MHB Algebras - k-algebra and R-algebra - reconciling two definitions ....

[Math Amateur]

I am trying to get a full understanding of the notion of an algebra ... I have thus consulted two books - Cohn: "Introduction to Ring Theory and Dummit and Foote: Abstract Algebra.

Cohn defines a k-algebra (or linear algebra) as follows:View attachment 4765
View attachment 4766If we want to amend the above definition to an R-Algebra where R is a commutative ring with identity then presumably we just change the field k with the ring R ...

But then with the definition above amended to be an R-algebra how do we reconcile the above definition with Dummit and Foote's definition of an R-Algebra which reads as follows:View attachment 4767Can someone please clarify how exactly these two definitions are saying the same thing ...

Also ... just as an aside that someone may help with ... in Cohn's definition of a k-algebra he refers to k as an operator domain ... can someone please clarify why he has called the field an operator domain ... what is the meaning of this reference ...

Peter[I should acknowledge that Mathbalarka has given me some help with the notion of an algebra in a previous post ... my thanks to him ... and my apologies to him if he has already answered the above question ... ]***EDIT***

I am somewhat perplexed by the statement

$$r \cdot a = a \cdot r = f(r)a
$$

in the note after D&F's definition ... seems like it is saying that $$r = f(r)$$ ... but that cannot be right, surely ... can someone please clarify this apparent confusion for me ...*** EDIT 2 ***

Seems after reflection that$$ r \cdot a$$ is the action of $$R$$ on the module $$A$$ ... while the $$f(r)a$$ is the element of the module $$A$$ that results ... can someone confirm that that is correct ...Peter

 

[Deveno]

Suppose we have a ring $A$ together with a ring-homomorphism that preserves unity $f: R \to A$.

Let us define an action of $R$ upon $A$ by:

$r \cdot a = f(r)a$.

Then:

$(r + s)\cdot a = f(r+s)a = [f(r) + f(s)]a$ (since $f$ is a ring-homomorphism)

$= f(r)a + f(s)a$ (since $A$ is a ring and we have the distributive laws)

$= r\cdot a + s \cdot a$ (by definition). This is LA.2.

Similarly:

$r\cdot (a + b) = f(r)(a + b) = f(r)a + f(r)b = r \cdot a + r \cdot b$ (again, using the distributive laws). This is LA.1.

Now $r \cdot (s \cdot a) = r \cdot (f(s)a) = f(r)(f(s)a) = (f(r)f(s))a$ (the associative law for ring multiplication)

$= f(rs)a = (rs)\cdot a$ (since $f$ is a ring-homomorphism). This is LA.3.

Moving on: $1_R\cdot a = f(1_R)a = 1_Aa = a$ (since $f$ preserves unity). This is LA.4.

Finally:

$r\cdot (ab) = f(r)(ab) = (f(r)a)b$ (associativity of multiplication in $A$)

$= (r\cdot a)b$. This is 'half" of LA.5.

Now is where we need that $f(R) \subseteq Z(A)$:

If $f(r) \in Z(A)$, then $(f(r)a)b = (af(r))b = a(f(r)b) = a(r \cdot b)$, so that:

$r\cdot (ab) = a(r\cdot b)$ as well-this is the "other half" of LA.5.

On the other hand, suppose LA.1-LA.5 hold for a ring with unity $R$, and a ring with unity $A$ and a map:

$R\times A \to A$ which we write as $(r,a) \mapsto r\cdot a$.

Let $f: R \to A$ be given by $f(r) = r\cdot 1_A$.

Then $f(r + s) = (r+s)\cdot 1_A = r\cdot 1_A + s\cdot 1_A$ (by LA.2)

$= f(r) + f(s)$, so $f$ is an abelian group homomorphism.

Also, $f(rs) = (rs)\cdot 1_A = r\cdot (s \cdot 1_A)$ (by LA.3)

$= r\cdot(s\cdot(1_A1_A))$ (since $1_A$ is a multiplicative identity)

$= r \cdot (1_A(s\cdot 1_A))$ (by LA.5-the "2nd half")

$= (r\cdot 1_A)(s \cdot 1_A)$ (by LA.5, again, the "first half")

$= f(r)f(s)$, so $f$ is a ring-homomorphism $R \to A$.

Now $f(1_R) = 1_R \cdot 1_A = 1_A$ (by LA.4), so $f$ preserves unity.

Finally, we have:

$f(r)a = (r\cdot 1_A)a = r\cdot(1_Aa)$ (by LA.5)

$= r\cdot(a1_A) = a(r\cdot 1_A)$ (by LA.5 again)

$= af(r)$, that is $f(r) \in Z(A)$.

So the two definitions are just different ways around the block.

EDIT: you may notice that LA.1 is conspicuously absent from showing the "LA" axioms imply the "homomorphism" axioms. It's there, but it's "hidden" in the fact that $R$ and $A$ are both RINGS.

$r\cdot (a+b) = (r\cdot a) + (r \cdot b)$ means that the map $r \mapsto r\cdot (-)$ is a map from $(R,+) \to \text{Hom}(A,A)$ (here the hom-set is considering $A$ as merely an abelian group).

This is part of what makes $A$ an $R$-module. You can think of an algebra as a happy coincidence of two distinct structures: an $R$-bimodule and an extension ring of a quotient of $R$, where the quotient is central. In many cases, the quotient is an injection of $R$ into $A$, that is $A$ is an extension ring of $R$.

Now, some rings have "wonky" multiplicative structures, and so the $R$-bimodule structure may not be as "nice" as we would like. Fields, however, are very "nice" rings, and $k$-modules (which are "naturally" bimodules) are simply vector spaces, which posses some nice invariants, like dimension.

It helps to have some examples to guide one's intuitions. Here are a couple of $k$-algebras for you to think about:

1) Let $V$ be a finite-dimensional vector space over a field $k$. Let $A = \text{Hom}_k(V,V)$ denote the space of ALL $k$-linear transformations $T: V \to V$.

We can define a ring-structure on $A$, by setting, for $S,T \in A$:

$(S + T)(v) = S(v) + T(v)$ ("point-wise" addition)
$(ST)(v) = S(T(v))$ (composition).

This ring is unital, with identity $1_A$ given by:

$1_A(v) = v$, for all $v \in V$.

We have a homomorphism $k \to A$ given by $a \mapsto a1_A$ (this map sends $v \mapsto av$). These linear maps are called "scalar maps", and form the ENTIRE CENTER of $A$.

2) Consider $A= k[X]$, polynomials in $X$ with coefficients in $k$. The elements $\{X^n: n\in \Bbb N\}$ form a $k$-basis, and we define the multiplication as the usual multiplication of polynomials. The map $k \to Z(A)$ is, in this case, the mapping of the number $a$ to the constant polynomial $f(x) = a$. This algebra is commutative (unlike the example above), so we need not check if the image of $k$ lies in the center-$A = Z(A)$.

3) Let $k$ be any field, and let $A$ be any extension ring (which may ALSO be a field, but need not be) with $k \subseteq Z(A)$. The last condition is automatic if $A$ is commutative. For example, $\Bbb C$ is an $\Bbb R$-algebra.

 

[Math Amateur]

Suppose we have a ring $A$ together with a ring-homomorphism that preserves unity $f: R \to A$.

Let us define an action of $R$ upon $A$ by:

$r \cdot a = f(r)a$.

Then:

$(r + s)\cdot a = f(r+s)a = [f(r) + f(s)]a$ (since $f$ is a ring-homomorphism)

$= f(r)a + f(s)a$ (since $A$ is a ring and we have the distributive laws)

$= r\cdot a + s \cdot a$ (by definition). This is LA.2.

Similarly:

$r\cdot (a + b) = f(r)(a + b) = f(r)a + f(r)b = r \cdot a + r \cdot b$ (again, using the distributive laws). This is LA.1.

Now $r \cdot (s \cdot a) = r \cdot (f(s)a) = f(r)(f(s)a) = (f(r)f(s))a$ (the associative law for ring multiplication)

$= f(rs)a = (rs)\cdot a$ (since $f$ is a ring-homomorphism). This is LA.3.

Moving on: $1_R\cdot a = f(1_R)a = 1_Aa = a$ (since $f$ preserves unity). This is LA.4.

Finally:

$r\cdot (ab) = f(r)(ab) = (f(r)a)b$ (associativity of multiplication in $A$)

$= (r\cdot a)b$. This is 'half" of LA.5.

Now is where we need that $f(R) \subseteq Z(A)$:

If $f(r) \in Z(A)$, then $(f(r)a)b = (af(r))b = a(f(r)b) = a(r \cdot b)$, so that:

$r\cdot (ab) = a(r\cdot b)$ as well-this is the "other half" of LA.5.

On the other hand, suppose LA.1-LA.5 hold for a ring with unity $R$, and a ring with unity $A$ and a map:

$R\times A \to A$ which we write as $(r,a) \mapsto r\cdot a$.

Let $f: R \to A$ be given by $f(r) = r\cdot 1_A$.

Then $f(r + s) = (r+s)\cdot 1_A = r\cdot 1_A + s\cdot 1_A$ (by LA.2)

$= f(r) + f(s)$, so $f$ is an abelian group homomorphism.

Also, $f(rs) = (rs)\cdot 1_A = r\cdot (s \cdot 1_A)$ (by LA.3)

$= r\cdot(s\cdot(1_A1_A))$ (since $1_A$ is a multiplicative identity)

$= r \cdot (1_A(s\cdot 1_A))$ (by LA.5-the "2nd half")

$= (r\cdot 1_A)(s \cdot 1_A)$ (by LA.5, again, the "first half")

$= f(r)f(s)$, so $f$ is a ring-homomorphism $R \to A$.

Now $f(1_R) = 1_R \cdot 1_A = 1_A$ (by LA.4), so $f$ preserves unity.

Finally, we have:

$f(r)a = (r\cdot 1_A)a = r\cdot(1_Aa)$ (by LA.5)

$= r\cdot(a1_A) = a(r\cdot 1_A)$ (by LA.5 again)

$= af(r)$, that is $f(r) \in Z(A)$.

So the two definitions are just different ways around the block.

Thanks so much for the help, Deveno ...

Just working through your post carefully now ...

Hopefully I will have a much better understanding of algebras when I am through ...

Thanks again for your help and support ... it's much appreciated ...

Peter

 

[Math Amateur]

Suppose we have a ring $A$ together with a ring-homomorphism that preserves unity $f: R \to A$.

Let us define an action of $R$ upon $A$ by:

$r \cdot a = f(r)a$.

Then:

$(r + s)\cdot a = f(r+s)a = [f(r) + f(s)]a$ (since $f$ is a ring-homomorphism)

$= f(r)a + f(s)a$ (since $A$ is a ring and we have the distributive laws)

$= r\cdot a + s \cdot a$ (by definition). This is LA.2.

Similarly:

$r\cdot (a + b) = f(r)(a + b) = f(r)a + f(r)b = r \cdot a + r \cdot b$ (again, using the distributive laws). This is LA.1.

Now $r \cdot (s \cdot a) = r \cdot (f(s)a) = f(r)(f(s)a) = (f(r)f(s))a$ (the associative law for ring multiplication)

$= f(rs)a = (rs)\cdot a$ (since $f$ is a ring-homomorphism). This is LA.3.

Moving on: $1_R\cdot a = f(1_R)a = 1_Aa = a$ (since $f$ preserves unity). This is LA.4.

Finally:

$r\cdot (ab) = f(r)(ab) = (f(r)a)b$ (associativity of multiplication in $A$)

$= (r\cdot a)b$. This is 'half" of LA.5.

Now is where we need that $f(R) \subseteq Z(A)$:

If $f(r) \in Z(A)$, then $(f(r)a)b = (af(r))b = a(f(r)b) = a(r \cdot b)$, so that:

$r\cdot (ab) = a(r\cdot b)$ as well-this is the "other half" of LA.5.

On the other hand, suppose LA.1-LA.5 hold for a ring with unity $R$, and a ring with unity $A$ and a map:

$R\times A \to A$ which we write as $(r,a) \mapsto r\cdot a$.

Let $f: R \to A$ be given by $f(r) = r\cdot 1_A$.

Then $f(r + s) = (r+s)\cdot 1_A = r\cdot 1_A + s\cdot 1_A$ (by LA.2)

$= f(r) + f(s)$, so $f$ is an abelian group homomorphism.

Also, $f(rs) = (rs)\cdot 1_A = r\cdot (s \cdot 1_A)$ (by LA.3)

$= r\cdot(s\cdot(1_A1_A))$ (since $1_A$ is a multiplicative identity)

$= r \cdot (1_A(s\cdot 1_A))$ (by LA.5-the "2nd half")

$= (r\cdot 1_A)(s \cdot 1_A)$ (by LA.5, again, the "first half")

$= f(r)f(s)$, so $f$ is a ring-homomorphism $R \to A$.

Now $f(1_R) = 1_R \cdot 1_A = 1_A$ (by LA.4), so $f$ preserves unity.

Finally, we have:

$f(r)a = (r\cdot 1_A)a = r\cdot(1_Aa)$ (by LA.5)

$= r\cdot(a1_A) = a(r\cdot 1_A)$ (by LA.5 again)

$= af(r)$, that is $f(r) \in Z(A)$.

So the two definitions are just different ways around the block.

Well ... ... that was an extremely clear exposition ... thanks ...

Leaves me thinking ... now why didn't I see that ... ... :)

Thanks again, Peter

PS just working carefully through your post again ...

 

[Math Amateur]

Suppose we have a ring $A$ together with a ring-homomorphism that preserves unity $f: R \to A$.

Let us define an action of $R$ upon $A$ by:

$r \cdot a = f(r)a$.

Then:

$(r + s)\cdot a = f(r+s)a = [f(r) + f(s)]a$ (since $f$ is a ring-homomorphism)

$= f(r)a + f(s)a$ (since $A$ is a ring and we have the distributive laws)

$= r\cdot a + s \cdot a$ (by definition). This is LA.2.

Similarly:

$r\cdot (a + b) = f(r)(a + b) = f(r)a + f(r)b = r \cdot a + r \cdot b$ (again, using the distributive laws). This is LA.1.

Now $r \cdot (s \cdot a) = r \cdot (f(s)a) = f(r)(f(s)a) = (f(r)f(s))a$ (the associative law for ring multiplication)

$= f(rs)a = (rs)\cdot a$ (since $f$ is a ring-homomorphism). This is LA.3.

Moving on: $1_R\cdot a = f(1_R)a = 1_Aa = a$ (since $f$ preserves unity). This is LA.4.

Finally:

$r\cdot (ab) = f(r)(ab) = (f(r)a)b$ (associativity of multiplication in $A$)

$= (r\cdot a)b$. This is 'half" of LA.5.

Now is where we need that $f(R) \subseteq Z(A)$:

If $f(r) \in Z(A)$, then $(f(r)a)b = (af(r))b = a(f(r)b) = a(r \cdot b)$, so that:

$r\cdot (ab) = a(r\cdot b)$ as well-this is the "other half" of LA.5.

On the other hand, suppose LA.1-LA.5 hold for a ring with unity $R$, and a ring with unity $A$ and a map:

$R\times A \to A$ which we write as $(r,a) \mapsto r\cdot a$.

Let $f: R \to A$ be given by $f(r) = r\cdot 1_A$.

Then $f(r + s) = (r+s)\cdot 1_A = r\cdot 1_A + s\cdot 1_A$ (by LA.2)

$= f(r) + f(s)$, so $f$ is an abelian group homomorphism.

Also, $f(rs) = (rs)\cdot 1_A = r\cdot (s \cdot 1_A)$ (by LA.3)

$= r\cdot(s\cdot(1_A1_A))$ (since $1_A$ is a multiplicative identity)

$= r \cdot (1_A(s\cdot 1_A))$ (by LA.5-the "2nd half")

$= (r\cdot 1_A)(s \cdot 1_A)$ (by LA.5, again, the "first half")

$= f(r)f(s)$, so $f$ is a ring-homomorphism $R \to A$.

Now $f(1_R) = 1_R \cdot 1_A = 1_A$ (by LA.4), so $f$ preserves unity.

Finally, we have:

$f(r)a = (r\cdot 1_A)a = r\cdot(1_Aa)$ (by LA.5)

$= r\cdot(a1_A) = a(r\cdot 1_A)$ (by LA.5 again)

$= af(r)$, that is $f(r) \in Z(A)$.

So the two definitions are just different ways around the block.

EDIT: you may notice that LA.1 is conspicuously absent from showing the "LA" axioms imply the "homomorphism" axioms. It's there, but it's "hidden" in the fact that $R$ and $A$ are both RINGS.

$r\cdot (a+b) = (r\cdot a) + (r \cdot b)$ means that the map $r \mapsto r\cdot (-)$ is a map from $(R,+) \to \text{Hom}(A,A)$ (here the hom-set is considering $A$ as merely an abelian group).

This is part of what makes $A$ an $R$-module. You can think of an algebra as a happy coincidence of two distinct structures: an $R$-bimodule and an extension ring of a quotient of $R$, where the quotient is central. In many cases, the quotient is an injection of $R$ into $A$, that is $A$ is an extension ring of $R$.

Now, some rings have "wonky" multiplicative structures, and so the $R$-bimodule structure may not be as "nice" as we would like. Fields, however, are very "nice" rings, and $k$-modules (which are "naturally" bimodules) are simply vector spaces, which posses some nice invariants, like dimension.

It helps to have some examples to guide one's intuitions. Here are a couple of $k$-algebras for you to think about:

1) Let $V$ be a finite-dimensional vector space over a field $k$. Let $A = \text{Hom}_k(V,V)$ denote the space of ALL $k$-linear transformations $T: V \to V$.

We can define a ring-structure on $A$, by setting, for $S,T \in A$:

$(S + T)(v) = S(v) + T(v)$ ("point-wise" addition)
$(ST)(v) = S(T(v))$ (composition).

This ring is unital, with identity $1_A$ given by:

$1_A(v) = v$, for all $v \in V$.

We have a homomorphism $k \to A$ given by $a \mapsto a1_A$ (this map sends $v \mapsto av$). These linear maps are called "scalar maps", and form the ENTIRE CENTER of $A$.

2) Consider $A= k[X]$, polynomials in $X$ with coefficients in $k$. The elements $\{X^n: n\in \Bbb N\}$ form a $k$-basis, and we define the multiplication as the usual multiplication of polynomials. The map $k \to Z(A)$ is, in this case, the mapping of the number $a$ to the constant polynomial $f(x) = a$. This algebra is commutative (unlike the example above), so we need not check if the image of $k$ lies in the center-$A = Z(A)$.

3) Let $k$ be any field, and let $A$ be any extension ring (which may ALSO be a field, but need not be) with $k \subseteq Z(A)$. The last condition is automatic if $A$ is commutative. For example, $\Bbb C$ is an $\Bbb R$-algebra.

Hi Deveno,

Just been reflecting on your post ...

You begin by assuming the D&F definition of an R-algebra is true ... with the aim of showing the Cohn definition to follow ... so you write:

" ... ... Suppose we have a ring $A$ together with a ring-homomorphism that preserves unity $f: R \to A$. ... ..."

BUT ... then you write:

" ... ... Let us define an action of $R$ upon $A$ by:

$r \cdot a = f(r)a$. ... ... "


and then derive the conditions of the Cohn definition ... ... So ... then ... ? ... it seems that the Cohn definition only follows from the D&F definition if we assume the action as defined ...My question is, then, shouldn't the action be added to the D&F definition ... that is shouldn't the action you have defined be part of the D&F definition of an R-algebra ...



Hope you can clarify this issue/question ...

Peter

 

[Math Amateur]

Suppose we have a ring $A$ together with a ring-homomorphism that preserves unity $f: R \to A$.

Let us define an action of $R$ upon $A$ by:

$r \cdot a = f(r)a$.

Then:

$(r + s)\cdot a = f(r+s)a = [f(r) + f(s)]a$ (since $f$ is a ring-homomorphism)

$= f(r)a + f(s)a$ (since $A$ is a ring and we have the distributive laws)

$= r\cdot a + s \cdot a$ (by definition). This is LA.2.

Similarly:

$r\cdot (a + b) = f(r)(a + b) = f(r)a + f(r)b = r \cdot a + r \cdot b$ (again, using the distributive laws). This is LA.1.

Now $r \cdot (s \cdot a) = r \cdot (f(s)a) = f(r)(f(s)a) = (f(r)f(s))a$ (the associative law for ring multiplication)

$= f(rs)a = (rs)\cdot a$ (since $f$ is a ring-homomorphism). This is LA.3.

Moving on: $1_R\cdot a = f(1_R)a = 1_Aa = a$ (since $f$ preserves unity). This is LA.4.

Finally:

$r\cdot (ab) = f(r)(ab) = (f(r)a)b$ (associativity of multiplication in $A$)

$= (r\cdot a)b$. This is 'half" of LA.5.

Now is where we need that $f(R) \subseteq Z(A)$:

If $f(r) \in Z(A)$, then $(f(r)a)b = (af(r))b = a(f(r)b) = a(r \cdot b)$, so that:

$r\cdot (ab) = a(r\cdot b)$ as well-this is the "other half" of LA.5.

On the other hand, suppose LA.1-LA.5 hold for a ring with unity $R$, and a ring with unity $A$ and a map:

$R\times A \to A$ which we write as $(r,a) \mapsto r\cdot a$.

Let $f: R \to A$ be given by $f(r) = r\cdot 1_A$.

Then $f(r + s) = (r+s)\cdot 1_A = r\cdot 1_A + s\cdot 1_A$ (by LA.2)

$= f(r) + f(s)$, so $f$ is an abelian group homomorphism.

Also, $f(rs) = (rs)\cdot 1_A = r\cdot (s \cdot 1_A)$ (by LA.3)

$= r\cdot(s\cdot(1_A1_A))$ (since $1_A$ is a multiplicative identity)

$= r \cdot (1_A(s\cdot 1_A))$ (by LA.5-the "2nd half")

$= (r\cdot 1_A)(s \cdot 1_A)$ (by LA.5, again, the "first half")

$= f(r)f(s)$, so $f$ is a ring-homomorphism $R \to A$.

Now $f(1_R) = 1_R \cdot 1_A = 1_A$ (by LA.4), so $f$ preserves unity.

Finally, we have:

$f(r)a = (r\cdot 1_A)a = r\cdot(1_Aa)$ (by LA.5)

$= r\cdot(a1_A) = a(r\cdot 1_A)$ (by LA.5 again)

$= af(r)$, that is $f(r) \in Z(A)$.

So the two definitions are just different ways around the block.

EDIT: you may notice that LA.1 is conspicuously absent from showing the "LA" axioms imply the "homomorphism" axioms. It's there, but it's "hidden" in the fact that $R$ and $A$ are both RINGS.

$r\cdot (a+b) = (r\cdot a) + (r \cdot b)$ means that the map $r \mapsto r\cdot (-)$ is a map from $(R,+) \to \text{Hom}(A,A)$ (here the hom-set is considering $A$ as merely an abelian group).

This is part of what makes $A$ an $R$-module. You can think of an algebra as a happy coincidence of two distinct structures: an $R$-bimodule and an extension ring of a quotient of $R$, where the quotient is central. In many cases, the quotient is an injection of $R$ into $A$, that is $A$ is an extension ring of $R$.

Now, some rings have "wonky" multiplicative structures, and so the $R$-bimodule structure may not be as "nice" as we would like. Fields, however, are very "nice" rings, and $k$-modules (which are "naturally" bimodules) are simply vector spaces, which posses some nice invariants, like dimension.

It helps to have some examples to guide one's intuitions. Here are a couple of $k$-algebras for you to think about:

1) Let $V$ be a finite-dimensional vector space over a field $k$. Let $A = \text{Hom}_k(V,V)$ denote the space of ALL $k$-linear transformations $T: V \to V$.

We can define a ring-structure on $A$, by setting, for $S,T \in A$:

$(S + T)(v) = S(v) + T(v)$ ("point-wise" addition)
$(ST)(v) = S(T(v))$ (composition).

This ring is unital, with identity $1_A$ given by:

$1_A(v) = v$, for all $v \in V$.

We have a homomorphism $k \to A$ given by $a \mapsto a1_A$ (this map sends $v \mapsto av$). These linear maps are called "scalar maps", and form the ENTIRE CENTER of $A$.

2) Consider $A= k[X]$, polynomials in $X$ with coefficients in $k$. The elements $\{X^n: n\in \Bbb N\}$ form a $k$-basis, and we define the multiplication as the usual multiplication of polynomials. The map $k \to Z(A)$ is, in this case, the mapping of the number $a$ to the constant polynomial $f(x) = a$. This algebra is commutative (unlike the example above), so we need not check if the image of $k$ lies in the center-$A = Z(A)$.

3) Let $k$ be any field, and let $A$ be any extension ring (which may ALSO be a field, but need not be) with $k \subseteq Z(A)$. The last condition is automatic if $A$ is commutative. For example, $\Bbb C$ is an $\Bbb R$-algebra.

Hi Deveno,

Further reflecting on your post above ... I need some further help ...

I am having real trouble following exactly what you mean when you write:

" ... ... EDIT: you may notice that LA.1 is conspicuously absent from showing the "LA" axioms imply the "homomorphism" axioms. It's there, but it's "hidden" in the fact that $R$ and $A$ are both RINGS.

$r\cdot (a+b) = (r\cdot a) + (r \cdot b)$ means that the map $r \mapsto r\cdot (-)$ is a map from $(R,+) \to \text{Hom}(A,A)$ (here the hom-set is considering $A$ as merely an abelian group).

This is part of what makes $A$ an $R$-module. ... ... ..."

What do you mean by "the map $r \mapsto r\cdot (-)$ is a map from $(R,+) \to \text{Hom}(A,A)$"?

Can you expand or explain further ...

Not quite sure where the map $r \mapsto r\cdot (-)$ came from, nor its meaning and relevance to LA.1...

Hope you can find a way to expand what you have said ...

Peter

 

[Deveno]

Hi Deveno,

Just been reflecting on your post ...

You begin by assuming the D&F definition of an R-algebra is true ... with the aim of showing the Cohn definition to follow ... so you write:

" ... ... Suppose we have a ring $A$ together with a ring-homomorphism that preserves unity $f: R \to A$. ... ..."

BUT ... then you write:

" ... ... Let us define an action of $R$ upon $A$ by:

$r \cdot a = f(r)a$. ... ... "


and then derive the conditions of the Cohn definition ... ... So ... then ... ? ... it seems that the Cohn definition only follows from the D&F definition if we assume the action as defined ...My question is, then, shouldn't the action be added to the D&F definition ... that is shouldn't the action you have defined be part of the D&F definition of an R-algebra ...



Hope you can clarify this issue/question ...

Peter

Dummit & Foote write:"If $A$ is an $R$-algebra it is easy to check that a natural left and right (unital) $R$-module structure defined by $r\cdot a = a \cdot r = f(r)a$, where $f(r)a$ is just the multiplication in the ring $A$..."

They later go on to write that other actions ARE possible, and so they ARE including this specific action in the definition:

"...but unless otherwise stated, this natural action on an algebra will be assumed."

Hi Deveno,

Further reflecting on your post above ... I need some further help ...

I am having real trouble following exactly what you mean when you write:

" ... ... EDIT: you may notice that LA.1 is conspicuously absent from showing the "LA" axioms imply the "homomorphism" axioms. It's there, but it's "hidden" in the fact that $R$ and $A$ are both RINGS.

$r\cdot (a+b) = (r\cdot a) + (r \cdot b)$ means that the map $r \mapsto r\cdot (-)$ is a map from $(R,+) \to \text{Hom}(A,A)$ (here the hom-set is considering $A$ as merely an abelian group).

This is part of what makes $A$ an $R$-module. ... ... ..."

What do you mean by "the map $r \mapsto r\cdot (-)$ is a map from $(R,+) \to \text{Hom}(A,A)$"?

Can you expand or explain further ...

Not quite sure where the map $r \mapsto r\cdot (-)$ came from, nor its meaning and relevance to LA.1...

Hope you can find a way to expand what you have said ...

Peter

To explain this, I am going to have to take a BIG detour.

Let's say you have a set, say $X$. there is a "natural" way to form a monoid associated with $X$ called the monoid of transformations of X, let's call it $T(X)$. What we do is form the set of all FUNCTIONS (set-morphisms) $X \to X$.

The operation of functional composition is associative, and possesses an identity, the identity function $1_X$ given by:

$1_X(x) = x$ for all $x \in X$.

Thus we have a monoid structure on $T(X)$.

Now given a monoid, say $M$, we have, for every $a \in M$, a function $L_a: M \to M$ given by:

$L_a(m) = am$ (left-multiplication by $a$).

Thus we have a mapping $M \to T(M)$ given by $a \mapsto L_a$.

Furthermore, this mapping is actually a monoid-homomorphism:

$L_{ab} = L_a \circ L_b$
$L_e = 1_M$

A monoid-homomorphism $M \to T(X)$ for some set $X$ is called an ACTION of $M$ on $X$, and the action is also called a representation of $M$ in $X$. So we have a fundamental result:

Every monoid $M$ acts on itself as a representation in $T(M)$ (technically, this is a left action-right actions also exist), via left-multiplication.

We can associate with groups a similar construction by limiting $T(X)$ to the sub-monoid of units (invertible functions). This sub-monoid has a special name: the permutations of $X$, or $\text{Sym}(X)$.

It turns out that every group can also be represented as a subgroup of $\text{Sym}(G)$, this is known as Cayley's Theorem, and the associated action (via left-multiplication) of $G$ upon itself is the left regular representation of $G$.

The homomorphism $G \to \text{Sym}(G)$ (the $G$ on the right is the underlying SET of $G$) can be abstracted to a homomorphism:

$G \to \text{Sym}(X)$, which yields the concept of a group $G$ acting on a set $X$.

This is often viewed as a mapping $G \times X \to X$, where $(g,x) \mapsto g\cdot x$. In other words, a way to "multiply" set-elements by group-elements. This sort of "special" group/set structure is called a $G$-set, and can be viewed as a certain kind of "compatibility" of the set $X$ with the group $G$.

Note how this hybrid structure mixes a "more complicated structure" with a "simpler" one.

Now with an ABELIAN group $A$, we can make a RING related to $A$, ALSO in a "natural" way.

We start with $T(A)$, the monoid of transformations of $A$, and restrict our attention to the sub-monoid of these transformations of $A$ that are also group-homomorphisms (note that homomorphisms are "composable", thus closed under our monoid multiplication of composition, and the identity transformation is also fortuitously a group homomorphism). Let's call this sub-monoid $E(A)$.

We can "create" an addition on $E(A)$ by defining:

$(\phi + \psi)(a) = \phi(a) + \psi(a)$ (where the + on the RHS is the abelian group operation).

Let's see that $\phi + \psi \in E(A)$:

$(\phi + \psi)(a + b) = \phi(a + b) + \psi(a + b)$ (by our definition)
$= \phi(a) + \phi(b) + \psi(a) + \psi(b)$ (because $\phi,\psi$ are both group-homomorphisms)
$= \phi(a) + \psi(a) + \phi(b) + \psi(b)$ (because addition in $A$ is COMMUTATIVE)
$= (\phi + \psi)(a) + (\phi + \psi)(b)$ (again, by definition).

So, we at least have a commutative semi-group structure $(E(A),+)$.

The 0-map $z(a) = 0$ for all $a \in A$ is an abelian group homomorphism (albeit a trivial one), and clearly functions as an identity:

$\phi + z = z + \phi = \phi$, for all $\phi \in E(A)$.

If we define $-\phi$ by: $(-\phi)(a) = -(\phi(a))$, for all $a \in A$, since $A$ is abelian, $-\phi \in E(A)$, and it is straight-forward to see that:

$\phi + -\phi = z$, for any $\phi \in E(A)$.

So now we have $(E(A),+)$ is an abelian group, and $(E(A),\circ)$ is a monoid. So all that's missing to show $E(A)$ is a ring, are the distributive laws. Here's the left one:

$\phi\circ (\psi + \chi) = (\phi \circ \psi) + (\phi \circ \chi)$

Proof:

Let $a \in A$. Then:

$[\phi \circ (\psi + \chi)](a) = \psi((\psi + \chi)(a)) = \psi(\psi(a) + \chi(a))$

$= \psi(\psi(a)) + \psi(\chi(a)) = (\psi\circ\psi)(a) + (\psi\circ\chi)(a)$

$= [(\phi \circ \psi) + (\phi \circ \chi)](a)$.

So now, we have created a ring $E(A)$ out of an abelian group $A$, using certain "special maps" $A \to A$ (I hope you are seeing the pattern, here). This ring is called the ring of endomorphism of A.

Now, certainly any ring $R$ has an associated abelian group, namely $(R,+)$. This suggests we can form a ring-homomorphism:

$R \to E(R)$ via the abelian group homomorphisms $L_r$ for $r \in R$ (why are they group-homomorphisms? the distributive laws!).

Abstracting from THIS, we call a ring-homomorphism $R \to E(A)$ an ACTION of $R$ on the abelian group $A$. The more usual name for an abelian group with a (left) $R$-action is... (left) $R$-module.

We can view this action in 2 ways:

1) a map $R \times A \to A$ (the usual "scalar multiplication").

2) a map $R \to E(A)$, which sends $r$ to the MAP that sends $a \mapsto r\cdot a$.

(2) is the map I mean when I write $r\cdot (-)$ (the "-" stands for "fill in the blank"). When I write $\text{Hom}(A,A)$, I am referring to $E(A)$.

To say that we have a map $(R,+) \to E(A)$ means that we have an ABELIAN GROUP HOMOMORPHISM, which is to say:

($\ast$): $(r + s)\cdot (-) = r\cdot (-) + s\cdot (-)$, that is, for every $a \in A$:

$(r + s)\cdot a = r\cdot a + s\cdot a$

Note that the + in (*) on the LHS is the addition in $R$, but the + on the RHS is the (pointwise) addition in the ring of endomorphisms, $E(A)$. On the line below, this translates to a + in $R$ on the LHS, and a + in $A$ on the RHS.

Let me lay out the parallels:

Monoid acting on Set relates to Monoid of Transformations (of set)
Group acting on Set relates to Group of Permutations (of set)
Ring acting on Abelian Group relates to Ring of Endomorphisms (of abelian group) (= $R$-module)
Field acting on Abelian Group relates to Ring of Endomorphisms (of abelian group) (= vector space).

In each case we can view the action as either:

1) $A \times B \to B$ (obeying certain rules, for "consistency's sake"-these are often presented as axioms, without motivation)
2) $A \to H(B)$ where we have an $A$-morphism to some canonical $A$-structure $H(B)$ based on certain maps $B \to B$ (this IS the motivation for the "form" of the axioms).

There is another important example of this phenomenon, which involves a weaker structure acting on a stronger structure:

Group acting on Vector Space (relates to General Linear Group = Group of units of ring of endomophisms of vector space),

which forms the basis for "Group Representation Theory".

The basic idea is to use facts about the $B$-structures (which are often, by their nature, easier to investigate) to gain insight into the $A$-structures. Often, new facets of behavior emerge in the "hybrids", yielding new $A$- and $B$-type information (such as in the orbit-stabilizer theorem).

 

[Math Amateur]

Dummit & Foote write:"If $A$ is an $R$-algebra it is easy to check that a natural left and right (unital) $R$-module structure defined by $r\cdot a = a \cdot r = f(r)a$, where $f(r)a$ is just the multiplication in the ring $A$..."

They later go on to write that other actions ARE possible, and so they ARE including this specific action in the definition:

"...but unless otherwise stated, this natural action on an algebra will be assumed."
To explain this, I am going to have to take a BIG detour.

Let's say you have a set, say $X$. there is a "natural" way to form a monoid associated with $X$ called the monoid of transformations of X, let's call it $T(X)$. What we do is form the set of all FUNCTIONS (set-morphisms) $X \to X$.

The operation of functional composition is associative, and possesses an identity, the identity function $1_X$ given by:

$1_X(x) = x$ for all $x \in X$.

Thus we have a monoid structure on $T(X)$.

Now given a monoid, say $M$, we have, for every $a \in M$, a function $L_a: M \to M$ given by:

$L_a(m) = am$ (left-multiplication by $a$).

Thus we have a mapping $M \to T(M)$ given by $a \mapsto L_a$.

Furthermore, this mapping is actually a monoid-homomorphism:

$L_{ab} = L_a \circ L_b$
$L_e = 1_M$

A monoid-homomorphism $M \to T(X)$ for some set $X$ is called an ACTION of $M$ on $X$, and the action is also called a representation of $M$ in $X$. So we have a fundamental result:

Every monoid $M$ acts on itself as a representation in $T(M)$ (technically, this is a left action-right actions also exist), via left-multiplication.

We can associate with groups a similar construction by limiting $T(X)$ to the sub-monoid of units (invertible functions). This sub-monoid has a special name: the permutations of $X$, or $\text{Sym}(X)$.

It turns out that every group can also be represented as a subgroup of $\text{Sym}(G)$, this is known as Cayley's Theorem, and the associated action (via left-multiplication) of $G$ upon itself is the left regular representation of $G$.

The homomorphism $G \to \text{Sym}(G)$ (the $G$ on the right is the underlying SET of $G$) can be abstracted to a homomorphism:

$G \to \text{Sym}(X)$, which yields the concept of a group $G$ acting on a set $X$.

This is often viewed as a mapping $G \times X \to X$, where $(g,x) \mapsto g\cdot x$. In other words, a way to "multiply" set-elements by group-elements. This sort of "special" group/set structure is called a $G$-set, and can be viewed as a certain kind of "compatibility" of the set $X$ with the group $G$.

Note how this hybrid structure mixes a "more complicated structure" with a "simpler" one.

Now with an ABELIAN group $A$, we can make a RING related to $A$, ALSO in a "natural" way.

We start with $T(A)$, the monoid of transformations of $A$, and restrict our attention to the sub-monoid of these transformations of $A$ that are also group-homomorphisms (note that homomorphisms are "composable", thus closed under our monoid multiplication of composition, and the identity transformation is also fortuitously a group homomorphism). Let's call this sub-monoid $E(A)$.

We can "create" an addition on $E(A)$ by defining:

$(\phi + \psi)(a) = \phi(a) + \psi(a)$ (where the + on the RHS is the abelian group operation).

Let's see that $\phi + \psi \in E(A)$:

$(\phi + \psi)(a + b) = \phi(a + b) + \psi(a + b)$ (by our definition)
$= \phi(a) + \phi(b) + \psi(a) + \psi(b)$ (because $\phi,\psi$ are both group-homomorphisms)
$= \phi(a) + \psi(a) + \phi(b) + \psi(b)$ (because addition in $A$ is COMMUTATIVE)
$= (\phi + \psi)(a) + (\phi + \psi)(b)$ (again, by definition).

So, we at least have a commutative semi-group structure $(E(A),+)$.

The 0-map $z(a) = 0$ for all $a \in A$ is an abelian group homomorphism (albeit a trivial one), and clearly functions as an identity:

$\phi + z = z + \phi = \phi$, for all $\phi \in E(A)$.

If we define $-\phi$ by: $(-\phi)(a) = -(\phi(a))$, for all $a \in A$, since $A$ is abelian, $-\phi \in E(A)$, and it is straight-forward to see that:

$\phi + -\phi = z$, for any $\phi \in E(A)$.

So now we have $(E(A),+)$ is an abelian group, and $(E(A),\circ)$ is a monoid. So all that's missing to show $E(A)$ is a ring, are the distributive laws. Here's the left one:

$\phi\circ (\psi + \chi) = (\phi \circ \psi) + (\phi \circ \chi)$

Proof:

Let $a \in A$. Then:

$[\phi \circ (\psi + \chi)](a) = \psi((\psi + \chi)(a)) = \psi(\psi(a) + \chi(a))$

$= \psi(\psi(a)) + \psi(\chi(a)) = (\psi\circ\psi)(a) + (\psi\circ\chi)(a)$

$= [(\phi \circ \psi) + (\phi \circ \chi)](a)$.

So now, we have created a ring $E(A)$ out of an abelian group $A$, using certain "special maps" $A \to A$ (I hope you are seeing the pattern, here). This ring is called the ring of endomorphism of A.

Now, certainly any ring $R$ has an associated abelian group, namely $(R,+)$. This suggests we can form a ring-homomorphism:

$R \to E(R)$ via the abelian group homomorphisms $L_r$ for $r \in R$ (why are they group-homomorphisms? the distributive laws!).

Abstracting from THIS, we call a ring-homomorphism $R \to E(A)$ an ACTION of $R$ on the abelian group $A$. The more usual name for an abelian group with a (left) $R$-action is... (left) $R$-module.

We can view this action in 2 ways:

1) a map $R \times A \to A$ (the usual "scalar multiplication").

2) a map $R \to E(A)$, which sends $r$ to the MAP that sends $a \mapsto r\cdot a$.

(2) is the map I mean when I write $r\cdot (-)$ (the "-" stands for "fill in the blank"). When I write $\text{Hom}(A,A)$, I am referring to $E(A)$.

To say that we have a map $(R,+) \to E(A)$ means that we have an ABELIAN GROUP HOMOMORPHISM, which is to say:

($\ast$): $(r + s)\cdot (-) = r\cdot (-) + s\cdot (-)$, that is, for every $a \in A$:

$(r + s)\cdot a = r\cdot a + s\cdot a$

Note that the + in (*) on the LHS is the addition in $R$, but the + on the RHS is the (pointwise) addition in the ring of endomorphisms, $E(A)$. On the line below, this translates to a + in $R$ on the LHS, and a + in $A$ on the RHS.

Let me lay out the parallels:

Monoid acting on Set relates to Monoid of Transformations (of set)
Group acting on Set relates to Group of Permutations (of set)
Ring acting on Abelian Group relates to Ring of Endomorphisms (of abelian group) (= $R$-module)
Field acting on Abelian Group relates to Ring of Endomorphisms (of abelian group) (= vector space).

In each case we can view the action as either:

1) $A \times B \to B$ (obeying certain rules, for "consistency's sake"-these are often presented as axioms, without motivation)
2) $A \to H(B)$ where we have an $A$-morphism to some canonical $A$-structure $H(B)$ based on certain maps $B \to B$ (this IS the motivation for the "form" of the axioms).

There is another important example of this phenomenon, which involves a weaker structure acting on a stronger structure:

Group acting on Vector Space (relates to General Linear Group = Group of units of ring of endomophisms of vector space),

which forms the basis for "Group Representation Theory".

The basic idea is to use facts about the $B$-structures (which are often, by their nature, easier to investigate) to gain insight into the $A$-structures. Often, new facets of behavior emerge in the "hybrids", yielding new $A$- and $B$-type information (such as in the orbit-stabilizer theorem).

Thanks for the extensive help, Deveno

Just working carefully through your post now ...

Thanks again ...

Peter

 

[Math Amateur]

Dummit & Foote write:"If $A$ is an $R$-algebra it is easy to check that a natural left and right (unital) $R$-module structure defined by $r\cdot a = a \cdot r = f(r)a$, where $f(r)a$ is just the multiplication in the ring $A$..."

They later go on to write that other actions ARE possible, and so they ARE including this specific action in the definition:

"...but unless otherwise stated, this natural action on an algebra will be assumed."
To explain this, I am going to have to take a BIG detour.

Let's say you have a set, say $X$. there is a "natural" way to form a monoid associated with $X$ called the monoid of transformations of X, let's call it $T(X)$. What we do is form the set of all FUNCTIONS (set-morphisms) $X \to X$.

The operation of functional composition is associative, and possesses an identity, the identity function $1_X$ given by:

$1_X(x) = x$ for all $x \in X$.

Thus we have a monoid structure on $T(X)$.

Now given a monoid, say $M$, we have, for every $a \in M$, a function $L_a: M \to M$ given by:

$L_a(m) = am$ (left-multiplication by $a$).

Thus we have a mapping $M \to T(M)$ given by $a \mapsto L_a$.

Furthermore, this mapping is actually a monoid-homomorphism:

$L_{ab} = L_a \circ L_b$
$L_e = 1_M$

A monoid-homomorphism $M \to T(X)$ for some set $X$ is called an ACTION of $M$ on $X$, and the action is also called a representation of $M$ in $X$. So we have a fundamental result:

Every monoid $M$ acts on itself as a representation in $T(M)$ (technically, this is a left action-right actions also exist), via left-multiplication.

... ... ...

Hi Deveno,

Working through your post I found the level of abstraction a bit of a challenge ... so I started to use an illustrative example ... but eventually started to lose my way and my confidence regarding the example ... hope you can help ...

First, you write:

" ... ... Let's say you have a set, say $X$. there is a "natural" way to form a monoid associated with $X$ called the monoid of transformations of X, let's call it $T(X)$. ... ... ...
... ... ...
Thus we have a monoid structure on $T(X)$."So ... for my illustrative example here I took $$X = \{ a, b \}$$ ... ... and proceeded as in Figure 1 directly below ... ... https://www.physicsforums.com/attachments/4769So we can see in Figure 1 that T(X) is a monoid, given that function composition is closed, associative and has an identity ...You then write:" ... ... Now given a monoid, say $M$, we have, for every $a \in M$, a function $L_a: M \to M$ given by:

$L_a(m) = am$ (left-multiplication by $a$).

... ... ...

A monoid-homomorphism $M \to T(X)$ for some set $X$ is called an ACTION of $M$ on $X$, and the action is also called a representation of $M$ in $X$. So we have a fundamental result:

Every monoid $M$ acts on itself as a representation in $T(M)$ (technically, this is a left action-right actions also exist), via left-multiplication. ... ... "


I could not see how to use $$T(X)$$ from my Figure 1 in this, especially as you mention a monoid M and talk about a monoid-homomorphism $M \to T(X)$ ...

So ... I took $$M \equiv \mathbb{Z}$$ which is a monoid under multiplication ... ...

We then clearly have for every $$a \in \mathbb{Z}$$ that

$$L_a \ : \ M \rightarrow M$$ where $$L_a(m) = am$$ for all $$m \in \mathbb{Z}$$ ...

Then we consider the mapping given by $$\phi \ : \ \mathbb{Z} \rightarrow T(\mathbb{Z}) $$ given by $$\phi (a) = L_a$$ ...

We have $$\phi (ab) = L_ab$$ and $$\phi (a) \circ \phi (b) = L_a L_b$$ Now $$L_ab(m) = abm
$$
and

$$L_a \circ L_b (m) = L_a ( L_b(m)) = L_a (bm) = abm$$ so $$\phi (ab) = \phi (a) \circ \phi (b)$$ where $$a, b \in \mathbb{Z}$$

or in simpler notation $$L_ab = L_a \circ L_b $$BUT ... now we need to show or illustrate a monoid homomorphism $$M \rightarrow T(X)$$ ...

In terms of our example, so far, then we need to form and illustrate/demonstrate a monoid homomorphism

$$\mathbb{Z} \rightarrow T(X) \equiv \{ T_1, T_2, T_3, T_4 \}$$ But ... at this point I am feeling a little lost ... need something like mapping $$\mathbb{Z}$$ onto something modulo 4 ? ... hmmm ... not sure what to do ...

Are you able to help me continue the example ... I would be particularly grateful if you could illustrate a representation of $M$ in $X$ and exactly how every monoid $M$ acts on itself as a representation in $T(M)$ ... ...

I really need a continuous and tangible example to make sure I have understood your post ...

Hope you can help further ...

Peter

 

[Deveno]

Let's use $\Bbb N$ instead of $\Bbb Z$, so we don't have to worry about negative integers.

Our set is also a monoid, but we'll regard it as just a set of four functions: $\{a,b\} \to \{a,b\}$.

For $k \in \Bbb N$, we'll define $k\cdot T_j = (T_j)^k = T_j \circ \cdots \circ T_j$ ($k$-fold composition).

So, for example, with $k = 4$ and $j = 2$ we have:

$4\cdot T_2 = T_2 \circ T_2 \circ T_2 \circ T_2$, and this is:

$T_2(T_2(T_2(T_2(a)))) = T_2(T_2(T_2(b))) = T_2(T_2(a)) = T_2(b) = a$
$T_2(T_2(T_2(T_2(b)))) = T_2(T_2(T_2(a))) = T_2(T_2(b)) = T_2(b) = b$

In other words $4\cdot T_2 = T_1$.

Here is the "complete" action of $4$ on $X$:

$4\cdot T_1 = T_1$
$4\cdot T_2 = T_1$
$4\cdot T_3 = T_3$
$4\cdot T_4 = T_4$.

So, to ensure we have a monoid-homomorphism, we have to check that:

$k\cdot (m \cdot T_j) = (km)\cdot T_j$
$1 \cdot T_j = T_j$.

The second is obvious. To see the first, note that:

$k\cdot(m\cdot T_j) = k\cdot (T_j)^m = ((T_j)^m)^k = (T_j)^{mk} = (T_j)^{km} = (km)\cdot T_j$

For example, $2\cdot (3\cdot T_2) = 2\cdot (T_2\circ T_2\circ T_2)$

$= (T_2 \circ T_2 \circ T_2)\circ (T_2 \circ T_2\circ T_2) = (T_j)^6$.

(if you consider $0 \in \Bbb N$, then use the convention that by definition:

$(T_j)^0 = T_1 = 1_{\{a,b\}}$, so that $0 \cdot T_j = T_1$ for all $j$).