgroup$
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
edited Dec 7 '18 at 8:55
egreg
180k1485202
180k1485202
asked Dec 7 '18 at 8:35
JasmineJasmine
283
283
$begingroup$
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
|
show 2 more comments
$begingroup$
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
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
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 $(a,b)+(c,d)=(a+c,b+d)$.
- $(a,b)(c,d)=(ac,bd)$
- $(a,b)(c,d)=(ac,ad+bc)$
- $(a,b)(c,d)=(ac-bd,ad+bc)$
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $(a,b)+(c,d)=(a+c,b+d)$.
- $(a,b)(c,d)=(ac,bd)$
- $(a,b)(c,d)=(ac,ad+bc)$
- $(a,b)(c,d)=(ac-bd,ad+bc)$
$endgroup$
add a comment |
$begingroup$
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 $(a,b)+(c,d)=(a+c,b+d)$.
- $(a,b)(c,d)=(ac,bd)$
- $(a,b)(c,d)=(ac,ad+bc)$
- $(a,b)(c,d)=(ac-bd,ad+bc)$
$endgroup$
add a comment |
$begingroup$
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 $(a,b)+(c,d)=(a+c,b+d)$.
- $(a,b)(c,d)=(ac,bd)$
- $(a,b)(c,d)=(ac,ad+bc)$
- $(a,b)(c,d)=(ac-bd,ad+bc)$
$endgroup$
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 $(a,b)+(c,d)=(a+c,b+d)$.
- $(a,b)(c,d)=(ac,bd)$
- $(a,b)(c,d)=(ac,ad+bc)$
- $(a,b)(c,d)=(ac-bd,ad+bc)$
answered Dec 7 '18 at 8:59
egregegreg
180k1485202
180k1485202
add a comment |
add a comment |
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$begingroup$
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