Definitions of Algebras in Cohn and in Golan

[Math Amateur]

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

In Cohn's book, in Chapter 2: Linear Algebras and Artinian Rings, we find the definition of an algebra ... ... but in Jonathan Golan's book ["The Linear Algebra a Beginning Graduate Student Ought to Know"] we find a (apparently) 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:https://www.physicsforums.com/attachments/3276In Golan's book the definition of a \(\displaystyle k\)-algebra (Golan actually uses \(\displaystyle F\) to stand for the field) is in Chapter 4: Algebras Over a Field and reads as follows (pages 33-34):
https://www.physicsforums.com/attachments/3277
https://www.physicsforums.com/attachments/3278

Can someone please help me reconcile these two definitions of a \(\displaystyle k\)-algebra?

Help would be appreciated.

Peter

 

[Deveno]

They are equivalent, under certain assumptions.

Suppose we have a ring with a multiplication $k \times A \to A$ that satisfies (LA.1)-(LA.5).

Then $(A,+)$ is an abelian group and (LA.1)-(LA.4) say that $A$ is a $k$-module, that is, a vector space. (LA.5) says that the scalar multiplication is compatible with the ring multiplication. Note that (LA.5) and (3) are the same condition, and that (1) and (2) are just the distributive laws for a ring.

Since rings are typically held to be associative, Cohn's definition of linear algebra is actually a definition of an associative $k$-algebra.

On the other hand, suppose we have a vector space $V$ over $k$ satisfying (1)-(3). Then $V$ automatically satisfies (LA.1)-(LA.4), being a $k$-module, and (3) is (LA.5). (1) and (2) say that $V$ is a (non-associative) ring. If we stipulate (4) as well, then we recover Cohn's definition.

Note that (1)-(3) say that the multiplication of $V$ is a $k$-bilinear map (some texts use the term $k$-balanced map, although technically this is slightly more general than $k$-bilinear, a balanced map $B$ need not satisfy:

$B(\alpha r,s) = \alpha B(r,s)$ (this is "part" of (3))

but merely:

$B(\alpha r,s) = B(r,\alpha s)$ (this is the "other part")).

Loosely speaking, an algebra is a ring which is a vector space, or a vector space which is a ring (with the requisite compatibility condition). It really doesn't matter if the ring structure is added to the vector space, or the vector space structure is added to the ring.

There is a third characterization of an (associative) $k$-algebra:

Suppose $A$ is an (associative) ring, and we have a ring-homomorphism $\phi: k \to Z(A)$. Since $\phi$ is a ring-homomorphism, we have:

$\phi(\alpha + \beta)r = \phi(\alpha)r + \phi(\beta)r$ (LA.2) -using the homomorphism property of $\phi$ and right-distrbutivity
$\phi(\alpha)(r + s) = \phi(\alpha)r + \phi(\alpha)s$ (LA.1)-using left-distributivity
$(\phi(\alpha\beta))r = \phi(\alpha)(\phi(\beta)r)$ (LA.3) -using associativity of $A$, and the homomorphism property of $\phi$
$\phi(\alpha)(rs) = (\phi(\alpha)r)s = r(\phi(\alpha)s)$ (LA.5) -since $\phi(\alpha) \in Z(A)$ and thus commutes with all of $A$, and using associativity of $A$.

Note that to recover (LA.4) we need to add the additional requirement that $\phi(1_k)(-): A \to A$ is the identity map on $A$. If $A$ is already a unital ring, and $\phi$ preserves the unity, this is not necessary (texts differ on how "ring" is defined).

In this case, the obvious $k$-action on $A$ is: $\alpha\cdot r = \phi(\alpha)r$. In particular, if $A$ is an extension ring of $k$, we may take $\phi$ to be the inclusion map.

 

[Math Amateur]

They are equivalent, under certain assumptions.

Suppose we have a ring with a multiplication $k \times A \to A$ that satisfies (LA.1)-(LA.5).

Then $(A,+)$ is an abelian group and (LA.1)-(LA.4) say that $A$ is a $k$-module, that is, a vector space. (LA.5) says that the scalar multiplication is compatible with the ring multiplication. Note that (LA.5) and (3) are the same condition, and that (1) and (2) are just the distributive laws for a ring.

Since rings are typically held to be associative, Cohn's definition of linear algebra is actually a definition of an associative $k$-algebra.

On the other hand, suppose we have a vector space $V$ over $k$ satisfying (1)-(3). Then $V$ automatically satisfies (LA.1)-(LA.4), being a $k$-module, and (3) is (LA.5). (1) and (2) say that $V$ is a (non-associative) ring. If we stipulate (4) as well, then we recover Cohn's definition.

Note that (1)-(3) say that the multiplication of $V$ is a $k$-bilinear map (some texts use the term $k$-balanced map, although technically this is slightly more general than $k$-bilinear, a balanced map $B$ need not satisfy:

$B(\alpha r,s) = \alpha B(r,s)$ (this is "part" of (3))

but merely:

$B(\alpha r,s) = B(r,\alpha s)$ (this is the "other part")).

Loosely speaking, an algebra is a ring which is a vector space, or a vector space which is a ring (with the requisite compatibility condition). It really doesn't matter if the ring structure is added to the vector space, or the vector space structure is added to the ring.

There is a third characterization of an (associative) $k$-algebra:

Suppose $A$ is an (associative) ring, and we have a ring-homomorphism $\phi: k \to Z(A)$. Since $\phi$ is a ring-homomorphism, we have:

$\phi(\alpha + \beta)r = \phi(\alpha)r + \phi(\beta)r$ (LA.2) -using the homomorphism property of $\phi$ and right-distrbutivity
$\phi(\alpha)(r + s) = \phi(\alpha)r + \phi(\alpha)s$ (LA.1)-using left-distributivity
$(\phi(\alpha\beta))r = \phi(\alpha)(\phi(\beta)r)$ (LA.3) -using associativity of $A$, and the homomorphism property of $\phi$
$\phi(\alpha)(rs) = (\phi(\alpha)r)s = r(\phi(\alpha)s)$ (LA.5) -since $\phi(\alpha) \in Z(A)$ and thus commutes with all of $A$, and using associativity of $A$.

Note that to recover (LA.4) we need to add the additional requirement that $\phi(1_k)(-): A \to A$ is the identity map on $A$. If $A$ is already a unital ring, and $\phi$ preserves the unity, this is not necessary (texts differ on how "ring" is defined).

In this case, the obvious $k$-action on $A$ is: $\alpha\cdot r = \phi(\alpha)r$. In particular, if $A$ is an extension ring of $k$, we may take $\phi$ to be the inclusion map.

Thanks Deveno ... still reading, thinking and reflecting on what you have said ...

BUT ... could you clarify a source of uncertainty and puzzlement for me ... ...

At the start of Cohn's definition of a \(\displaystyle k\)-algebra, we read that ...

... ... by a \(\displaystyle k\)-algebra "we understand a ring \(\displaystyle A\) with a mapping from \(\displaystyle k \times A\) to \(\displaystyle A\), denoted by \(\displaystyle ( \alpha, r ) \to \alpha r\) such that ... ... ... "
However at the start of Golan's definition we read:

" ... ... A vector space over a field \(\displaystyle \mathcal{F}\) is an \(\displaystyle \mathcal{F}\)-algebra if and only if there exists a function \(\displaystyle ( v, w) \to v \bullet w \text{ from } V \times V\) to \(\displaystyle V\) such that ... ... "Now it puzzles me that the two mappings do not seem to be the same or at least similar ... how do we explain/reconcile the differences ... ... of course it may be that they are quite different ... ?

Can you explain the role of these functions (mappings, actions ...) and their role in the definition of an algebra ... are the actions the same, similar, different ... ... ? [they seem quite different ... ?]

Peter

 

[mathbalarka]

It's actually quite patently obvious. [The notations also confused me at first, hehe]

The usual $\cdot$ (juxtaposition) in Cohn is really the action of $k$ over $A$. On the other hand $\bullet$ in Golan's definition is the ring multiplication.

Let's see how this all works out : Cohn never explicitly says that $A$ is an $k$-module, but the axioms LA.1. to LA.4. are precisely the module axioms, thus gives a vector space structure over $A$. That goes with Golans already defined vector space $V$.

Now note that a vector space may or may not be a ring -- it's usually an abelian group. But axioms (1), (2) and (4) in Golan says that there is a binary operator $\bullet : V \times V \to V$ that is associative and distributes over addition. That gives $V$ a ring structure (compare that with Cohns already defined ring $A$!).

All we are left with is LA.5. of Cohn and (3) of Golan. But these are really equivalent -- Cohn just doesn't explicitly writes out the $k$-action and vector multiplication differently, but prefers them both to be juxtaposition.

 

[Deveno]

Here is an exercise, to test how well you get these ideas:

Let $A$ be the $\Bbb R$-algebra $\text{Mat}_n(\Bbb R)$ of $n \times n$ matrices with real entries.

Explicitly describe a field embedding $\eta: \Bbb R \to A$, and demonstrate that $\eta(\Bbb R) \subseteq Z(A)$.

For extra credit, show that $\eta(\Bbb R) = Z(A)$ (the way I would do this is to show that if: $(a_{ij}) \not\in \eta(\Bbb R)$ that you can produce SOME $n \times n$ matrix it doesn't commute with. The elementary matrices may prove useful in this regard).

Another exercise: verify $k[x]$ is a $k$-algebra. What is $\eta(k)$?

 

[Math Amateur]

It's actually quite patently obvious. [The notations also confused me at first, hehe]

The usual $\cdot$ (juxtaposition) in Cohn is really the action of $k$ over $A$. On the other hand $\bullet$ in Golan's definition is the ring multiplication.

Let's see how this all works out : Cohn never explicitly says that $A$ is an $k$-module, but the axioms LA.1. to LA.4. are precisely the module axioms, thus gives a vector space structure over $A$. That goes with Golans already defined vector space $V$.

Now note that a vector space may or may not be a ring -- it's usually an abelian group. But axioms (1), (2) and (4) in Golan says that there is a binary operator $\bullet : V \times V \to V$ that is associative and distributes over addition. That gives $V$ a ring structure (compare that with Cohns already defined ring $A$!).

All we are left with is LA.5. of Cohn and (3) of Golan. But these are really equivalent -- Cohn just doesn't explicitly writes out the $k$-action and vector multiplication differently, but prefers them both to be juxtaposition.

Thanks Mathbalarka ... I will study what you say and work carefully through the definitions again ...

I do appreciate your assistance in understanding these ideas ... ...

Peter

- - - Updated - - -

Here is an exercise, to test how well you get these ideas:

Let $A$ be the $\Bbb R$-algebra $\text{Mat}_n(\Bbb R)$ of $n \times n$ matrices with real entries.

Explicitly describe a field embedding $\eta: \Bbb R \to A$, and demonstrate that $\eta(\Bbb R) \subseteq Z(A)$.

For extra credit, show that $\eta(\Bbb R) = Z(A)$ (the way I would do this is to show that if: $(a_{ij}) \not\in \eta(\Bbb R)$ that you can produce SOME $n \times n$ matrix it doesn't commute with. The elementary matrices may prove useful in this regard).

Another exercise: verify $k[x]$ is a $k$-algebra. What is $\eta(k)$?

Thanks for these exercises, Deveno ... I will work on them later today when I get a chance to do so ...

Thanks for your considerable help in understanding k-algebras ...

Then I move on in Cohn to Chapter 2, section 2.3: Artinian Rings: The Semisimple Case ... which sounds very interesting indeed ... ... (just had a quick look at some of the Theorems :-) ... ... )

Peter

 

[Math Amateur]

Here is an exercise, to test how well you get these ideas:

Let $A$ be the $\Bbb R$-algebra $\text{Mat}_n(\Bbb R)$ of $n \times n$ matrices with real entries.

Explicitly describe a field embedding $\eta: \Bbb R \to A$, and demonstrate that $\eta(\Bbb R) \subseteq Z(A)$.

For extra credit, show that $\eta(\Bbb R) = Z(A)$ (the way I would do this is to show that if: $(a_{ij}) \not\in \eta(\Bbb R)$ that you can produce SOME $n \times n$ matrix it doesn't commute with. The elementary matrices may prove useful in this regard).

Another exercise: verify $k[x]$ is a $k$-algebra. What is $\eta(k)$?

Thanks again for your help, Deveno ... ...

In the above post you write:

"Here is an exercise, to test how well you get these ideas:

Let $A$ be the $\Bbb R$-algebra $\text{Mat}_n(\Bbb R)$ of $n \times n$ matrices with real entries.

Explicitly describe a field embedding $\eta: \Bbb R \to A$, and demonstrate that $\eta(\Bbb R) \subseteq Z(A)$. ... ... "
Here is my attempt at this exercise ... ...I will work with \(\displaystyle \text{Mat}_2 ( \mathbb{R} ) \), since it makes things easier to post, notationally speaking, and the exercise 'scales up' quite naturally ...Define a field embedding (injective mapping with a field as a domain) \(\displaystyle \eta\) as follows:

\(\displaystyle \eta \ : \ \mathbb{R} \to \text{ Mat}_2 ( \mathbb{R} ) \)

where \(\displaystyle \eta (k) = \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix}\) for all \(\displaystyle k \in \mathbb{R}\)

\(\displaystyle \eta\) is obviously an embedding.

Now we need to show that \(\displaystyle \eta (k) \subseteq Z ( \text{Mat}_2 ( \mathbb{R} ) )\)

So consider any particular matrix in \(\displaystyle \eta ( \mathbb{R} ) \), say \(\displaystyle \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix}\) where \(\displaystyle k\) is some real number ... ...

Further let \(\displaystyle \begin{bmatrix} m_{11} & m_{12} \\ m_{21} & m_{22} \end{bmatrix}\) be any matrix in \(\displaystyle \text{Mat}_2 ( \mathbb{R} )\)Then we have:

\(\displaystyle \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix} \begin{bmatrix} m_{11} & m_{12} \\ m_{21} & m_{22} \end{bmatrix}\)

\(\displaystyle = \begin{bmatrix} km_{11} & km_{12} \\ km_{21} & km_{22} \end{bmatrix}\) ... ... by the definition of matrix multiplicationand we also have:\(\displaystyle \begin{bmatrix} m_{11} & m_{12} \\ m_{21} & m_{22} \end{bmatrix} \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix}\)

\(\displaystyle = \begin{bmatrix} m_{11}k & m_{12}k \\ m_{21}k & m_{22}k \end{bmatrix} \) ... ... by the definition of matrix multiplication

\(\displaystyle = \begin{bmatrix} km_{11} & km_{12} \\ km_{21} & km_{22} \end{bmatrix}\) ... ... since multiplication in \(\displaystyle \mathbb{R}\) is commutative

Thus \(\displaystyle \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix}\) commutes with any matrix in \(\displaystyle \text{Mat}_2 ( \mathbb{R} )\)So \(\displaystyle \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix} \in Z ( \text{Mat}_2 ( \mathbb{R} ) )\)But \(\displaystyle \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix}\) was an arbitrary matrix in \(\displaystyle \eta (R)\) So then \(\displaystyle \eta (\mathbb{R}) \subseteq Z ( \text{Mat}_2 ( \mathbb{R} ) )\)

Can someone please critique my argument/analysis?
Note that I am having troubles with Deveno's second part to the exercise: namely

" ... ... show that $\eta(\Bbb R) = Z(A)$ ... ..."

but suspect my knowledge regarding linear algebra and matrix algebra in particular is perhaps letting me down here ...

Can someone please help?

Peter

 

[mathbalarka]

Yes, your map looks good. Indeed, the natural embedding $\Bbb{R} \hookrightarrow \text{Mat}_n(\Bbb R)$ comes from identifying the reals (or more generally any field element) with the corresponding scalar matrix. The computation of $f(\Bbb R) \subseteq Z(A)$ also looks correct to me.

Note that I am having troubles with Deveno's second part to the exercise

I have no idea, I don't do noncommutative rings :p Looking at Deveno's hint, however, it looks like that'd require trick.

 

[Deveno]

Here is a hint. By the way, to prove that the matrix $kI$ commutes with any matrix, one doesn't even need the matrix multiplication details:

$(kI)M = k(IM) = kM = k(MI) = M(kI)$ by (LA.5)

My hint is as follows:

First assume that $M$ is diagonal. Since $M \neq kI$, say that:

$M = \text{diag}(d_1,d_2,\dots,d_n)$

We must have for some $i < j$, that $d_i \neq d_j$. Consider $ME_{ij}$ and $E_{ij}M$. Show these cannot be equal.

if $M$ is not diagonal, it must have some off-diagonal non-zero entry, say $m_{ij}$.

Consider $ME_{ji}$ and $E_{ji}M$, in particular the $i,i$-th entry.

It may be helpful to recall that $BE_{ij}$ takes the $i$-th column of $B$ and moves it to column $j$, and zeroes out everything else, while $E_{ij}B$ takes the $j$-th row of $B$ and moves it to row $i$, making all other rows 0.