Let's take $k$ to be a field, eg $$\mathbb{R}, \mathbb{C}, \mathbb{F}_p$$
Here is a more instructive example of a field to keep in mind.
Let $x$ be an indeterminant. Set
$$
\mathbb{F}_p(x) := \left\lbrace \frac{p(x)}{q(x)} \mid p(x), q(x) \text{ polynomials}
\right\rbrace
$$
So
$$
p(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_0,~ a_i \in \mathbb{F}_p
$$
We all remember from linear algebra that a $k$-vector space is completely determined
by a basis.
Precisely, any $v \in V$ can be written as
$$
v = \sum a_i e_i
$$
for unique $a_i \in k$ where $e_i$ are the basis vectors.
In particular, if $V$ and $W$ are two vector spaces each possessing bases with
the same number of elements we get an $k$-linear isomorphism
$$
A : V \overset{\sim}{\to} W
$$
A single number determines a (finite-dimensional) vector space.
To classify $k$-vector spaces, we really just need to understand the
vector space $k$. If $\dim V = n$, then
$$
V = \underbrace{k \oplus \cdots \oplus k}_n
$$
We can build everything from $k$ up to sums (and isomorphisms).
Let's enrich the structure a bit. Suppose instead of a single vector
space $V$, we consider pairs $(V,T)$ where $T : V \to V$ is a linear
map.
Is there is a single number that determines such pairs completely?
Are there basic building blocks?
There is natural universal object in algebra we can extract from such
pairs: the polynomial ring $k[x]$
$$
k[x] = \lbrace a_n x^n + \cdots + a_0 \mid a_i \in k \rbrace
$$
Given a polynomial $a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$, we
can plug in $T$ to obtain a linear map
$$
a_n T^n + \cdots + a_0 I : V \to V
$$
Conversely, any module $M$ over $k[x]$ is a just a pair a $k$-vector space
$M$ with an action of $x$ on $M$.
In particular, we can take $k[x]$ itself. Multiplication by $x$ is a $k$-linear.
The module $k[x]$ is our simplest building block.
But not everything is a sum of copies of $k[x]$'s.
Suppose that $p(T)$ is the minimal polynomial of $$T: V \to V.$$ Then we know that
multiplication by $p(x)$ is identically $0$.
Thus, the action of $k[x]$ factors through the quotient
$$
k[x] \to k[x]/(p(x))
$$
where we set polynomials of the form $q(x)p(x) = 0$.
But we still get our classification.
Any $(V,T)$ is isomorphic to
$$
k[x]^{\oplus a} \oplus \bigoplus \left(\frac{k[x]}{(q_i(x)^{e_i})}\right)^{\oplus b_i}
$$
where $q_i(x)$ are polynomials that cannot be factored.
Perhaps we need to allow for more building blocks?
No!!
We need more ways to build!
For example, each factor of the form $k[x]/(p(x))$ can be understood in terms of a
linear map between $k[x]$ and itself:
$$
k[x] \overset{p(x)}{\to} k[x]
$$
The cokernel of multiplication by $p(x)$ is exactly $k[x]/(p(x))$ and there is
no kernel.
We express this by decorating the map
$$
0 \to k[x] \overset{p(x)}{\to} k[x] \to k[x]/(p(x)) \to 0
$$
Then composing any pair of composable maps in the diagram will yield the zero
map.
Revisiting our classification: any finitely-generated $k[x]$-module is the cokernel of
a $k[x]$-linear map
$$
k[x]^{\oplus a} \overset{M}{\to} k[x]^{\oplus b}
$$
Here $M$ can be written a matrix with entries in $k[x]$.
How about two commuting linear endomorphisms of $V$?
We have to add one more indeterminant to our polynomial ring $k[x,y]$ and we
have to allow our exact complexes to be one arrow longer:
$$
0 \to F_{-2} \overset{d_{-1}}{\to} F_{-1} \overset{d_0}{\to} F_0 \overset{\phi}{\to} M \to 0
$$
where $F_i = k[x]^{n_i}$ and
$$
\phi \circ d_0 = d_0 \circ d_{-1} = d_{-1} \circ d_0 = 0
$$
And moreover $\operatorname{ker} d_i = \operatorname{im} d_{i-1}$.
Theorem(Hilbert). More generally, any finitely-generated $k[x_1,\ldots,x_n]$-module $M$ fits into an
exact complex of the form
$$
0 \to F_{-n} \to \cdots \to F_{-1} \to F_0 \to M \to 0
$$
The structure of these resolutions can be very complex but they are ultimately a
finite amount of linear algebraic data.
Suppose instead that we only cared about $$T : V \to V$$ satisfying $T^2 = 0$.
So
$k[x]/(x^2)$ are our basic building blocks with which we want to understand all
other modules.
In particular, let's try to understand the module $k$ where $x$ acts by $0$.
We could seemingly start with what we know
$$
0 \to k[x] \overset{x}{\to} k[x] \to k \to 0
$$
and just mod out by $x^2$ everywhere
$$
0 \to \frac{k[x]}{(x^2)} \overset{x}{\to} \frac{k[x]}{(x^2)} \to k \to 0
$$
but something has gone wrong.
We have lost exactness. The map
$$
\frac{k[x]}{(x^2)} \overset{x}{\to} \frac{k[x]}{(x^2)}
$$
is no longer injective.
Can we fix it?
The kernel is $kx \subset k[x]/(x^2)$. So we need surject onto that
$$
\frac{k[x]}{(x^2)} \overset{x}{\to}\frac{k[x]}{(x^2)} \overset{x}{\to} \frac{k[x]}{(x^2)}
$$
There is still kernel of this map.
The kernel is $kx \subset k[x]/(x^2)$. So we need surject onto that
$$
\frac{k[x]}{(x^2)} \overset{x}{\to}\frac{k[x]}{(x^2)} \overset{x}{\to} \frac{k[x]}{(x^2)}
$$
There is still kernel of this map.
The kernel is $kx \subset k[x]/(x^2)$. So we need surject onto that
$$
\frac{k[x]}{(x^2)} \overset{x}{\to}\frac{k[x]}{(x^2)} \overset{x}{\to} \frac{k[x]}{(x^2)}
$$
There is no finite resolution of $k$ as $k[x]/(x^2)$-module.
Things get much worse.
Try taking the resolution of $k$ as a $k[x,y,z]/(xy,yz)$-module.
Theorem(Auslander-Buchsbaum-Serre). Suppose $R = k[x_1,\ldots,x_n]/(f_1,\ldots,f_s)$.
Then $k$ has a finite resolution as an $R$-module if and only if there is a there is a non-vanshing minor
of size $\operatorname{codim} R$ in the matrix $(\partial f_i/\partial x_j)$.
So, for singular $R$, there is no finite amount of linear algebra data that builds
an $R$-module $M$ from $R$ in general.
When $p = 0$ in our field $k$, there is a magic map: the Frobenius map.
$$
x \mapsto x^p
$$
For example,
$$
(x+y)^p = x^p + \sum \binom{p}{j} x^jy^{p-j} + y^p \\
= x^p + y^p
$$
In characteristic $p$, given a $k$-vector space $V$, we get a new one $V^{(p)}$ by declaring that
$$
a \cdot v := a^p v
$$
This is not the same in general.
Pull back out the field $\mathbb{F}_p(x)$ from the start. Note that
$$
\mathbb{F}_p(x)^{(p)} \not \cong \mathbb{F}_p(x)
$$
but
$$
\mathbb{F}_p^{(p)} = \mathbb{F}_p
$$
For a ring $R$ with characteristic $p$, we say it is $F$-finite if $R^{(p)}$ is a
finitely-generated $R$-module.
For example, if $R = \mathbb{F}_p[x]$, then
$$
R^{(p)} = \mathbb{F}_p[x^p] \oplus x \mathbb{F}_p[x^p] \oplus \cdots \oplus
x^{p-1} \mathbb{F}_p[x^p]
$$
We've seen that if $R$ are our basic building blocks then we cannot construct
all modules using a finite amount of linear algebra data in general.
Can we do better with $R^{(p)}$?
Theorem(B.-Lank). For $F$-finite rings, yes.
Independently, Iyengar-Mukhopadhyay-Pollitz also.
Let's look at our problematic example: $$\mathbb{F}_p[x]/(x^2)$$
Then
$$
\mathbb{F}_p[x]/(x^2)^{(p)} \cong \mathbb{F}_p \oplus \mathbb{F}_p x
$$
and $x$ acting via multiplication of $x^p$ acts by $0$. So the above
is an isomorphism of modules.
We've got $\mathbb{F}_p$ as a summand of $R^{(p)}$ already.
In general, $R^{(p^e)}$ generates the derived category $D^b(\operatorname{mod} R)$
for any $F$-finite Noetherian ring $R$.
In other words, building finite complexes out of $R^{(p^e)}$ gives all modules.
(Indeed all complexes.) Moreover, there is a uniform bound on the length
of the complex necessary.
Using $R^{(p^e)}$ as our basic building blocks, everything is build from a finite
amount of linear algebraic data, restoring the nice situation for regular rings.
More generally, high enough Frobenius pushforwards of bounded complexes of
finite rank vector bundles generate $D^b(\operatorname{coh} X)$ for an $F$-finite
Noetherian scheme.
Thanks for your attention.
