# Endomorphisms of a countably infinite vector space.

Let $$F$$ be a field, $$V$$ a countably infinite $$F$$-space. Let $$A=end_F(V)$$ be the endomorphism ring of $$V$$. Let $$I=\{\phi \in A:rank_F(\phi)<\infty \}$$. It is easy to show this is an ideal of $$A$$. However, the problem asks us to prove $$I$$ to be a maximal ideal.

The intuition is that if we take any function in $$I^c$$, and include it into $$I$$, we should be able to generate the identity. However, I can't see why this would possibly be true. For example, if we let $$\phi$$ be such that it kills any even indexed element of a vector, i.e. $$\phi(1,0,1,0,1,\cdots )=(0,0,0,\cdots)$$ Then, the dimension of both its kernel and its image are infinite. So, I can't think of any way to compose this with a function in $$I$$ to give us the identity. If we trust the problem statement, then there must be, but in this case, and in the general case, I'm not sure I trust the problem statement.

• Do you really mean that (the cardinality of) $V$ is countably infinite? Don't you rather mean its dimension? Oct 1 at 4:13
• As for your definition of $\phi$, you probably rather mean: $V$ is identified with the space of sequences in $F$ with only a finite number of non-zero terms, and if a sequence has only one non-zero terms then its image by $\phi$ is either itself or $0,$ depending on the parity of the index of that term. Oct 1 at 4:23
• Note $(1,0,1,0,\cdots)\not\in V$, adopting Anne's interpretation. I assume you have $$\phi(a_0,a_1,a_2,a_3,\cdots)=(0,a_1,0,a_3,\cdots).$$ If you define forward and backwards shift operators $$f(a_0,a_1,a_2,\cdots)=(0,a_0,a_1,\cdots), \quad b(a_0,a_1,a_2,\cdots)=(a_1,a_2,a_3,\cdots)$$ then the identity endomorphism is $\phi+b\circ\phi\circ f$. Oct 1 at 4:28

I assume you mean that $$V$$ is countable-dimensional. Using elements of $$I$$ is actually a red herring; when generating a two-sided ideal you get to multiply on the left and the right by elements of $$A$$ and that's what actually does the job here. In other words it is actually already true that if $$\phi : V \to V$$ is any endomorphism of infinite rank then the two-sided ideal $$(\phi)$$ generated by $$\phi$$ is all of $$A$$, and this suffices.
To see this we can argue as follows. Pick any complement of $$\text{ker}(\phi)$$ in $$V$$ and let $$f : V \to V$$ be any injection with image exactly this complement (which exists because, if $$f$$ has infinite rank, its kernel must have infinite codimension); then $$\phi \circ f : V \to V$$ is injective. Similarly, let $$g : V \to V$$ be any surjection which restricts to an isomorphism from $$\text{im}(\phi)$$ to $$V$$ (which exists because $$\text{im}(\phi)$$ is countable-dimensional and so has the same dimension as $$V$$); then $$g \circ \phi$$ is surjective. Combining, we get that $$g \circ \phi \circ f$$ is an automorphism, and so after composing one more time on either side with its inverse, we get the identity.