Theorem For every prime $p$, there is a Sidon set $A \subset \{1,...,p^2-1\}$ with $|A| = p$.
A Sidon set, is a set of integers where every 2-sum $a_i+a_j$ is different.
Proof Let $\theta \in \mathbb{F}_{p^2}$ be a generator of the multiplicative group $\mathbb{F}_{p^2}$, so $$\mathbb{F}_{p^2} = \{0,\theta,...,\theta^{p^2 -1 } =1 \}$$ **First, what does this exactly mean? Why do the exponents range the residue class mod $p^2$?
We select $A$ from these exponents. Let $\mathbb{F}_p = \{c_1,...,c_p\}$ be the subfield of $\mathbb{F}_{p^2}$. Trivially, $\theta$ is not an element of $\mathbb{F}_p$, as it is a generator of a bigger group Why? so $\theta + c_j \ne 0$ (I'm assuming in $\mathbb{F}_{p^2}$ and why does this follow?
Let $A = \{a_1,...a_p\}$ where each $a_i$ is an exponent of $\theta$ in the finite field, such that $$\theta^{a_j} = \theta +c_j$$ How does this follow?
If we suppose that $a_i + a_j = a_k +a_l$, then $$\theta^{a_i + a_j} = (\theta +c_i)(\theta + c_j) = (\theta +c_k)(\theta + c_l) = \theta^{a_k + a_l}$$ $$\theta(c_i+c_j) + c_ic_j = \theta(c_k + c_l) + c_k c_l$$ Since $\theta$ is not in $\mathbb{F}_p$, it cannot be a solution of a linear equation with coefficients in $\mathbb{F}_p$, so $$c_i + c_j = c_k + c_l, c_ic_j = c_kc_l$$ If we let the two values be $b, d$, then these are solutions of the quadratic of the quadratic equation $X^2 - bX + d = 0$ so the two sets are the same and $\{a_i,a_j\} = \{a_k,a_l\}$.
I would appreciate some help understanding the critical parts! Also, now I have to show that the construction of $|A|$ can be extended by putting $p^2$ into the set, which I hope I can see after I fully understand the proof.
I hope I can answer all of these questions.
Ok, since $\Bbb{F}_{p^2}$ is a field, the multiplicative group (should be denoted something like $(\Bbb{F}_{p^2})^\star$), is exactly the non-zero elements of $\Bbb{F}_{p^2}$. Note that this group has order $(p^2-1)$. An element is a generator of this group precisely if when we raise exponents from $1$ to $(p^2-1)$ -(the order of the group), the whole group is generated.
Ok, if $\theta$ were an element of $\Bbb{F}_p$ then as this is a field, which means it will be closed under multiplication, then all of $\{\theta, \theta^2, ..., \theta^{(p^2-1)}\}$ are all elements in $\Bbb{F}_p$. BUT we picked these elements to be distinct, (they are the elements of $\Bbb{F}_{p^2}$) and $\Bbb{F}_p$ only has $p$ elements. So we cannot have $\theta \in \Bbb{F}_p$.
Then $\theta + c_j \ne 0$, and $\Bbb{F}_p$ is a subfield of $\Bbb{F}_{p^2}$ so the $0$ in $\Bbb{F}_p$ is the $0$ in $\Bbb{F}_{p^2}$. And this follows since if $\theta + c_j = 0 \implies \theta = -c_j \in \Bbb{F}_p$- BUT we just showed that $\theta \not \in \Bbb{F}_p$. So we cannot have $\theta +c_j = 0$ as the proof says.
This doesn't follow, we are letting this happen, this is a definition of $A$- that's all. We can do this precisely because we have shown that for any $j \in \{1, ...,p\}$, we have that $\theta + c_j \ne 0$, so these $\theta + c_j$ are non-zero elements of $\Bbb{F}_{p^2}$ and hence can be made with exponents $a_j$ of the generator $\theta$ we picked for $(\Bbb{F}_{p^2})^\star$.
I am hoping that this explains everything! I do not quite understand what you mean regarding the extension of $|A|$ by putting $p^2$ into the set.- However if you do need help with this after reading through this answer then try to explain to me exactly what you mean and I'll see what I can do :)