A polynomial $\phi$ of degree $d$ is a vector of $d+1$ coefficients:
\begin{align} \phi &= [\phi_0, \phi_1, \phi_2, \dots, \phi_d] \end{align}
For example, $\phi = [1, 10, 9]$ is a degree 2 polynomial. Also, $\phi’ = [1, 10, 9, 0, 0, 0]$ is also a degree 2 polynomial, since the zero coefficients at the end do not count. But $\phi’’ = [1, 10, 9, 0, 0, 0, 1]$ is a degree 6 polynomial, since the last nonzero coefficient is $\phi_6 = 3$.
“A list of numbers? That makes no sense!”
Don’t panic!
You are probably more familiar to polynomials expressed as function of a variable $X$:
\begin{align}
\phi(X) &= \phi_0 + \phi_1\cdot X + \phi_2\cdot X^2 + \cdots + \phi_d\cdot X^d]\\
&= \sum_{i=0}^{d+1} \phi_i X^i
\end{align}
For example, $\phi = [1, 10, 9]$ and $\phi(X) = 9X^2 + 10X + 1$ are one and the same thing.
Note: The degree is defined as the index $i$ of the last nonzero coefficient: $\deg(\phi)=i$ s.t. $\forall j > i, \phi_j = 0$.
The basics of polynomials
Roots of polynomials
We say $z$ is a root of $\phi(X)$ if $\phi(z) = 0$. In this case, $\exists q(X)$ such that $\phi(X) = q(X)(Xz)$.
But what if $z$ is also a root $q(X)$? We can capture this notion as follows: we say $z$ has a multiplicity $k$ if $\exists q’(X)$ such that $\phi(X) = q’(X) (Xz)^k$.
The polynomial remainder theorem
This theorem says that:
\begin{align} \phi(a) = y\Leftrightarrow \exists q(X), \phi(X) &= q(X)(Xa) + \phi(a) \end{align}
This property is leveraged by certain cryptosystems^{1}.
Dividing polynomials
Division of polynomials conceptually resembles division of integers.
Specifically, dividing a polynomial $a(X)$ by $b(X)$ gives a quotient $q(X)$ and a remainder $r(X)$ such that:
\[a(X) = q(X) b(X) + r(X)\]Importantly, $\deg{r} < \deg{b}$ and, if $\deg{a} \ge \deg{b}$, then $\deg{q} = \deg{a}  \deg{b}$. Otherwise, $q(X) = 0$.
Lagrange interpolation
Given $n$ pairs $(x_i, y_i)_{i\in[n]}$, one can compute or interpolate a degree $\le n1$ polynomial $\phi(X)$ such that: \(\phi(x_i)=y_i,\forall i\in[n]\)
Specifically, the Lagrange interpolation formula says that: \begin{align} \phi(X) &= \sum_{i\in[n]} y_i \cdot \lagr_i(X),\ \text{where}\ \lagr_i(X) = \prod_{j\in[n],j\ne i} \frac{Xx_j}{x_ix_j} \end{align}
This formula is intimidating at first, but there’s a very simple intuition behind it. The key idea is that $\lagr_i(X)$ is defined so that it has two properties:
 $\lagr_i(x_i) = 1,\forall i\in[n]$
 $\lagr_i(x_j) = 0,\forall j \in [n]\setminus{i}$
You can actually convince yourself that $\lagr_i(X)$ has these properties by plugging in $x_i$ and $x_j$ to see what happens.
Important: The $\lagr_i(X)$ polynomials are dependent on the set of $x_i$’s only (and thus on $n$)! Specifically each $\lagr_i(X)$ has degree $n1$ and has a root at each $x_j$ when $j\ne i$! In this sense, a better notation for them would be $\lagr_i^{[x_i, n]}(X)$ or $\lagr_i^{[n]}(X)$ to indicate this dependence.
Furthermore, consider the following example with $n=3$ pairs. Then, by the Lagrange formula, we have:
\[\phi(X) = y_1 \lagr_1(X) + y_2 \lagr_2(X) + y_3 \lagr_3(X)\]Next, by applying the two key properties of $\lagr_i(X)$ from above, you can easily check that $\phi(x_i) = y_i,\forall i\in[3]$:
\begin{align}
\phi(x_1) &= y_1 \lagr_1(x_1) + y_2 \lagr_2(x_1) + y_3 \lagr_3(x_1) = y_1 \cdot 1 + y_2 \cdot 0 + y_3 \cdot 0 = y_1\\
\phi(x_2) &= y_1 \lagr_1(x_2) + y_2 \lagr_2(x_2) + y_3 \lagr_3(x_2) = y_1 \cdot 0 + y_2 \cdot 1 + y_3 \cdot 0 = y_2\\
\phi(x_3) &= y_1 \lagr_1(x_3) + y_2 \lagr_2(x_3) + y_3 \lagr_3(x_3) = y_1 \cdot 0 + y_2 \cdot 0 + y_3 \cdot 1 = y_3
\end{align}
An important detail is that the degree of the interpolated $\phi(X)$ is $\le n1$ and not necessarily exactly equal to $n1$. To see this, consider interpolating the polynomial $\phi(X)$ such that $\phi(i) = i$ for all $i\in [n]$. In other words, $x_i = y_i = i$.
The inspired reader might notice that the polynomial $\phi(X) = X$ could satisfy our constraints.
But is this what the Lagrange interpolation will return?
After all, the interpolated $\phi(X)$ is a sum of degree $n1$ polynomials $\lagr_i(X)$, so could it have degree 1?
Well, it turns out, yes, because things cancel out.
To see, this take a simple example, with $n=3$:
\begin{align}
\phi(X) &=\sum_{i\in [3]} i \cdot \lagr_i(X) = \sum_{i\in [3]} i \cdot \prod_{j\in[3]\setminus{i}} \frac{X  j}{i  j}\\
&= 1\cdot \frac{X2}{12}\frac{X3}{13} + 2\cdot \frac{X1}{21}\frac{X3}{23} + 3\cdot\frac{X1}{31}\frac{X2}{32}\\
&= \frac{X2}{1}\frac{X3}{2} + 2\cdot \frac{X1}{1}\frac{X3}{1} + 3\cdot \frac{X1}{2}\frac{X2}{1}\\
&= \frac{1}{2}(X2)(X3)  2(X1)(X3) + \frac{3}{2}(X1)(X2)\\
&= \frac{1}{2}[(X2)(X3) + 3(X1)(X2)]  2(X1)(X3)\\
&= \frac{1}{2}[(X2)(4X6)]  2(X1)(X3)\\
&= (X2)(2X3)  2(X1)(X3)\\
&= (2X^2  4X  3X + 6)  2(X^2  4X +3)\\
&= (2X^2  7X + 6)  2X^2 + 8X  6\\
&= X
\end{align}

Evaluation proofs in KZG polynomial commitments leverage the polynomial remainder theorem. ↩