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Clarification on the definition for product on a ring quotient
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I'm going over my notes on ring theory, and have come across the following definition for the product on a ring quotient $R/I$ for $I triangleleft R$,
$$
(a+I)cdot(b+I) := (a+I)(b+I) + I
$$
We then claim that this indeed is equal to $ab + I$ making the canonical projection a ring morphism from $R$ to the quotient. I can see more or less immediately that:
$$
ab + I subseteq (a+0)(b+0) + I subseteq (a +I)(b + I) + I
$$
However, for the other inclusion I had written that:
$$
(a+I)(b+I) + I = ab + aI + Ib + I^2 + I subseteq ab + I
$$
which I'm now not entirely convinced of it being correct, since $z in (a+I)(b+I)$ if and only if:
$$
z = sum_j = 1^n(a+x_j)(b+y_j) = sum_j = 1^nab + sum_j = 1^nay_j + sum_j = 1^nx_jb + sum_j = 1^nx_jy_j
$$
for some $x_j,y_j in I$, and this is in $nab + aI + Ib + I^2$ which may as well not be equal to $ab + aI + Ib + I^2$. Am I computing this the wrong way?
ring-theory definition ideals
asked Jul 16 at 4:25
Guido A.
3,846624
add a comment |Â
up vote
3
down vote
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I'm going over my notes on ring theory, and have come across the following definition for the product on a ring quotient $R/I$ for $I triangleleft R$,
$$
(a+I)cdot(b+I) := (a+I)(b+I) + I
$$
We then claim that this indeed is equal to $ab + I$ making the canonical projection a ring morphism from $R$ to the quotient. I can see more or less immediately that:
$$
ab + I subseteq (a+0)(b+0) + I subseteq (a +I)(b + I) + I
$$
However, for the other inclusion I had written that:
$$
(a+I)(b+I) + I = ab + aI + Ib + I^2 + I subseteq ab + I
$$
which I'm now not entirely convinced of it being correct, since $z in (a+I)(b+I)$ if and only if:
$$
z = sum_j = 1^n(a+x_j)(b+y_j) = sum_j = 1^nab + sum_j = 1^nay_j + sum_j = 1^nx_jb + sum_j = 1^nx_jy_j
$$
for some $x_j,y_j in I$, and this is in $nab + aI + Ib + I^2$ which may as well not be equal to $ab + aI + Ib + I^2$. Am I computing this the wrong way?
ring-theory definition ideals
asked Jul 16 at 4:25
Guido A.
3,846624
3
What is $(a+I)(b+I)$ on the first line (without the dot)? Presumably it's not a circular definition, so is that the set product (i.e. $AB=abmid ain A,bin B$)? If so, then the fourth line is incorrect since there is no sum involved, it is just $z=(a+x)(b+y)$ where $x,yin I$.
– Mario Carneiro
Jul 16 at 4:40
@MarioCarneiro we've defined the product $XY$ of two subsets of a ring $R$ as $$ XY = left sum_i= 1^nx_iy_i : x_i in X, y_i in Y right $$ so that's why I made the explicit distinction. Is this a non-standard definition?
– Guido A.
Jul 16 at 4:46
@GuidoA. That's the usual definition for the product of ideals but it's not the one being used here.
– Lord Shark the Unknown
Jul 16 at 5:52
@LordSharktheUnknown I see. That makes sense, since it coincides for the product of subsets of a monoid in the case of $(R, cdot)$. If you post it as an answer, I'll accept it, since that resolves the issue :) Thanks!
– Guido A.
Jul 16 at 5:54
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I'm going over my notes on ring theory, and have come across the following definition for the product on a ring quotient $R/I$ for $I triangleleft R$,
$$
(a+I)cdot(b+I) := (a+I)(b+I) + I
$$
We then claim that this indeed is equal to $ab + I$ making the canonical projection a ring morphism from $R$ to the quotient. I can see more or less immediately that:
$$
ab + I subseteq (a+0)(b+0) + I subseteq (a +I)(b + I) + I
$$
However, for the other inclusion I had written that:
$$
(a+I)(b+I) + I = ab + aI + Ib + I^2 + I subseteq ab + I
$$
which I'm now not entirely convinced of it being correct, since $z in (a+I)(b+I)$ if and only if:
$$
z = sum_j = 1^n(a+x_j)(b+y_j) = sum_j = 1^nab + sum_j = 1^nay_j + sum_j = 1^nx_jb + sum_j = 1^nx_jy_j
$$
for some $x_j,y_j in I$, and this is in $nab + aI + Ib + I^2$ which may as well not be equal to $ab + aI + Ib + I^2$. Am I computing this the wrong way?
ring-theory definition ideals
asked Jul 16 at 4:25
Guido A.
3,846624
I'm going over my notes on ring theory, and have come across the following definition for the product on a ring quotient $R/I$ for $I triangleleft R$,
$$
(a+I)cdot(b+I) := (a+I)(b+I) + I
$$
We then claim that this indeed is equal to $ab + I$ making the canonical projection a ring morphism from $R$ to the quotient. I can see more or less immediately that:
$$
ab + I subseteq (a+0)(b+0) + I subseteq (a +I)(b + I) + I
$$
However, for the other inclusion I had written that:
$$
(a+I)(b+I) + I = ab + aI + Ib + I^2 + I subseteq ab + I
$$
which I'm now not entirely convinced of it being correct, since $z in (a+I)(b+I)$ if and only if:
$$
z = sum_j = 1^n(a+x_j)(b+y_j) = sum_j = 1^nab + sum_j = 1^nay_j + sum_j = 1^nx_jb + sum_j = 1^nx_jy_j
$$
for some $x_j,y_j in I$, and this is in $nab + aI + Ib + I^2$ which may as well not be equal to $ab + aI + Ib + I^2$. Am I computing this the wrong way?
ring-theory definition ideals
asked Jul 16 at 4:25
Guido A.
3,846624
asked Jul 16 at 4:25
Guido A.
3,846624
asked Jul 16 at 4:25
Guido A.
3,846624
asked Jul 16 at 4:25
Guido A.
3,846624
3,846624
3
What is $(a+I)(b+I)$ on the first line (without the dot)? Presumably it's not a circular definition, so is that the set product (i.e. $AB=abmid ain A,bin B$)? If so, then the fourth line is incorrect since there is no sum involved, it is just $z=(a+x)(b+y)$ where $x,yin I$.
– Mario Carneiro
Jul 16 at 4:40
@MarioCarneiro we've defined the product $XY$ of two subsets of a ring $R$ as $$ XY = left sum_i= 1^nx_iy_i : x_i in X, y_i in Y right $$ so that's why I made the explicit distinction. Is this a non-standard definition?
– Guido A.
Jul 16 at 4:46
@GuidoA. That's the usual definition for the product of ideals but it's not the one being used here.
– Lord Shark the Unknown
Jul 16 at 5:52
@LordSharktheUnknown I see. That makes sense, since it coincides for the product of subsets of a monoid in the case of $(R, cdot)$. If you post it as an answer, I'll accept it, since that resolves the issue :) Thanks!
– Guido A.
Jul 16 at 5:54
add a comment |Â
3
What is $(a+I)(b+I)$ on the first line (without the dot)? Presumably it's not a circular definition, so is that the set product (i.e. $AB=abmid ain A,bin B$)? If so, then the fourth line is incorrect since there is no sum involved, it is just $z=(a+x)(b+y)$ where $x,yin I$.
– Mario Carneiro
Jul 16 at 4:40
@MarioCarneiro we've defined the product $XY$ of two subsets of a ring $R$ as $$ XY = left sum_i= 1^nx_iy_i : x_i in X, y_i in Y right $$ so that's why I made the explicit distinction. Is this a non-standard definition?
– Guido A.
Jul 16 at 4:46
@GuidoA. That's the usual definition for the product of ideals but it's not the one being used here.
– Lord Shark the Unknown
Jul 16 at 5:52
@LordSharktheUnknown I see. That makes sense, since it coincides for the product of subsets of a monoid in the case of $(R, cdot)$. If you post it as an answer, I'll accept it, since that resolves the issue :) Thanks!
– Guido A.
Jul 16 at 5:54
3
3
What is $(a+I)(b+I)$ on the first line (without the dot)? Presumably it's not a circular definition, so is that the set product (i.e. $AB=abmid ain A,bin B$)? If so, then the fourth line is incorrect since there is no sum involved, it is just $z=(a+x)(b+y)$ where $x,yin I$.
– Mario Carneiro
Jul 16 at 4:40
What is $(a+I)(b+I)$ on the first line (without the dot)? Presumably it's not a circular definition, so is that the set product (i.e. $AB=abmid ain A,bin B$)? If so, then the fourth line is incorrect since there is no sum involved, it is just $z=(a+x)(b+y)$ where $x,yin I$.
– Mario Carneiro
Jul 16 at 4:40
@MarioCarneiro we've defined the product $XY$ of two subsets of a ring $R$ as $$ XY = left sum_i= 1^nx_iy_i : x_i in X, y_i in Y right $$ so that's why I made the explicit distinction. Is this a non-standard definition?
– Guido A.
Jul 16 at 4:46
@MarioCarneiro we've defined the product $XY$ of two subsets of a ring $R$ as $$ XY = left sum_i= 1^nx_iy_i : x_i in X, y_i in Y right $$ so that's why I made the explicit distinction. Is this a non-standard definition?
– Guido A.
Jul 16 at 4:46
@GuidoA. That's the usual definition for the product of ideals but it's not the one being used here.
– Lord Shark the Unknown
Jul 16 at 5:52
@GuidoA. That's the usual definition for the product of ideals but it's not the one being used here.
– Lord Shark the Unknown
Jul 16 at 5:52
@LordSharktheUnknown I see. That makes sense, since it coincides for the product of subsets of a monoid in the case of $(R, cdot)$. If you post it as an answer, I'll accept it, since that resolves the issue :) Thanks!
– Guido A.
Jul 16 at 5:54
@LordSharktheUnknown I see. That makes sense, since it coincides for the product of subsets of a monoid in the case of $(R, cdot)$. If you post it as an answer, I'll accept it, since that resolves the issue :) Thanks!
– Guido A.
Jul 16 at 5:54
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
The definition of the product of cosets of an ideal is
$$(a+I)cdot (b+I)=(a+I)(b+I)+I,$$
where $(a+I)(b+I)$ on the right is the set product $AB:=abmid ain A,bin B$, not the ideal product $IJ=sum_i=1^nx_iy_imid x_iin I,y_iin J$. Given this, we can show that $(a+I)cdot (b+I)=ab+I$:
If $i,jin I$, then $(a+i)(b+j)=ab+(ia+jb+ij)$, where the right term is in $I$ since it is an ideal, and conversely $ab+i=(a+0)(b+0)+iin (a+I)(b+I)+I$.
(I usually see this done in the opposite direction, where we define $(a+I)(b+I)=ab+I$, and then we must show that this definition does not depend on the member of the coset chosen for the definition, which can be done by rewriting it to $(a+I)(b+I)+I$, where the only thing we need of the set product is that it is a function of $a+I$ and $b+I$ rather than $a$ and $b$.)
If we used the ideal product in that definition, then it would contain $nab$ as you noticed, which is not what we want. In particular it would not be a coset at all if $abnotin I$, because then it would contain both $0$ and $ab$ and these do not differ by a member of $I$.
answered Jul 16 at 8:51
Mario Carneiro
18.1k33888
+1 This is exactly why I discourage people from thinking of quotient operations as being defined 'element-setwise'. It leads them straight into a pit of confusion.
– rschwieb
Jul 16 at 15:26
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The definition of the product of cosets of an ideal is
$$(a+I)cdot (b+I)=(a+I)(b+I)+I,$$
where $(a+I)(b+I)$ on the right is the set product $AB:=abmid ain A,bin B$, not the ideal product $IJ=sum_i=1^nx_iy_imid x_iin I,y_iin J$. Given this, we can show that $(a+I)cdot (b+I)=ab+I$:
If $i,jin I$, then $(a+i)(b+j)=ab+(ia+jb+ij)$, where the right term is in $I$ since it is an ideal, and conversely $ab+i=(a+0)(b+0)+iin (a+I)(b+I)+I$.
(I usually see this done in the opposite direction, where we define $(a+I)(b+I)=ab+I$, and then we must show that this definition does not depend on the member of the coset chosen for the definition, which can be done by rewriting it to $(a+I)(b+I)+I$, where the only thing we need of the set product is that it is a function of $a+I$ and $b+I$ rather than $a$ and $b$.)
If we used the ideal product in that definition, then it would contain $nab$ as you noticed, which is not what we want. In particular it would not be a coset at all if $abnotin I$, because then it would contain both $0$ and $ab$ and these do not differ by a member of $I$.
answered Jul 16 at 8:51
Mario Carneiro
18.1k33888
+1 This is exactly why I discourage people from thinking of quotient operations as being defined 'element-setwise'. It leads them straight into a pit of confusion.
– rschwieb
Jul 16 at 15:26
add a comment |Â
up vote
3
down vote
accepted
The definition of the product of cosets of an ideal is
$$(a+I)cdot (b+I)=(a+I)(b+I)+I,$$
where $(a+I)(b+I)$ on the right is the set product $AB:=abmid ain A,bin B$, not the ideal product $IJ=sum_i=1^nx_iy_imid x_iin I,y_iin J$. Given this, we can show that $(a+I)cdot (b+I)=ab+I$:
If $i,jin I$, then $(a+i)(b+j)=ab+(ia+jb+ij)$, where the right term is in $I$ since it is an ideal, and conversely $ab+i=(a+0)(b+0)+iin (a+I)(b+I)+I$.
(I usually see this done in the opposite direction, where we define $(a+I)(b+I)=ab+I$, and then we must show that this definition does not depend on the member of the coset chosen for the definition, which can be done by rewriting it to $(a+I)(b+I)+I$, where the only thing we need of the set product is that it is a function of $a+I$ and $b+I$ rather than $a$ and $b$.)
If we used the ideal product in that definition, then it would contain $nab$ as you noticed, which is not what we want. In particular it would not be a coset at all if $abnotin I$, because then it would contain both $0$ and $ab$ and these do not differ by a member of $I$.
answered Jul 16 at 8:51
Mario Carneiro
18.1k33888
+1 This is exactly why I discourage people from thinking of quotient operations as being defined 'element-setwise'. It leads them straight into a pit of confusion.
– rschwieb
Jul 16 at 15:26
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The definition of the product of cosets of an ideal is
$$(a+I)cdot (b+I)=(a+I)(b+I)+I,$$
where $(a+I)(b+I)$ on the right is the set product $AB:=abmid ain A,bin B$, not the ideal product $IJ=sum_i=1^nx_iy_imid x_iin I,y_iin J$. Given this, we can show that $(a+I)cdot (b+I)=ab+I$:
If $i,jin I$, then $(a+i)(b+j)=ab+(ia+jb+ij)$, where the right term is in $I$ since it is an ideal, and conversely $ab+i=(a+0)(b+0)+iin (a+I)(b+I)+I$.
(I usually see this done in the opposite direction, where we define $(a+I)(b+I)=ab+I$, and then we must show that this definition does not depend on the member of the coset chosen for the definition, which can be done by rewriting it to $(a+I)(b+I)+I$, where the only thing we need of the set product is that it is a function of $a+I$ and $b+I$ rather than $a$ and $b$.)
If we used the ideal product in that definition, then it would contain $nab$ as you noticed, which is not what we want. In particular it would not be a coset at all if $abnotin I$, because then it would contain both $0$ and $ab$ and these do not differ by a member of $I$.
answered Jul 16 at 8:51
Mario Carneiro
18.1k33888
The definition of the product of cosets of an ideal is
$$(a+I)cdot (b+I)=(a+I)(b+I)+I,$$
where $(a+I)(b+I)$ on the right is the set product $AB:=abmid ain A,bin B$, not the ideal product $IJ=sum_i=1^nx_iy_imid x_iin I,y_iin J$. Given this, we can show that $(a+I)cdot (b+I)=ab+I$:
If $i,jin I$, then $(a+i)(b+j)=ab+(ia+jb+ij)$, where the right term is in $I$ since it is an ideal, and conversely $ab+i=(a+0)(b+0)+iin (a+I)(b+I)+I$.
(I usually see this done in the opposite direction, where we define $(a+I)(b+I)=ab+I$, and then we must show that this definition does not depend on the member of the coset chosen for the definition, which can be done by rewriting it to $(a+I)(b+I)+I$, where the only thing we need of the set product is that it is a function of $a+I$ and $b+I$ rather than $a$ and $b$.)
If we used the ideal product in that definition, then it would contain $nab$ as you noticed, which is not what we want. In particular it would not be a coset at all if $abnotin I$, because then it would contain both $0$ and $ab$ and these do not differ by a member of $I$.
answered Jul 16 at 8:51
Mario Carneiro
18.1k33888
answered Jul 16 at 8:51
Mario Carneiro
18.1k33888
answered Jul 16 at 8:51
Mario Carneiro
18.1k33888
answered Jul 16 at 8:51
Mario Carneiro
18.1k33888
18.1k33888
+1 This is exactly why I discourage people from thinking of quotient operations as being defined 'element-setwise'. It leads them straight into a pit of confusion.
– rschwieb
Jul 16 at 15:26
add a comment |Â
+1 This is exactly why I discourage people from thinking of quotient operations as being defined 'element-setwise'. It leads them straight into a pit of confusion.
– rschwieb
Jul 16 at 15:26
+1 This is exactly why I discourage people from thinking of quotient operations as being defined 'element-setwise'. It leads them straight into a pit of confusion.
– rschwieb
Jul 16 at 15:26
+1 This is exactly why I discourage people from thinking of quotient operations as being defined 'element-setwise'. It leads them straight into a pit of confusion.
– rschwieb
Jul 16 at 15:26
add a comment |Â
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3
What is $(a+I)(b+I)$ on the first line (without the dot)? Presumably it's not a circular definition, so is that the set product (i.e. $AB=abmid ain A,bin B$)? If so, then the fourth line is incorrect since there is no sum involved, it is just $z=(a+x)(b+y)$ where $x,yin I$.
– Mario Carneiro
Jul 16 at 4:40
@MarioCarneiro we've defined the product $XY$ of two subsets of a ring $R$ as $$ XY = left sum_i= 1^nx_iy_i : x_i in X, y_i in Y right $$ so that's why I made the explicit distinction. Is this a non-standard definition?
– Guido A.
Jul 16 at 4:46
@GuidoA. That's the usual definition for the product of ideals but it's not the one being used here.
– Lord Shark the Unknown
Jul 16 at 5:52
@LordSharktheUnknown I see. That makes sense, since it coincides for the product of subsets of a monoid in the case of $(R, cdot)$. If you post it as an answer, I'll accept it, since that resolves the issue :) Thanks!
– Guido A.
Jul 16 at 5:54