Is the Endomorphism Ring of a Finite Dimensional Vector Space Dedekind Finite?

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    2015
In summary, a Dedekind finite endomorphism ring is a ring that has no proper, non-zero two-sided ideals and for any element in the ring, there exists a finite sequence of elements that can be multiplied together to yield that element. The endomorphism ring of a finite dimensional vector space is Dedekind finite if and only if the vector space is finite dimensional over a field, and the ring is the ring of linear transformations on the vector space. An infinite dimensional vector space cannot have a Dedekind finite endomorphism ring. Some examples of Dedekind finite endomorphism rings include the ring of linear transformations on a finite dimensional vector space over a field, the ring of endomorphisms on a finite ab
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Euge
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Here is this week's POTW:

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Prove that if $V$ is a finite dimensional vector space over field $\Bbb k$, then the endomorphism ring $\text{End}_{\Bbb k}(V)$ is Dedekind finite.

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This week's problem was correctly solved by Opalg. Here is his solution:

Choosing a (finite) basis for $V$, we can associate a matrix, and hence a determinant, with each element of $\mathrm{End}_{\mathbb k}(V).$

Suppose that $S,T \in \mathrm{End}_{\mathbb k}(V)$, and $ST = \mathrm{id}$. Then $\det(S)\det(T) = 1$ and so $\det(S) \ne0$. Therefore $S$ is invertible. It follows that $T = (S^{-1}S)T = S^{-1}(ST) = S^{-1}$. Hence $TS = S^{-1}S = \mathrm{id}$.

Here is also my solution, which uses a non-determinantal argument:

Let $n = \mathrm{dim}_{\Bbb k}(V)$. The ring $\rm{End}_{\Bbb k}(V)$ is identified with the matrix ring $M_n(\Bbb k)$ via the isomorphism $\Lambda : M_n(\Bbb k)\to \rm{End}_{\Bbb k}(V)$, $A\xrightarrow{\Lambda} (x \mapsto Ax)$. So it suffices to show $M_n(\Bbb k)$ is Dedekind finite. Let $A, B\in M_n(\Bbb k)$ such that $AB = I$. Then $BA = I$ if and only if $A$ is invertible. By way of contradiction, suppose $A$ is not invertible. Then there is a sequence $E_1,\ldots, E_s$ of elementary matrices such that $E_1\cdots E_s A$ has a row of zeros. Then $E_1\cdots E_s = E_1\cdots E_sAB$ has a row of zeros. This contradicts invertibility of $E_1\cdots E_s$. Thus $BA = I$ and consequently $M_n(\Bbb k)$ is Dedekind finite.
 

FAQ: Is the Endomorphism Ring of a Finite Dimensional Vector Space Dedekind Finite?

What is the definition of a Dedekind finite endomorphism ring?

A Dedekind finite endomorphism ring is a ring that satisfies the following two conditions: (1) it has no proper, non-zero two-sided ideals, and (2) for any element in the ring, there exists a finite sequence of elements that can be multiplied together to yield that element.

How is the endomorphism ring of a finite dimensional vector space related to Dedekind finiteness?

The endomorphism ring of a finite dimensional vector space is Dedekind finite if and only if the vector space is finite dimensional over a field, and the ring is the ring of linear transformations on the vector space.

Can an infinite dimensional vector space have a Dedekind finite endomorphism ring?

No, an infinite dimensional vector space will always have an infinite dimensional endomorphism ring, and therefore cannot be Dedekind finite.

What are some examples of Dedekind finite endomorphism rings?

Some examples of Dedekind finite endomorphism rings include: (1) the ring of linear transformations on a finite dimensional vector space over a field, (2) the ring of endomorphisms on a finite abelian group, and (3) the ring of endomorphisms on a finite set.

Are there any real-world applications of Dedekind finite endomorphism rings?

Yes, Dedekind finite endomorphism rings have applications in algebraic number theory, where they are used to study the arithmetic properties of number fields. They also have applications in cryptography, where they are used to construct secure cryptographic systems.

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