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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.

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    $\begingroup$ Do you really mean that (the cardinality of) $V$ is countably infinite? Don't you rather mean its dimension? $\endgroup$
    – Anne Bauval
    Oct 1 at 4:13
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    $\begingroup$ 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. $\endgroup$
    – Anne Bauval
    Oct 1 at 4:23
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    $\begingroup$ 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 $\phi+b\circ\phi\circ f$. $\endgroup$
    – coiso
    Oct 1 at 4:28

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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.

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