Equivalence of Definitions of Algebras in D&F and Cooperstein | Peter

[Math Amateur]

I am reading Dummit and Foote's book "Abstract Algebra" (3rd Edition) and am focused on Chapter 15: Commutative Rings and Algebraic Geometry ... ...

On page 657 D&F give a definition of a k-algebra ... as follows:View attachment 5661I have to say I do not find that this definition gives me a good intuitive idea of the nature of an algebra over a field ... ... I much prefer the definition given by Cooperstein in his book "Advanced Linear Algebra" (Second Edition) where he defines what he calls an "associative algebra over a field F" (which I take to be the same as D&F's k-algebra ... ... is that right?) ... ... Cooperstein's definition is as follows:

View attachment 5662
I find Cooperstein's definition more clear regarding the nature of an algebra ... ... but I am currently working from D&F and wish to fully understand the approach to an algebra ... but I am unclear on exactly why these two definitions are the same or equivalent ... ... Can someone please help me to formally and rigorously prove that these two definitions are equivalent ... ... Help will be much appreciated ...

Peter

 

[Deveno]

Re: k-algebra/associative algebra - equivalence of definitions of algebras in D&F and Cooperstein

Suppose a(n associative) ring (with identity) $R$ is a $k$-algebra over a field $k$.

Define $\mu: R\times R \to R$ by $\mu(r,s) = rs$. (where the RHS is the ring product in $R$).

It is clear that $\mu$ is bi-additive, because of the distributive laws in a ring, and also $\mu$ is associative, because $R$ is an associative ring (some texts do not require this associativity condition, for example, with Lie rings). This takes care of (M1) and (M2).

Since $k \subseteq Z(R)$, for any $\gamma \in k$, and $r,s \in R$, we have:

$\gamma(rs) = (\gamma r)s = (r\gamma)s = r(\gamma s)$, since any element of $k$ commutes with all of $R$.

This dispatches (M3).

However, we're not quite done, we need to show that $R$ is a $k$-vector space. Clearly $(R,+)$ is an abelian group (since $R$ is a ring), so we can use that as our vector addition.

For scalar multiplication, we define:

$\gamma \cdot r = \gamma r$, where the RHS is just multiplication in $R$. Again, the distributive laws of the ring $R$ ensure that:

$(\gamma + \delta)\cdot r = (\gamma + \delta)r = \gamma r + \delta r = \gamma \cdot r + \delta \cdot r$, and

$\gamma\cdot (r + s) = \gamma(r + s) = \gamma r + \gamma s = \gamma\cdot r + \gamma\cdot s$.

The last equation/axiom is particularly interesting, because it tells us that the map $R \to R$, given by:

$r \mapsto \gamma\cdot r$ is an abelian group homomorphism from $R \to R$, in other words, this map is an element of : $\text{Hom}_{\Bbb Z}(R,R) = \text{End}_{\Bbb Z}(R)$

(the $\Bbb Z$ in the subscript is to indicate this is merely an abelian group homomorphism, not $R$-linear).

If we call this map $\phi_{\gamma}$, then the map $\phi: \gamma \mapsto \phi_{\gamma}$, which sends $\gamma$ to "scalar multiplication by $\gamma$" is (by the equation before the last one) an abelian group homomorphism from $k$ to $\text{End}_{\Bbb Z}(R)$, that is:

$\phi_{\gamma + \delta} = \phi_{\gamma} + \phi_{\delta}$, where the sum of endomorphisms on the right is defined *pointwise*:

$[\phi_{\gamma} + \phi_{\delta}](r) = \phi_{\gamma}(r) + \phi_{\delta}(r)$

But this is not all, we have that if we take the "multiplication" in $\text{End}_{\Bbb Z}(R)$ to be functional composition, this gives the endomorphisms a ring structure, and then we have a ring-homomorphism from $k \to \text{End}_{\Bbb Z}(R)$, that is:

$\phi_{\gamma\delta} = \phi_{\gamma} \circ \phi_{\delta}$.

To unpack what this means, we examine the action of these functions on a "generic" element of $R$, say, $r$.

Now, $\phi_{\gamma\delta}(r) = (\gamma\delta)\cdot r = (\gamma\delta)r$, while:

$(\phi_{\gamma} \circ \phi_{\delta})(r) = \phi_{\gamma}(\phi_{\delta}(r)) = \gamma\cdot(\phi_{\delta}(r))$

$= \gamma\cdot(\delta\cdot r) = \gamma(\delta r)$.

These are equal by the associativity of $R$'s multiplication ($\mu$).

Furthermore, this discussion is tantamount to showing the $k$-module axiom:

$\gamma\cdot(\delta\cdot r) = (\gamma\delta)\cdot r$ holds.

Now if $R$ has an identity $1_R$, and this identity is also the identity $1_k$, then we have:

$1_k \cdot r = 1_kr = 1_Rr = r$, which is the only vector space axiom we have not established, so $R$ is a $k$-vector space.

That shows D&F's definition implies Cooperstein's.

On the other hand, suppose $A$ is an associative $\Bbb F$-algebra with identity. Axioms (M1) and (M2) tell us we can impose a ring-structure on $(A,+,\mu)$.

Taking $b = 1_A$, we have, by (M3):

$(\gamma 1_A)a = \gamma(1_A a) = \gamma a = \gamma(a 1_A) = a(\gamma 1_A)$,

so for any $\gamma \in \Bbb F$, the element $\gamma 1_A$ is in the center of $A$. It is not hard (using vector space axioms-verify this!) to see that:

$\gamma \mapsto \gamma 1_A$ is a ring-homomorphism of $\Bbb F$ into $A$, and since we may take $\gamma = 1_{\Bbb F}$, we know it is a non-zero homomorphism, and thus injective, because $\Bbb F$ is a FIELD (and ring homomorphisms from a field can have only two possible kernels, the entire field, or solely the zero-element of the field).

In this way we can regard the image of this mapping as an embedding of $\Bbb F$ in $A$, and that this embedding is in the center of $A$.

Finally, from the vector space axiom:

$1_{\Bbb F}a = a$, taking $a = 1_A$, we have:

$1_{\Bbb F}1_A = 1_A$, which shows the image of $1_{\Bbb F}$ in our embedding *is* the identity of $A$, and we have a $k$-algebra (with $k = \Bbb F1_A$) in the sense of D&F.

 

[Math Amateur]

Re: k-algebra/associative algebra - equivalence of definitions of algebras in D&F and Cooperstein

Suppose a(n associative) ring (with identity) $R$ is a $k$-algebra over a field $k$.

Define $\mu: R\times R \to R$ by $\mu(r,s) = rs$. (where the RHS is the ring product in $R$).

It is clear that $\mu$ is bi-additive, because of the distributive laws in a ring, and also $\mu$ is associative, because $R$ is an associative ring (some texts do not require this associativity condition, for example, with Lie rings). This takes care of (M1) and (M2).

Since $k \subseteq Z(R)$, for any $\gamma \in k$, and $r,s \in R$, we have:

$\gamma(rs) = (\gamma r)s = (r\gamma)s = r(\gamma s)$, since any element of $k$ commutes with all of $R$.

This dispatches (M3).

However, we're not quite done, we need to show that $R$ is a $k$-vector space. Clearly $(R,+)$ is an abelian group (since $R$ is a ring), so we can use that as our vector addition.

For scalar multiplication, we define:

$\gamma \cdot r = \gamma r$, where the RHS is just multiplication in $R$. Again, the distributive laws of the ring $R$ ensure that:

$(\gamma + \delta)\cdot r = (\gamma + \delta)r = \gamma r + \delta r = \gamma \cdot r + \delta \cdot r$, and

$\gamma\cdot (r + s) = \gamma(r + s) = \gamma r + \gamma s = \gamma\cdot r + \gamma\cdot s$.

The last equation/axiom is particularly interesting, because it tells us that the map $R \to R$, given by:

$r \mapsto \gamma\cdot r$ is an abelian group homomorphism from $R \to R$, in other words, this map is an element of : $\text{Hom}_{\Bbb Z}(R,R) = \text{End}_{\Bbb Z}(R)$

(the $\Bbb Z$ in the subscript is to indicate this is merely an abelian group homomorphism, not $R$-linear).

If we call this map $\phi_{\gamma}$, then the map $\phi: \gamma \mapsto \phi_{\gamma}$, which sends $\gamma$ to "scalar multiplication by $\gamma$" is (by the equation before the last one) an abelian group homomorphism from $k$ to $\text{End}_{\Bbb Z}(R)$, that is:

$\phi_{\gamma + \delta} = \phi_{\gamma} + \phi_{\delta}$, where the sum of endomorphisms on the right is defined *pointwise*:

$[\phi_{\gamma} + \phi_{\delta}](r) = \phi_{\gamma}(r) + \phi_{\delta}(r)$

But this is not all, we have that if we take the "multiplication" in $\text{End}_{\Bbb Z}(R)$ to be functional composition, this gives the endomorphisms a ring structure, and then we have a ring-homomorphism from $k \to \text{End}_{\Bbb Z}(R)$, that is:

$\phi_{\gamma\delta} = \phi_{\gamma} \circ \phi_{\delta}$.

To unpack what this means, we examine the action of these functions on a "generic" element of $R$, say, $r$.

Now, $\phi_{\gamma\delta}(r) = (\gamma\delta)\cdot r = (\gamma\delta)r$, while:

$(\phi_{\gamma} \circ \phi_{\delta})(r) = \phi_{\gamma}(\phi_{\delta}(r)) = \gamma\cdot(\phi_{\delta}(r))$

$= \gamma\cdot(\delta\cdot r) = \gamma(\delta r)$.

These are equal by the associativity of $R$'s multiplication ($\mu$).

Furthermore, this discussion is tantamount to showing the $k$-module axiom:

$\gamma\cdot(\delta\cdot r) = (\gamma\delta)\cdot r$ holds.

Now if $R$ has an identity $1_R$, and this identity is also the identity $1_k$, then we have:

$1_k \cdot r = 1_kr = 1_Rr = r$, which is the only vector space axiom we have not established, so $R$ is a $k$-vector space.

That shows D&F's definition implies Cooperstein's.

On the other hand, suppose $A$ is an associative $\Bbb F$-algebra with identity. Axioms (M1) and (M2) tell us we can impose a ring-structure on $(A,+,\mu)$.

Taking $b = 1_A$, we have, by (M3):

$(\gamma 1_A)a = \gamma(1_A a) = \gamma a = \gamma(a 1_A) = a(\gamma 1_A)$,

so for any $\gamma \in \Bbb F$, the element $\gamma 1_A$ is in the center of $A$. It is not hard (using vector space axioms-verify this!) to see that:

$\gamma \mapsto \gamma 1_A$ is a ring-homomorphism of $\Bbb F$ into $A$, and since we may take $\gamma = 1_{\Bbb F}$, we know it is a non-zero homomorphism, and thus injective, because $\Bbb F$ is a FIELD (and ring homomorphisms from a field can have only two possible kernels, the entire field, or solely the zero-element of the field).

In this way we can regard the image of this mapping as an embedding of $\Bbb F$ in $A$, and that this embedding is in the center of $A$.

Finally, from the vector space axiom:

$1_{\Bbb F}a = a$, taking $a = 1_A$, we have:

$1_{\Bbb F}1_A = 1_A$, which shows the image of $1_{\Bbb F}$ in our embedding *is* the identity of $A$, and we have a $k$-algebra (with $k = \Bbb F1_A$) in the sense of D&F.

... thanks so much for the help, Deveno ... really wish to understand algebras ...

Just working through your post in detail and reflecting on what you have said ...

Thanks again,

Peter