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

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

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

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 \(\displaystyle k = R\) being a commutative ring with \(\displaystyle 1_R\)) 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:

\(\displaystyle f \ : \ R \to A\) mapping \(\displaystyle 1_A \to 1_B\)

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

[ Note that the center of \(\displaystyle A\) is as follows:

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

\(\displaystyle f \ : \ R \to A\) by \(\displaystyle \alpha \mapsto \alpha 1_A\)

Now, we need to show that \(\displaystyle f\) is a ring homomorphism ... ... that is that:

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

where \(\displaystyle \alpha , \beta \in R\)

and

\(\displaystyle ( \alpha \beta ) f = ( \alpha f ) ( \beta f )
\)

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

\(\displaystyle ( \alpha + \beta ) f
\)

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

= \(\displaystyle \alpha 1_A + \beta 1_A\) ... ... by L.A.2

\(\displaystyle = \alpha f + \beta f \) ... ... Definition of fFurther we have:

\(\displaystyle ( \alpha \beta ) f
\)

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

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

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

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

\(\displaystyle 1_R f = 1_A\)

But,

\(\displaystyle 1_R f = 1_R 1_A\) ... ... Definition of f

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

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

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

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

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

\(\displaystyle xy \)

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

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

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

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

\(\displaystyle = yx\)

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


***NOTE***

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

 

[mathbalarka]

\(\displaystyle ( \alpha + \beta ) f
\)

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

= \(\displaystyle \alpha 1_A + \beta 1_A\) ... ... by L.A.2

\(\displaystyle = \alpha f + \beta f \) ... ... Definition of f

This part is good.

Further we have:

\(\displaystyle ( \alpha \beta ) f
\)

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

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

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

\(\displaystyle = ( \alpha f) ( \beta f)\) ... ... 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 \(\displaystyle f\) maps identity \(\displaystyle 1_R\) to identity \(\displaystyle 1_A\), that is

\(\displaystyle 1_R f = 1_A\)

But,

\(\displaystyle 1_R f = 1_R 1_A\) ... ... Definition of f

\(\displaystyle = 1_A\) ... ... by L.A.4

This looks good.

Now, given the above:

\(\displaystyle xy \)

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

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

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

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

\(\displaystyle = yx\)

Hence \(\displaystyle x \in Z(A)\) 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 \(\displaystyle f(R) \subseteq Z(A)\)

where:

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

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

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

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

\(\displaystyle ( \alpha f ) a\)

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

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

\(\displaystyle = \alpha a\) ... ... since 1_A a = a
However we also have that:

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

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

\(\displaystyle = \alpha a\) ... ... since \(\displaystyle 1_A a = a\)
So we have that

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

So, we have \(\displaystyle xa = ax\) for all \(\displaystyle a \in A\); that is \(\displaystyle x \in Z(A)\)
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 \(\displaystyle f(R) \subseteq Z(A)\)

where:

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

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

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

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

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

\(\displaystyle ( \alpha f ) a\)

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

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

\(\displaystyle = \alpha a\) ... ... since 1_A a = a
However we also have that:

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

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

\(\displaystyle = \alpha a\) ... ... since \(\displaystyle 1_A a = a\)
So we have that

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

So, we have \(\displaystyle xa = ax\) for all \(\displaystyle a \in A\); that is \(\displaystyle x \in Z(A)\)
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 \(\displaystyle ( \alpha \beta ) f = ( \alpha f ) ( \beta f )
\) where \(\displaystyle \alpha , \beta \in R \) we proceed as follows: \(\displaystyle ( \alpha f ) ( \beta f )\)

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

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

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

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

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

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

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

= \(\displaystyle ( \alpha \beta) f \)

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 \(\displaystyle ( \alpha \beta ) f = ( \alpha f ) ( \beta f )
\) where \(\displaystyle \alpha , \beta \in R \) we proceed as follows: \(\displaystyle ( \alpha f ) ( \beta f )\)

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

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

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

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

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

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

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

= \(\displaystyle ( \alpha \beta) f \)

Can someone please confirm that my analysis above is correct?

Peter

This is also good.