Serre’s (modularity) conjecture, first made clear in their 1975 paper, Valeurs propres des opérateurs de Hecke modulo l, states the following:
Conjecture. (Serre) Let $\rho : \text{Gal} (\overline{ \mathbf Q}/\mathbf Q) \to \text{GL}_2 (\overline{\mathbf F_p})$ be a continuous, odd, irreducible Galois representation. Then there exists normalized eigenform $f \in S_{k (\rho)} (N (\rho), \epsilon (\rho); \overline{\mathbf F_p})$. with associated Galois representation $\rho_f$ such that $\rho_f \simeq \rho$. Furthermore, $N(\rho) $ and $k (\rho)$ are the minimial weight and level for which there exists such a form $f$.
The whole point of this log is to go through the nuts and bolts to really appreciate this conjecture (which is now a theorem!) and see some of its applications. In fact, a direct application is FLT, so this gives another proof.
Modular Forms
Let $\mathfrak h$ denote the upper half plane, i.e. $\mathfrak h = \{ z \in \mathbf C \colon \text{Im} (z) > 0 \}$. Take $z \in \mathfrak h$ and let $a,b,c,d \in \mathbf R$ with $ad -bc > 0$. Then $\frac{az + b}{cz + d} \in \mathfrak h$ as $\text{Im}\left(\frac{az+b}{cz+d}\right)=\frac{(ad-bc)y}{|cz+d|^2}$ with $ad-bc>0$ and $|cz+d|^2>0$, so we have $\text{Im}\left(\frac{az+b}{cz+d}\right)>0$, so $\frac{az+b}{cz+d}\in\mathfrak{h}.$ Thereby $\text{SL}_2 (\mathbf Z) = \Gamma (1) \curvearrowright \mathfrak h$ via $(\gamma, z) \mapsto \frac{az+ b}{cz+d}$ where $\gamma = \begin{pmatrix} a & b \\ c & d\end{pmatrix} $. Thereby we get an equivalence relation on $\mathfrak h$ via $z \sim z^\prime $ if there exists $\gamma \in \Gamma (1)$ such that $\gamma z = z^\prime$. We write $\Gamma(1)\backslash\mathfrak{h},$ to emphasize this equivalence relation, and the set $\mathfrak h/\Gamma(1)$ denotes the set of orbits of this action. Importantly, there is a bijection $j \colon \mathfrak h/\Gamma(1) \to \mathbf C$ and that $\mathfrak h /\Gamma(1)$ is isomorphic to a compact Riemann surface with one point missing. We can further extend the bijection $j$ to an isomorphism $j \colon \mathfrak h/\Gamma(1) \cup \{ \infty\} \to \mathbf P^1 (\mathbf C)$ of compact Riemann surfaces. To go into more detail, write $\mathfrak h^\ast = \mathfrak h \cup \mathbf P^1 (\mathbf Q)$, and note that $\Gamma(1) \curvearrowright \mathbf P^1 (\mathbf Q)$ via $$\left( \begin{pmatrix} a & b \\ c & d \end{pmatrix} , (x_0 \colon x_1) \right ) \mapsto (a x_0 + b x_1 \colon c x_0 + d x_1).$$ Defining $X(1) = \mathfrak h^\ast /\Gamma(1)$. Then $X(1) = \mathfrak h/\Gamma(1) \cup \{ \infty \}$ is a compact Riemann surface.
Note that $-I$, where $I$ is the identity of $\Gamma(1)$, acts trivially on the upper half-plane as $-I z = \frac{-z + 0}{0 -1} = z$, and the group $\text{PSL}_2 (\mathbf Z) := \text{SL}_2(\mathbf Z)/\{ \pm I \} = \Gamma(1)/\{ \pm I \}$ acts on $\mathfrak h$ in a natural way.
Definition. Let $k$ be a nonnegative integer. A holomorphic function $f$ on $\mathfrak h$ is a *weak modular form* of weight $k$ if $f (\gamma z ) = (cz + d)^k f(z) $ for all $z \in \mathfrak h$ where $\gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \Gamma(1)$ and $\gamma z = \frac{az + b} {cz + d}$.
If $k$ were odd, then for $\gamma = -I$ we would have $f(\gamma z) = f(z) = (-1)^k f(z) = - f(z)$, and so $f = 0$ . Therefore if $f \neq 0$, then $k$ must be even. So the weight of any nonzero weak modular form is even. In order to check that $f(z)$ is a weak modular form, it suffices to check that $f(z + 1) = f(z)$ and $f(\frac{-1}{z}) = z^k f(z)$ as $\Gamma(1)$ is generated by $T = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $S = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. As any weak modular form $f$ satisfies $f(z + 1) = f(z)$, so periodic of period $1$ as a function of its real part, we obtain a Fourier series. We can write $f(z) = \sum_{n \in \mathbf Z} a_n q^n$ where $q = e^{2\pi i z}$. If $a_n = 0$ for $n<0$ then $f$ is said to be holomorphic at infinity.
Definition. A weak modular form $f$ of weight $k$ is a modular form of weight $k$ if it is holomorphic at infinity. If, furthermore, we have that $a_0 = 0$ in its $q$-expansion, then $f$ is said to be a cusp form.
Let $k$ be a nonnegative even integer. We write $M_k (\Gamma(1))$ to denote the vector space of modular forms of weight $k$ for $\Gamma(1)$, and we write $S_k (\Gamma(1))$ to denote the space of cusp forms of weight $k$ over $\Gamma(1)$.
Lemma. Let $f$ be a modular form of weight $k$ and $g$ a modular form of weight $k^\prime$, both over $\Gamma(1)$. Then $fg$ is a modular form of weight $k+k^\prime$. Furthermore, $\bigoplus _{k = 0}^\infty M_k (\Gamma(1))$ is a graded algebra.
Proof. Write $h = fg$. Then $h( \gamma z) = f(\gamma z) g(\gamma z) = (cz + d)^k (cz + d)^{k ^\prime} f(z) g(z) = (cz + d)^{k + k^\prime} h(z)$. Now write $f(z) = \sum_{n \in \mathbf Z} a_n q^n $ and $g(z) = \sum_{n \in \mathbf Z } b_n q^n$. So, $h(z) = \sum_{n \in \mathbf Z} c_n q^n$ with $c_n = \sum_{m \in \mathbf Z} a_m b_{n-m} = \sum_{m \geq 0 } a_n b_{n-m} + \sum_{m < 0} a_n b_{n-m} $. If $n < 0$, then $ n-m < 0$, and as $f$ and $g$ are modular forms, we have that $a_n$ and $b_n$ both vanish, so $c_n = 0 + 0 = 0$. Therefore $h = fg$ is a modular form with weight $k + k^\prime$. We have that $\bigoplus _{k = 0}^\infty M_k (\Gamma(1))$ is a graded algebra as $f \in M_k (\Gamma(1))$ and $g \in M_{k^\prime} (\Gamma(1))$ implies $fg \in M_k(\Gamma(1))$, where addition is defined component wise and $M_0(\Gamma(1)) \cong \mathbf C$ acts as the scalars. The last claim follows from the following argument: Let $f \in M_0(\Gamma(1))$. Then $f(\gamma z) = f (z)$ for all $\gamma \in \Gamma (1)$, so $f$ is periodic under $T \colon z \mapsto z + 1$. Hence, $f(z) = F(q)$ where $q = e^{2 \pi iz}$, and holomorphic at the cusp implies that $F$ is holomorphic at $q = 0 $. Thereby $F$ is holomorphic on $\overline{\mathbf {D}}$. By maximum modulus principle, it must obtain its maximum in the interior, so $F$ (and hence $f$) must be constant.
Definition. An $n$-dimensional