MHB Definitions of Algebras in Cohn and in Dummit and Foote

[Math Amateur]

I am reading "Introduction to Ring Theory" by P. M. Cohn (Springer Undergraduate Mathematics Series)

In Chapter 2: Linear Algebras and Artinian Rings we find the definition of an algebra ... ... but in the chapter on module theory on page 342 of Dummit and Foote we find a different definition ... I cannot see how to reconcile these definitions ...

Cohn's definition of a $$k$$-algebra ($$k$$ is a field) reads as follows:View attachment 3275In Cohn's terms, then, presumably an $$R$$-algebra, where $$R$$ is a commutative ring with identity, would be a mapping $$R \times A \to A$$ denoted by $$( \alpha , r ) \to \alpha r$$ such that L.A.1 to L.A.5 hold with $$\alpha$$ and $$\beta \in R$$ instead of $$k$$.

Now, on page 342 Dummit and Foote define an R-algebra as follows:

"Definition. Let $$R$$ be a commutative ring with identity. An $$R$$-algebra is a ring $$A$$ with identity together with a ring homomorphism $$f \ : \ R \to A $$ mapping $$1_R$$ to $$1_A$$ such that the subring $$f(R)$$ of $$A$$ is contained in the center of $$A$$."

I cannot reconcile these two definitions ... can someone please help?

Peter

 

[mathbalarka]

Those two definitions are indeed equivalent.

For the $(\Rightarrow)$ case, note that in D-F's definition of an $R$-algebra, the ring homomorphism $f : R \to A$ gives a natural action of $R$ over $A$ defined as $(r, a) \mapsto f(r) \cdot a$. Can you show that this action satisfies axiom LA.1. to LA.5. in Cohn?

For the $(\Leftarrow)$ case, note that Cohn's axiom LA.1. to LA.4. are really $R$-module axioms, so that gives $A$ an $R$-module structure, with the action $(r, a) \to r \cdot a$. Define the map $f : R \to A$ as $r \mapsto r \cdot 1_A$, where $\cdot$ is our action of $R$ over $A$. Can you verify that $f$ is a ring homomorphism sending identity to identity? Use axiom LA.5. to verify that $f(R) \subseteq Z(A)$.

 

[Math Amateur]

Those two definitions are indeed equivalent.

For the $(\Rightarrow)$ case, note that in D-F's definition of an $R$-algebra, the ring homomorphism $f : R \to A$ gives a natural action of $R$ over $A$ defined as $(r, a) \mapsto f(r) \cdot a$. Can you show that this action satisfies axiom LA.1. to LA.5. in Cohn?

For the $(\Leftarrow)$ case, note that Cohn's axiom LA.1. to LA.4. are really $R$-module axioms, so that gives $A$ an $R$-module structure, with the action $(r, a) \to r \cdot a$. Define the map $f : R \to A$ as $r \mapsto r \cdot 1_A$, where $\cdot$ is our action of $R$ over $A$. Can you verify that $f$ is a ring homomorphism sending identity to identity? Use axiom LA.5. to verify that $f(R) \subseteq Z(A)$.

Thanks for the help and guidance Mathbalarka ... working on the post using your ideas ...

Thanks again,

Peter

 

[Math Amateur]

Those two definitions are indeed equivalent.

For the $(\Rightarrow)$ case, note that in D-F's definition of an $R$-algebra, the ring homomorphism $f : R \to A$ gives a natural action of $R$ over $A$ defined as $(r, a) \mapsto f(r) \cdot a$. Can you show that this action satisfies axiom LA.1. to LA.5. in Cohn?

For the $(\Leftarrow)$ case, note that Cohn's axiom LA.1. to LA.4. are really $R$-module axioms, so that gives $A$ an $R$-module structure, with the action $(r, a) \to r \cdot a$. Define the map $f : R \to A$ as $r \mapsto r \cdot 1_A$, where $\cdot$ is our action of $R$ over $A$. Can you verify that $f$ is a ring homomorphism sending identity to identity? Use axiom LA.5. to verify that $f(R) \subseteq Z(A)$.

In this post I will attempt to show that if we assume the conditions of Cohn's Definition (with $$k = R$$ being a commutative ring with $$1_R$$) then the conditions of Cohn's definition imply the conditions of D&F's definition ... ...

We need, then, to show that there is a ring homomorphism:

$$f \ : \ R \to A$$ mapping $$1_A \to 1_B$$

such that the subring $$f(R)$$ of $$A$$ is contained in the center of $$A$$, that is $$f(R) \subseteq Z(A)$$ ... ...

[ Note that the center of $$A$$ is as follows:

$$Z(A) = \{ z \in A \ | \ za = az \text{ for all } a \in A \}$$ ]Now (following Mathbalarka - see above post) we define

$$f \ : \ R \to A$$ by $$\alpha \mapsto \alpha 1_A$$

Now, we need to show that $$f$$ is a ring homomorphism ... ... that is that:

$$( \alpha + \beta ) f = \alpha f + \beta f
$$

where $$\alpha , \beta \in R$$

and

$$( \alpha \beta ) f = ( \alpha f ) ( \beta f )
$$

where, again, $$ \alpha , \beta \in R
$$Now to show this we proceed as follows:

$$( \alpha + \beta ) f
$$

$$= ( \alpha + \beta ) 1_A$$ ... ... Definition of f

= $$\alpha 1_A + \beta 1_A$$ ... ... by L.A.2

$$= \alpha f + \beta f $$ ... ... Definition of fFurther we have:

$$( \alpha \beta ) f
$$

$$= ( \alpha \beta ) 1_A$$ ... ... Definition of f

$$= \alpha ( \beta 1_A)$$ ... ... by L.A.3

$$= ( \alpha 1_A ) ( \beta 1_A)$$ ... ... since $$\alpha = \alpha 1_A$$ ( ? Is this correct ?)

$$= ( \alpha f) ( \beta f)$$ ... ... Definition of f
Now, we also need to show that $$f$$ maps identity $$1_R$$ to identity $$1_A$$, that is

$$1_R f = 1_A$$

But,

$$1_R f = 1_R 1_A$$ ... ... Definition of f

$$= 1_A$$ ... ... by L.A.4
Now we must show that $$f(R) \subseteq Z(A)$$, where:

$$Z(A) = \{ z \in A \ | \ za = az \text{ for all } a \in A \}$$So to proceed, let $$x \in f(R)$$

Now $$x \in f(R) \Longrightarrow \text{ there exists } \alpha \in R \text{ such that } x = \alpha f$$

Now also consider $$y \in f(R)$$ ...

$$y \in f(R) \Longrightarrow \text{ there exists } \beta \in R \text{ such that } y= \beta f$$Now, given the above:

$$xy $$

$$= ( \alpha f ) ( \beta f )$$ ... ... definition of $$x,y$$

$$= ( \alpha \beta ) f $$ ... ... since f is a homomorphism

$$= ( \beta \alpha ) f$$ ... ... since $$R$$ is commutative

$$= ( \beta f ) ( \alpha f)$$ ... ... since f is a homomorphism

$$= yx$$

Hence $$x \in Z(A)$$ as required ...Can someone please confirm that the above argument/analysis is correct?


***NOTE***

I am much concerned about my demonstration that $$f(R) \subseteq Z(A)$$ since I did not explicitly use L.A.5 as Mathbalarka advised ... ...

 

[mathbalarka]

$$( \alpha + \beta ) f
$$

$$= ( \alpha + \beta ) 1_A$$ ... ... Definition of f

= $$\alpha 1_A + \beta 1_A$$ ... ... by L.A.2

$$= \alpha f + \beta f $$ ... ... Definition of f

This part is good.

Further we have:

$$( \alpha \beta ) f
$$

$$= ( \alpha \beta ) 1_A$$ ... ... Definition of f

$$= \alpha ( \beta 1_A)$$ ... ... by L.A.3

$$= ( \alpha 1_A ) ( \beta 1_A)$$ ... ... since $$\alpha = \alpha 1_A$$ ( ? Is this correct ?)

$$= ( \alpha f) ( \beta f)$$ ... ... Definition of f

I don't follow your step $4$. $\alpha \cdot 1_A = f(\alpha)$ which is an element of $A$, NOT an element of $R$, so surely $\alpha$ can't be replaced by $\alpha \cdot 1_A$. That step is invalid. [Hint : You need to use axiom LA.5. to prove that $f(\alpha \beta) = f(\alpha) f(\beta)$]

Now, we also need to show that $$f$$ maps identity $$1_R$$ to identity $$1_A$$, that is

$$1_R f = 1_A$$

But,

$$1_R f = 1_R 1_A$$ ... ... Definition of f

$$= 1_A$$ ... ... by L.A.4

This looks good.

Now, given the above:

$$xy $$

$$= ( \alpha f ) ( \beta f )$$ ... ... definition of $$x,y$$

$$= ( \alpha \beta ) f $$ ... ... since f is a homomorphism

$$= ( \beta \alpha ) f$$ ... ... since $$R$$ is commutative

$$= ( \beta f ) ( \alpha f)$$ ... ... since f is a homomorphism

$$= yx$$

Hence $$x \in Z(A)$$ as required ...

You have only proved that element of $f(R)$ commute with each other, NOT $f(R) \subseteq Z(A)$. You need to show that $xf(y) = f(y)x$ for all $x \in A$ and $y \in R$ to accomplish this.

 

[Math Amateur]

This part is good.
I don't follow your step $4$. $\alpha \cdot 1_A = f(\alpha)$ which is an element of $A$, NOT an element of $R$, so surely $\alpha$ can't be replaced by $\alpha \cdot 1_A$. That step is invalid. [Hint : You need to use axiom LA.5. to prove that $f(\alpha \beta) = f(\alpha) f(\beta)$]
This looks good.
You have only proved that element of $f(R)$ commute with each other, NOT $f(R) \subseteq Z(A)$. You need to show that $xf(y) = f(y)x$ for all $x \in A$ and $y \in R$ to accomplish this.

Thanks for the critique and feedback Mathbalarka ... Getting tired now ... But will check all you have said tomorrow ...

Thanks again,

Peter

 

[Deveno]

(LA.5) states that:

$(\alpha\cdot x)y = x(\alpha\cdot y) = \alpha \cdot (xy)$ for all $\alpha \in R$, and all $x,y \in A$.

Now suppose $a \in (R)f$ (I am writing the mappings on the right, which is a bit odd in this scenario, because we have a LEFT action).

We want to show that for any $x \in A$:

$ax = xa$.

Since $a \in (R)f$, we have: $a = \alpha f$ for some $\alpha \in R$.

By definition, $\alpha f = \alpha \cdot 1_A$.

Applying (LA.5) we have:

$ax = (\alpha f)x = (\alpha \cdot 1_A)x = 1_A(\alpha \cdot x) = \alpha \cdot x$

$= \alpha \cdot (x1_A) = x(\alpha \cdot 1_A) = x(\alpha f) = xa$.

The idea is: "scalars" commute with "vectors" (or rather, the embeddings of scalars in $A$ afforded by $f$ do).

Note that this is an embedding (of necessity) when $R$ is a FIELD (as the only ring-homomorphisms from a field are injective, if we require "unity be preserved").

Note as well, that mathbalarka's construction require that $A$ be a UNITAL ring.

 

[Math Amateur]

This part is good.
I don't follow your step $4$. $\alpha \cdot 1_A = f(\alpha)$ which is an element of $A$, NOT an element of $R$, so surely $\alpha$ can't be replaced by $\alpha \cdot 1_A$. That step is invalid. [Hint : You need to use axiom LA.5. to prove that $f(\alpha \beta) = f(\alpha) f(\beta)$]
This looks good.
You have only proved that element of $f(R)$ commute with each other, NOT $f(R) \subseteq Z(A)$. You need to show that $xf(y) = f(y)x$ for all $x \in A$ and $y \in R$ to accomplish this.

Thanks again Mathbalarka ... indeed, you are correct ... how careless of me ...

Another attempt follows ...
We wish to show that $$f(R) \subseteq Z(A)$$

where:

$$Z(A) = \{ z \in A \ | \ za = az \text{ for all } a \in A \}$$
Now if we take $$x \in f(R)$$ ... ...

... then $$x \in Z(A)$$ if $$xa = ax \text{ for all } a \in A$$So ... ... consider any $$a \in A $$... ... Then proceed as follows ... ...$$x \in f(R) $$

$$\Longrightarrow$$ there exists $$\alpha \in R$$ such that $$\alpha f = x$$Now we want $$xa = ax$$ for all $$a \in A
$$

That is, we want to show $$( \alpha f ) a = a ( \alpha f )$$ for all $$a \in A
$$But we have that:

$$( \alpha f ) a$$

$$= ( \alpha 1_A )a$$ ... ... by definition of f

$$= \alpha (1_A a )$$ ... ... by L.A.5

$$= \alpha a$$ ... ... since 1_A a = a
However we also have that:

$$a ( \alpha f )
$$$$= a ( \alpha 1_A ) $$ ... ... by definition of f

$$= \alpha ( a 1_A)$$ ... ... by L.A.5

$$= \alpha a$$ ... ... since $$1_A a = a$$
So we have that

$$( \alpha f ) a = a ( \alpha f ) = \alpha a
$$ for all $$a \in A$$But $$\alpha f = x$$

So, we have $$xa = ax$$ for all $$a \in A$$; that is $$x \in Z(A)$$
Could someone please critique my demonstration that the subring f(R) of A is contained in the center of A?Peter

 

[Euge]

Thanks again Mathbalarka ... indeed, you are correct ... how careless of me ...

Another attempt follows ...
We wish to show that $$f(R) \subseteq Z(A)$$

where:

$$Z(A) = \{ z \in A \ | \ za = az \text{ for all } a \in A \}$$

Now if we take $$x \in f(R)$$ ... ...

... then $$x \in Z(A)$$ if $$xa = ax \text{ for all } a \in A$$So ... ... consider any $$a \in A $$... ... Then proceed as follows ... ...$$x \in f(R) $$

$$\Longrightarrow$$ there exists $$\alpha \in R$$ such that $$\alpha f = x$$Now we want $$xa = ax$$ for all $$a \in A
$$

That is, we want to show $$( \alpha f ) a = a ( \alpha f )$$ for all $$a \in A
$$But we have that:

$$( \alpha f ) a$$

$$= ( \alpha 1_A )a$$ ... ... by definition of f

$$= \alpha (1_A a )$$ ... ... by L.A.5

$$= \alpha a$$ ... ... since 1_A a = a
However we also have that:

$$a ( \alpha f )
$$$$= a ( \alpha 1_A ) $$ ... ... by definition of f

$$= \alpha ( a 1_A)$$ ... ... by L.A.5

$$= \alpha a$$ ... ... since $$1_A a = a$$
So we have that

$$( \alpha f ) a = a ( \alpha f ) = \alpha a
$$ for all $$a \in A$$But $$\alpha f = x$$

So, we have $$xa = ax$$ for all $$a \in A$$; that is $$x \in Z(A)$$
Could someone please critique my demonstration that the subring f(R) of A is contained in the center of A?Peter

Your proof is logically sound.

 

[Math Amateur]

This part is good.
I don't follow your step $4$. $\alpha \cdot 1_A = f(\alpha)$ which is an element of $A$, NOT an element of $R$, so surely $\alpha$ can't be replaced by $\alpha \cdot 1_A$. That step is invalid. [Hint : You need to use axiom LA.5. to prove that $f(\alpha \beta) = f(\alpha) f(\beta)$]
This looks good.
You have only proved that element of $f(R)$ commute with each other, NOT $f(R) \subseteq Z(A)$. You need to show that $xf(y) = f(y)x$ for all $x \in A$ and $y \in R$ to accomplish this.

Thanks again, Mathbalarka ...

In the above post you write:

" ... ... I don't follow your step $4$. $\alpha \cdot 1_A = f(\alpha)$ which is an element of $A$, NOT an element of $R$, so surely $\alpha$ can't be replaced by $\alpha \cdot 1_A$. That step is invalid. [Hint : You need to use axiom LA.5. to prove that $f(\alpha \beta) = f(\alpha) f(\beta)$] ... ... "
Indeed, I believe you are correct again!
Rethinking the proof of $$( \alpha \beta ) f = ( \alpha f ) ( \beta f )
$$ where $$ \alpha , \beta \in R $$ we proceed as follows: $$( \alpha f ) ( \beta f )$$

$$= ( \alpha 1_A ) ( \beta 1_A )$$ ... ... by definition of fNow let $$r = \alpha 1_A = \alpha f \in A$$so then we have:$$( \alpha f ) ( \beta f ) $$

= $$r ( \beta 1_A )$$ ... ... where $$r, 1_A \in A$$

$$= \beta ( r 1_A)$$ ... ... by L.A.5

$$= \beta ( \alpha 1_A 1_A)$$

= $$\beta ( \alpha 1_A ) $$ ... ... since $$1_A 1_A = 1_A $$

= $$( \beta \alpha ) 1_A$$ ... ... by L.A.3

= $$( \alpha \beta) 1_A$$ ... ... since $$R$$ is commutative

= $$( \alpha \beta) f $$

Can someone please confirm that my analysis above is correct?

Peter

 

[Euge]

Thanks again, Mathbalarka ...

In the above post you write:

" ... ... I don't follow your step $4$. $\alpha \cdot 1_A = f(\alpha)$ which is an element of $A$, NOT an element of $R$, so surely $\alpha$ can't be replaced by $\alpha \cdot 1_A$. That step is invalid. [Hint : You need to use axiom LA.5. to prove that $f(\alpha \beta) = f(\alpha) f(\beta)$] ... ... "
Indeed, I believe you are correct again!
Rethinking the proof of $$( \alpha \beta ) f = ( \alpha f ) ( \beta f )
$$ where $$ \alpha , \beta \in R $$ we proceed as follows: $$( \alpha f ) ( \beta f )$$

$$= ( \alpha 1_A ) ( \beta 1_A )$$ ... ... by definition of fNow let $$r = \alpha 1_A = \alpha f \in A$$so then we have:$$( \alpha f ) ( \beta f ) $$

= $$r ( \beta 1_A )$$ ... ... where $$r, 1_A \in A$$

$$= \beta ( r 1_A)$$ ... ... by L.A.5

$$= \beta ( \alpha 1_A 1_A)$$

= $$\beta ( \alpha 1_A ) $$ ... ... since $$1_A 1_A = 1_A $$

= $$( \beta \alpha ) 1_A$$ ... ... by L.A.3

= $$( \alpha \beta) 1_A$$ ... ... since $$R$$ is commutative

= $$( \alpha \beta) f $$

Can someone please confirm that my analysis above is correct?

Peter

This is also good.