# A Minimal left ideals in Hassani

1. Mar 22, 2016

### Geofleur

In the chapter on Algebras in Hassani's mathematical physics text, left ideals are defined as follows:

Let $\mathcal{A}$ be an algebra. A subspace $\mathcal{B}$ of $\mathcal{A}$ is called a left ideal of $\mathcal{A}$ if it contains $\mathbf{a}\mathbf{b}$ for all $\mathbf{a}\in \mathcal{A}$ and $\mathbf{b} \in \mathcal{B}$.

He then defines a minimal left ideal:

A left ideal $\mathcal{M}$ of an algebra $\mathcal{A}$ is called minimal if every left ideal of $\mathcal{A}$ contained in $\mathcal{M}$ coincides with $\mathcal{M}$.

Here is where I am confused. The set containing only the zero vector is a subspace of any vector space, because $\alpha \mathbf{0} + \beta \mathbf{0} = \mathbf{0}$ for any scalars $\alpha$ and $\beta$. Moreover, the set containing the zero vector is a subalgebra of any algebra, because $\mathbf{0} \mathbf{0} = \mathbf{0}$. In fact, the "zero set" is a left ideal of any algebra, because $\mathbf{a} \mathbf{0} = \mathbf{0}$ for any $\mathbf{a} \in \mathcal{A}$. But then the only minimal left ideal is just the zero set, because every left ideal has the zero vector as an element. This conclusion would make the whole concept of minimal ideals rather uninteresting. Am I going wrong somewhere here?

2. Mar 23, 2016

### andrewkirk

@Geofleur, your logic appears to be flawless. The subalgebra {0} is indeed a left ideal of A that is contained in any other left ideal M.

I'm pretty sure that the author just forgot to insert the words 'non-trivial' or 'nonzero' before 'left ideal'. If we compare his/her definition here with that on Wikipedia for minimal ideals of rings, we see that the 'nonzero' requirement is stipulated.

I suggest proceeding on the assumption that the author meant 'nonzero' but forgot to specify that.

See also this wiki page on Simple Algebras, which refers to 'minimal nonzero left ideals' of an algebra, which suggests that things only become interesting when one considers ideals that properly contain no nonzero ideals.

3. Mar 23, 2016

### Geofleur

OK, I'm glad that I am not totally off-base. On the other hand, what happens next in the book, when I make the assumption that a minimal left ideal cannot be the zero set, has me thoroughly puzzled. There comes a theorem (I've bolded the troublesome part):

Let $\mathcal{A}$ and $\mathcal{B}$ be algebras, $\phi : \mathcal{A} \rightarrow {\mathcal{B}}$ an epimorphism, and $\mathcal{L}$ a minimal left ideal of $\mathcal{A}$. Then $\phi(\mathcal{L})$ is a minimal left ideal of $\mathcal{B}$. In particular, any automorphism of an algebra is an isomorphism among its minimal ideals.

The problem is with this last statement, together with the comment: "The last statement of the theorem follows from the fact that ker $\phi$ is an ideal of $\mathcal{A}$."

Now, an automorphism is defined as an isomorphism of an algebra onto itself. An isomorphism is, among other things, an injective linear map. But the kernal of an injective linear map must be the zero set, for suppose $\mathbf{a} \in$ ker $\phi$. Then $\phi(\mathbf{a}) = \mathbf{0} = \phi(\mathbf{0})$. Hence, $\mathbf{a}=\mathbf{0}$ because $\phi$ is injective. So Hassani seems to be saying to think of the zero set as an ideal. If he does not want to consider the zero set as a minimal left ideal, I don't see how the comment is relevant to proving the last statement of the theorem. If he does want to consider the zero set as a minimal left ideal, then the concept of minimal left ideal is rendered trivial. Any ideas?

4. Mar 23, 2016

### micromass

Staff Emeritus
Agreed. Although this statement:

is still true if we define a minimal ideal as nonzero. His comment however

Is completely irrelevant.

Anyway, welcome to Hassani. I tried to like the book, but it is filled with stuff like this. It's way better to learn from actual math books.

5. Mar 23, 2016

### Geofleur

Thanks so much!