One-sided ideals in the algebra of endomorphisms of an infinite dimensional vector space $\newcommand\End{\operatorname{End}}$Let $V$ be an infinite-dimensional vector space over some field. It is well known that for each infinite cardinal $\kappa$ such that $\kappa<\dim V$ the subset of $\End(V)$ of all those linear maps $V\to V$ whose rank is strictly smaller than $\kappa$ is a non-zero and proper bilateral ideal in $\End(V)$ and that, moreover, we in this way obtain all non-zero proper bilateral ideals in that algebra.

What are the left ideals and the right ideals in that algebra?

For comparison, if $V$ is finite-dimensional, then the left ideals in $\End(V)$ are the  sets of the form $\{f\in\End(V):U\subseteq\ker f\}$ for some subspace $U$ of $V$, and the right ideals are the sets of the form $\{f\in\End(V):f(V)\subseteq W\}$ for some subspace $W$ of $V$. These two constructions work in the infinite-dimensional case too, of course, but surely do not exhaust the one-sided ideals — the bilateral ideals described above do not arise in this way, for example.
This is one thing that Jacobson left out in the second volume of his Basic Algebra, so the answer is probably not basic….
 A: $\newcommand{\L}{\mathcal{L}}$Call "linear filter" on $V$ a set $\L$ of subspaces of $V$ such that (a) $\{0\}\in\L$, (b) $W\in\L$, $W'\subset W$ implies $W'\in\L$, and (c) $W_1,W_2\in\L$ implies $W_1+W_2\in\L$.
Let $V,W$ be vector spaces (over a given field) and $\L$ a linear filter on $W$. Define $I_\L=\{f\in\mathrm{Hom}(V,W):f(V)\in\L\}$.
As suggested by Tom Goodwillie in a comment, it is natural to enlarge the setup to $\mathrm{Hom}(V,W)$.
Proposition:  Each such $I_\L$ is a $\mathrm{End}(V)$-submodule of $\mathrm{Hom}(V,W)$. Conversely, supposing $\dim(V)=\infty$, each $\mathrm{End}(V)$-submodule of $\mathrm{Hom}(V,W)$ has this form.
In particular, each right ideal of $\mathrm{End}(V)$ has this form.
[The assumption on infinite dimension is probably unnecessary, but this is not needed for the latter consequence, since the finite-dimensional case is standard.]
Example: consider subspaces $W_n$ with $W_0\subset W_1\subset\dots$: the set of homomorphisms whose image is contained in some $W_n$ has the given form. But even if $W$ has infinite countable dimension, not every linear filter has this form (if $\L$ is maximal among those linear filters on $W$ consisting of subspaces of infinite codimension, then it does not have this form).
Proof of the proposition. Let $J$ be a $\mathrm{End}(W)$-submodule of $\mathrm{Hom}(V,W)$. Define $\L=\{f(V):f\in J\}$. Clearly $\L$ satisfies (a) and (b).
To check (c), call "linear moiety" a subspace whose dimension equals its codimension. (I'm thus assuming $V$ has infinite dimension). I claim that (1) if $f\in J$ then there exists $f'\in J$ such that $\mathrm{Ker}(f')$ contains a moiety and $f'(V)=f(V)$, and (2) same but starting from a given moiety $M_0$, for every $f$ we can choose $f'$ such that $M_0\subset\mathrm{Ker}(f')$.
Actually passing from (1) to (2) is clear by precomposing by a linear automorphism mapping the given moiety to $M_0$. Fix a decomposition $V=M_1\oplus M_2\oplus M_3$. Let $p_i$ be the corresponding projections $V\to M_i$.
To prove (1), write $f_1=fp_1$ and $f_{23}=f(p_2+p_3)$. Both belong to $J$, and $f(V)=f_1(V)+f_{23}(V)$. Let $u$ be an automorphism mapping $M_1\oplus M_3$ into $M_1$. Then $g=f_{23}u$ also belongs to $M$ and vanishes on $M_1\oplus M_3$. Then $(f_1+g)(V)=f(V)$ (we use that $f_1$ vanishes on $M_2\oplus M_3$ while $g$ vanishes on $M_1\oplus M_3$). And $f_1+g$ vanishes on $M_3$.
Next we write $M'_2=M_2\oplus M_3$.
Now prove (c): given $W_1,W_2$, choose $f_1,f_2$ vanishing on $M'_2$, resp. $M_1$ with $f_i(V)=W_i$. Then $f_1+f_2$ has image $W_1+W_2$.
We have $J\subset I_\L$. Let us check this is an equality: let $W$ be an element of $\L$. So there exists $f$ vanishing on $M'_2$ such that $f(V)=W$. Let $g$ be an element of $\mathrm{Hom}(V,W)$ such that $g(V)=W$. If $g$ vanishes on a moiety, then there exists an automorphism $u$ such that $g=fu$ (just lift a basis of $W$ and complete) and hence $g\in J$. In general, write $g=g_1+g_2$ with each $g_i$ vanishing on a moiety and then $g_1,g_2\in J$ by the previous argument, so $g\in J$.

Here's not the analogue for left ideals.
Dually, call "linear cofilter" on $V$ a set $\L$ of subspaces of $V$ such that (a') $V\in\L$, (b') $W\in\L$, $W'\supset W$ implies $W'\in\L$, and (c') $W_1,W_2\in\L$ implies $W_1\cap W_2\in\L$.
Let $V,W$ be vector spaces (over a given field) and $\L$ a linear filter on $V$. Define $J_\L=\{f\in\mathrm{Hom}(V,W):\operatorname{Ker}f\in\L\}$.
Proposition:  Each such $I_\L$ is a $\mathrm{End}(W)$-submodule of $\mathrm{Hom}(V,W)$. Conversely, supposing $\dim(W)=\infty$, each $\mathrm{End}(W)$-submodule of $\mathrm{Hom}(V,W)$ has this form.
Indeed, let $I$ be a $\mathrm{End}(W)$-submodule of $\mathrm{Hom}(V,W)$. Define $\L=\{\operatorname{Ker}f:f\in I\}$. Clearly $\L$ satisfies (a'),(b').
Similarly as the previous case, one has to show that for every moiety $M$ of $W$ and $K\in\L$, there exists $f\in I$ such that $\operatorname{Ker}f=K$ and $f(V)\subset M$. This is dual to the statement (2) above and the verification is similar. One deduces (c') immediately.
Finally write $W=M\oplus M'$ (sum of moieties); if $K\in\L$, choose $f\in I$ with $\operatorname{Ker}f=K$ and $f(V)\subset M$. Let $g\in\mathrm{Hom}(V,W)$ be such that $\operatorname{Ker}g=K$. If $g(V)$ is contained in a moiety, then $g=uf$ for some automorphism $u$ of $W$, so $g\in I$. Otherwise write $g=g_1+g_2$ with both $g_1,g_2$ vanishing on $K$ and each being valued in a moiety. Then each $g_i$ belongs to $I$ by the previous argument, so $g\in I$. Hence $I=I_\L$.

Note: to completely claim a classification, one should clarify when $I_{\L_1}=I_{\L_2}$ for linear filters on $W$ (this might depend on $V$), resp. when $J_{\L_1}=J_{\L_2}$ for linear cofilters on $V$ (this might depend on $W$).
For instance if $\bar{\mathcal{L}}=\{K:K\subset\bigcup\mathcal{L}\}$ then for $V$ finite-dimensional we have $I_\mathcal{L}=I_{\bar{\mathcal{L}}}$. Similarly, if $W$ has uncountable dimension and $V$ has countable dimension, $\mathcal{L}\mapsto I_\mathcal{L}$ is not injective.
However if $\dim(V)\ge\dim(W)$, then $I=I_{\bar{\mathcal{L}}}$ determines $\mathcal{L}$ as the set of images of elements of $I$. To get injectivity when $\dim(V)<\dim(W)$, one can thus restrict to those linear filters consisting of subspaces of dimension $\le\dim(V)$ if $\dim(V)$ is infinite, and of subspaces of finite dimension if $\dim(V)$ is finite.
