Mixing
“distributive property” vs. “ring homomorphism”: comparing definitions
$begingroup$
The property of distributivity is defined using expressions like the following, from page one of "Introduction to Commutative Algebra" by Atiyah and MacDonald:
$$x(y + z) = xy + xz$$
$$(y + z)x = yx + zx.$$
On the other hand, homomorphisms are defined using expressions like the following, from the definition of ring homomorphism on page two of Atiyah-MacDonald:
$$f(x + y) = f(x) + f(y)$$
These expressions obviously bear some resemblance to one another, when viewed simply as strings of characters, but does this observation lead to any interesting examples or generalizations? Can the definitions be restated so as to have one subsuming the other?
abstract-algebra ring-theory notation ring-homomorphism
edited Jan 24 at 1:50
Antonios-Alexandros Robotis
10.5k41741
asked Jan 24 at 1:10
jessupjessup
233
$endgroup$
add a comment |
$begingroup$
The property of distributivity is defined using expressions like the following, from page one of "Introduction to Commutative Algebra" by Atiyah and MacDonald:
$$x(y + z) = xy + xz$$
$$(y + z)x = yx + zx.$$
On the other hand, homomorphisms are defined using expressions like the following, from the definition of ring homomorphism on page two of Atiyah-MacDonald:
$$f(x + y) = f(x) + f(y)$$
These expressions obviously bear some resemblance to one another, when viewed simply as strings of characters, but does this observation lead to any interesting examples or generalizations? Can the definitions be restated so as to have one subsuming the other?
abstract-algebra ring-theory notation ring-homomorphism
edited Jan 24 at 1:50
Antonios-Alexandros Robotis
10.5k41741
asked Jan 24 at 1:10
jessupjessup
233
$endgroup$
add a comment |
$begingroup$
The property of distributivity is defined using expressions like the following, from page one of "Introduction to Commutative Algebra" by Atiyah and MacDonald:
$$x(y + z) = xy + xz$$
$$(y + z)x = yx + zx.$$
On the other hand, homomorphisms are defined using expressions like the following, from the definition of ring homomorphism on page two of Atiyah-MacDonald:
$$f(x + y) = f(x) + f(y)$$
These expressions obviously bear some resemblance to one another, when viewed simply as strings of characters, but does this observation lead to any interesting examples or generalizations? Can the definitions be restated so as to have one subsuming the other?
abstract-algebra ring-theory notation ring-homomorphism
edited Jan 24 at 1:50
Antonios-Alexandros Robotis
10.5k41741
asked Jan 24 at 1:10
jessupjessup
233
$endgroup$
The property of distributivity is defined using expressions like the following, from page one of "Introduction to Commutative Algebra" by Atiyah and MacDonald:
$$x(y + z) = xy + xz$$
$$(y + z)x = yx + zx.$$
On the other hand, homomorphisms are defined using expressions like the following, from the definition of ring homomorphism on page two of Atiyah-MacDonald:
$$f(x + y) = f(x) + f(y)$$
These expressions obviously bear some resemblance to one another, when viewed simply as strings of characters, but does this observation lead to any interesting examples or generalizations? Can the definitions be restated so as to have one subsuming the other?
abstract-algebra ring-theory notation ring-homomorphism
abstract-algebra ring-theory notation ring-homomorphism
edited Jan 24 at 1:50
Antonios-Alexandros Robotis
10.5k41741
asked Jan 24 at 1:10
jessupjessup
233
edited Jan 24 at 1:50
Antonios-Alexandros Robotis
10.5k41741
asked Jan 24 at 1:10
jessupjessup
233
edited Jan 24 at 1:50
Antonios-Alexandros Robotis
10.5k41741
edited Jan 24 at 1:50
Antonios-Alexandros Robotis
10.5k41741
edited Jan 24 at 1:50
Antonios-Alexandros Robotis
10.5k41741
10.5k41741
asked Jan 24 at 1:10
jessupjessup
233
asked Jan 24 at 1:10
jessupjessup
233
asked Jan 24 at 1:10
jessupjessup
233
233
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
This is a good observation. The key here is that we want multiplication and our homomorphisms to be compatible with our addition operation. When we study groups, we have a set $G$ equipped with a single binary operation $Gtimes Gto G$ satisfying elementary properties: associativity, existence of inverses, and existence of an identity element. When we study rings $A$, we have two operations $+$ and $times$, namely addition and multiplication. Usually $(A,+,times)$ is required to be an Abelian group under $+$ and a (often commutative) monoid under $times$ (inverses might not exist under multiplication, but the other properties of a group are respected).
The natural question remains: how should the operations interact with each other? If there is no interaction between the multiplicative and additive structures of the ring, we might as well separately study $(A,+)$ and $(A,times)$ as an Abelian group and a monoid, respectively. So, we require that $times$ "preserve" the group structure of $(A,+)$. A sensible interpretation of this is the following: the map $phi: Atimes Ato A$ given by $(x,y)mapsto xtimes y$ should be "like a homomorphism of $(A,+)$ to itself." With suitable modification, it is. If we fix $x$, then the map
$$ phi_x:(A,+)to (A,+)$$
has
$$ phi_x(y+z)=phi(x,y+z)=x(y+z)=xy+xz=phi_x(y)+phi_x(z)$$
and if we fix $y$, the map
$$ phi^y:(A,+)to (A,+)$$
has
$$ phi^y(x+z)=phi(x+z,y)=(x+z)y=xy+zy=phi^y(x)+phi^y(z).$$
So we have a pair of group homomorphisms associated with right and left multiplication.
answered Jan 24 at 1:48
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
$endgroup$
add a comment |
$begingroup$
Well, you could word the left distributive law say that for each $x$, the operator $f_x$ defined by the formula $f_x(y)=xy$ is a homomorphism of the additive structure.
answered Jan 24 at 1:42
Lee MosherLee Mosher
50.6k33787
$endgroup$
add a comment |
$begingroup$
I do not think it is a mere coincidence.
Clearly the operation of addition is structural in a ring. What I mean by this is that it is a part of the ring structure by definition. So it seems natural that the functions, operations, constructions, etc., one usually considers in the Ring theory are "compatible" with this structure.
Next, I would argue that the compatibility condition like this
$$
f(x+y)=f(x)+f(y)
$$
is the most natural. Think about it this way: the image of a sum is the sum of images, so the sum in the 1-st ring should go to the sum in the 2-nd ring.
Now the two distributive laws:
$$
x(y+z)=xy+xz mbox{ and } (y+z)x=yx+zx,
$$
can be interpreted as follows. The function of the left multiplication by $x$, $L_x(r)=xr$, and the function of the right multiplication by $x$, $R_x(r)=rx$, are compatible with the addition as described above.
At the end I want to mention that the other explanation for axioms of any abstract algebraic concept (group, ring, field, homomorphism, ideal) is that there are several important examples satisfying these axioms.
answered Jan 24 at 1:55
Oleg SmirnovOleg Smirnov
658
$endgroup$
add a comment |
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3 Answers
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3 Answers
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$begingroup$
This is a good observation. The key here is that we want multiplication and our homomorphisms to be compatible with our addition operation. When we study groups, we have a set $G$ equipped with a single binary operation $Gtimes Gto G$ satisfying elementary properties: associativity, existence of inverses, and existence of an identity element. When we study rings $A$, we have two operations $+$ and $times$, namely addition and multiplication. Usually $(A,+,times)$ is required to be an Abelian group under $+$ and a (often commutative) monoid under $times$ (inverses might not exist under multiplication, but the other properties of a group are respected).
The natural question remains: how should the operations interact with each other? If there is no interaction between the multiplicative and additive structures of the ring, we might as well separately study $(A,+)$ and $(A,times)$ as an Abelian group and a monoid, respectively. So, we require that $times$ "preserve" the group structure of $(A,+)$. A sensible interpretation of this is the following: the map $phi: Atimes Ato A$ given by $(x,y)mapsto xtimes y$ should be "like a homomorphism of $(A,+)$ to itself." With suitable modification, it is. If we fix $x$, then the map
$$ phi_x:(A,+)to (A,+)$$
has
$$ phi_x(y+z)=phi(x,y+z)=x(y+z)=xy+xz=phi_x(y)+phi_x(z)$$
and if we fix $y$, the map
$$ phi^y:(A,+)to (A,+)$$
has
$$ phi^y(x+z)=phi(x+z,y)=(x+z)y=xy+zy=phi^y(x)+phi^y(z).$$
So we have a pair of group homomorphisms associated with right and left multiplication.
answered Jan 24 at 1:48
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
$endgroup$
add a comment |
$begingroup$
This is a good observation. The key here is that we want multiplication and our homomorphisms to be compatible with our addition operation. When we study groups, we have a set $G$ equipped with a single binary operation $Gtimes Gto G$ satisfying elementary properties: associativity, existence of inverses, and existence of an identity element. When we study rings $A$, we have two operations $+$ and $times$, namely addition and multiplication. Usually $(A,+,times)$ is required to be an Abelian group under $+$ and a (often commutative) monoid under $times$ (inverses might not exist under multiplication, but the other properties of a group are respected).
The natural question remains: how should the operations interact with each other? If there is no interaction between the multiplicative and additive structures of the ring, we might as well separately study $(A,+)$ and $(A,times)$ as an Abelian group and a monoid, respectively. So, we require that $times$ "preserve" the group structure of $(A,+)$. A sensible interpretation of this is the following: the map $phi: Atimes Ato A$ given by $(x,y)mapsto xtimes y$ should be "like a homomorphism of $(A,+)$ to itself." With suitable modification, it is. If we fix $x$, then the map
$$ phi_x:(A,+)to (A,+)$$
has
$$ phi_x(y+z)=phi(x,y+z)=x(y+z)=xy+xz=phi_x(y)+phi_x(z)$$
and if we fix $y$, the map
$$ phi^y:(A,+)to (A,+)$$
has
$$ phi^y(x+z)=phi(x+z,y)=(x+z)y=xy+zy=phi^y(x)+phi^y(z).$$
So we have a pair of group homomorphisms associated with right and left multiplication.
answered Jan 24 at 1:48
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
$endgroup$
add a comment |
$begingroup$
This is a good observation. The key here is that we want multiplication and our homomorphisms to be compatible with our addition operation. When we study groups, we have a set $G$ equipped with a single binary operation $Gtimes Gto G$ satisfying elementary properties: associativity, existence of inverses, and existence of an identity element. When we study rings $A$, we have two operations $+$ and $times$, namely addition and multiplication. Usually $(A,+,times)$ is required to be an Abelian group under $+$ and a (often commutative) monoid under $times$ (inverses might not exist under multiplication, but the other properties of a group are respected).
The natural question remains: how should the operations interact with each other? If there is no interaction between the multiplicative and additive structures of the ring, we might as well separately study $(A,+)$ and $(A,times)$ as an Abelian group and a monoid, respectively. So, we require that $times$ "preserve" the group structure of $(A,+)$. A sensible interpretation of this is the following: the map $phi: Atimes Ato A$ given by $(x,y)mapsto xtimes y$ should be "like a homomorphism of $(A,+)$ to itself." With suitable modification, it is. If we fix $x$, then the map
$$ phi_x:(A,+)to (A,+)$$
has
$$ phi_x(y+z)=phi(x,y+z)=x(y+z)=xy+xz=phi_x(y)+phi_x(z)$$
and if we fix $y$, the map
$$ phi^y:(A,+)to (A,+)$$
has
$$ phi^y(x+z)=phi(x+z,y)=(x+z)y=xy+zy=phi^y(x)+phi^y(z).$$
So we have a pair of group homomorphisms associated with right and left multiplication.
answered Jan 24 at 1:48
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
$endgroup$
This is a good observation. The key here is that we want multiplication and our homomorphisms to be compatible with our addition operation. When we study groups, we have a set $G$ equipped with a single binary operation $Gtimes Gto G$ satisfying elementary properties: associativity, existence of inverses, and existence of an identity element. When we study rings $A$, we have two operations $+$ and $times$, namely addition and multiplication. Usually $(A,+,times)$ is required to be an Abelian group under $+$ and a (often commutative) monoid under $times$ (inverses might not exist under multiplication, but the other properties of a group are respected).
The natural question remains: how should the operations interact with each other? If there is no interaction between the multiplicative and additive structures of the ring, we might as well separately study $(A,+)$ and $(A,times)$ as an Abelian group and a monoid, respectively. So, we require that $times$ "preserve" the group structure of $(A,+)$. A sensible interpretation of this is the following: the map $phi: Atimes Ato A$ given by $(x,y)mapsto xtimes y$ should be "like a homomorphism of $(A,+)$ to itself." With suitable modification, it is. If we fix $x$, then the map
$$ phi_x:(A,+)to (A,+)$$
has
$$ phi_x(y+z)=phi(x,y+z)=x(y+z)=xy+xz=phi_x(y)+phi_x(z)$$
and if we fix $y$, the map
$$ phi^y:(A,+)to (A,+)$$
has
$$ phi^y(x+z)=phi(x+z,y)=(x+z)y=xy+zy=phi^y(x)+phi^y(z).$$
So we have a pair of group homomorphisms associated with right and left multiplication.
answered Jan 24 at 1:48
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
answered Jan 24 at 1:48
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
answered Jan 24 at 1:48
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
answered Jan 24 at 1:48
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
10.5k41741
add a comment |
add a comment |
$begingroup$
Well, you could word the left distributive law say that for each $x$, the operator $f_x$ defined by the formula $f_x(y)=xy$ is a homomorphism of the additive structure.
answered Jan 24 at 1:42
Lee MosherLee Mosher
50.6k33787
$endgroup$
add a comment |
$begingroup$
Well, you could word the left distributive law say that for each $x$, the operator $f_x$ defined by the formula $f_x(y)=xy$ is a homomorphism of the additive structure.
answered Jan 24 at 1:42
Lee MosherLee Mosher
50.6k33787
$endgroup$
add a comment |
$begingroup$
Well, you could word the left distributive law say that for each $x$, the operator $f_x$ defined by the formula $f_x(y)=xy$ is a homomorphism of the additive structure.
answered Jan 24 at 1:42
Lee MosherLee Mosher
50.6k33787
$endgroup$
Well, you could word the left distributive law say that for each $x$, the operator $f_x$ defined by the formula $f_x(y)=xy$ is a homomorphism of the additive structure.
answered Jan 24 at 1:42
Lee MosherLee Mosher
50.6k33787
answered Jan 24 at 1:42
Lee MosherLee Mosher
50.6k33787
answered Jan 24 at 1:42
Lee MosherLee Mosher
50.6k33787
answered Jan 24 at 1:42
Lee MosherLee Mosher
50.6k33787
50.6k33787
add a comment |
add a comment |
$begingroup$
I do not think it is a mere coincidence.
Clearly the operation of addition is structural in a ring. What I mean by this is that it is a part of the ring structure by definition. So it seems natural that the functions, operations, constructions, etc., one usually considers in the Ring theory are "compatible" with this structure.
Next, I would argue that the compatibility condition like this
$$
f(x+y)=f(x)+f(y)
$$
is the most natural. Think about it this way: the image of a sum is the sum of images, so the sum in the 1-st ring should go to the sum in the 2-nd ring.
Now the two distributive laws:
$$
x(y+z)=xy+xz mbox{ and } (y+z)x=yx+zx,
$$
can be interpreted as follows. The function of the left multiplication by $x$, $L_x(r)=xr$, and the function of the right multiplication by $x$, $R_x(r)=rx$, are compatible with the addition as described above.
At the end I want to mention that the other explanation for axioms of any abstract algebraic concept (group, ring, field, homomorphism, ideal) is that there are several important examples satisfying these axioms.
answered Jan 24 at 1:55
Oleg SmirnovOleg Smirnov
658
$endgroup$
add a comment |
$begingroup$
I do not think it is a mere coincidence.
Clearly the operation of addition is structural in a ring. What I mean by this is that it is a part of the ring structure by definition. So it seems natural that the functions, operations, constructions, etc., one usually considers in the Ring theory are "compatible" with this structure.
Next, I would argue that the compatibility condition like this
$$
f(x+y)=f(x)+f(y)
$$
is the most natural. Think about it this way: the image of a sum is the sum of images, so the sum in the 1-st ring should go to the sum in the 2-nd ring.
Now the two distributive laws:
$$
x(y+z)=xy+xz mbox{ and } (y+z)x=yx+zx,
$$
can be interpreted as follows. The function of the left multiplication by $x$, $L_x(r)=xr$, and the function of the right multiplication by $x$, $R_x(r)=rx$, are compatible with the addition as described above.
At the end I want to mention that the other explanation for axioms of any abstract algebraic concept (group, ring, field, homomorphism, ideal) is that there are several important examples satisfying these axioms.
answered Jan 24 at 1:55
Oleg SmirnovOleg Smirnov
658
$endgroup$
add a comment |
$begingroup$
I do not think it is a mere coincidence.
Clearly the operation of addition is structural in a ring. What I mean by this is that it is a part of the ring structure by definition. So it seems natural that the functions, operations, constructions, etc., one usually considers in the Ring theory are "compatible" with this structure.
Next, I would argue that the compatibility condition like this
$$
f(x+y)=f(x)+f(y)
$$
is the most natural. Think about it this way: the image of a sum is the sum of images, so the sum in the 1-st ring should go to the sum in the 2-nd ring.
Now the two distributive laws:
$$
x(y+z)=xy+xz mbox{ and } (y+z)x=yx+zx,
$$
can be interpreted as follows. The function of the left multiplication by $x$, $L_x(r)=xr$, and the function of the right multiplication by $x$, $R_x(r)=rx$, are compatible with the addition as described above.
At the end I want to mention that the other explanation for axioms of any abstract algebraic concept (group, ring, field, homomorphism, ideal) is that there are several important examples satisfying these axioms.
answered Jan 24 at 1:55
Oleg SmirnovOleg Smirnov
658
$endgroup$
I do not think it is a mere coincidence.
Clearly the operation of addition is structural in a ring. What I mean by this is that it is a part of the ring structure by definition. So it seems natural that the functions, operations, constructions, etc., one usually considers in the Ring theory are "compatible" with this structure.
Next, I would argue that the compatibility condition like this
$$
f(x+y)=f(x)+f(y)
$$
is the most natural. Think about it this way: the image of a sum is the sum of images, so the sum in the 1-st ring should go to the sum in the 2-nd ring.
Now the two distributive laws:
$$
x(y+z)=xy+xz mbox{ and } (y+z)x=yx+zx,
$$
can be interpreted as follows. The function of the left multiplication by $x$, $L_x(r)=xr$, and the function of the right multiplication by $x$, $R_x(r)=rx$, are compatible with the addition as described above.
At the end I want to mention that the other explanation for axioms of any abstract algebraic concept (group, ring, field, homomorphism, ideal) is that there are several important examples satisfying these axioms.
answered Jan 24 at 1:55
Oleg SmirnovOleg Smirnov
658
answered Jan 24 at 1:55
Oleg SmirnovOleg Smirnov
658
answered Jan 24 at 1:55
Oleg SmirnovOleg Smirnov
658
answered Jan 24 at 1:55
Oleg SmirnovOleg Smirnov
658
658
add a comment |
add a comment |
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