## § Eisenstein Theorem for checking irreducibility

- Let $p(x) = a_0 + a_1 x + \dots + a_n x^n$
- If $p$ divides all coefficients except for the highest one ( $a_n$), $a_0$ is $p$-squarefree ( $p^2$ does not divide $a_0$), then $p(x)$ is irreducible.
- That is, $p | a0, p | a_1$, upto $p | a_{n-1}$, $p \not | a_n$, and finally $p^2 \not | a_0$.
- Then we must show that $p(x)$ is irreducible.
- Suppose for contradiction that $p(x) = q(x)r(x)$ where $q(x) = (b_0 + b_1 x+ \dots + b_k x^k)$ and $r(x) = (c_0 + c_1 x + \dots c_l x^l)$ (such that $k + l \geq n$, and $k > 0, l > 0$).
- See that $a_0 = b_0 c_0$. Since $p | a_0$, $p$ must divide one of $b_0, c_0$. Since $p^2$
*does not divide * $a_0$, $p$ cannot divide *both * $b_0, c_0$. WLOG, suppose $p$ divides $b_0$, and $p$ *does not divide * $c_0$. - Also see that since $a_n = (\sum_{i + j = n} b_i c_j)$, $p$ does not divide this coefficient $\sum_{i + j = n} b_i c_j$. Thus, at least one term in $\sum_{i + j = n} b_i c_j$ is not divisible by $p$.
- Now, we know that $p$ divides $b_0$, $p$ does not divide $c_0$. We will use this as a "domino" to show that $p$ divides $b_1$, $b_2$, and so on, all the way upto $b_k$. But this will imply that the final term $a_n$ will also be divisible by $p$, leading to contradiction.
- To show the domino effect, start with the coefficient of $x$, which is $a_1 = b_0 c_1 + b_1 c_0$. Since $a_1$ is divisible by $p$, $b_0$ is divisible by $p$, and $c_0$ is
*not * divisible by $p$, the whole equation reduces to $b_1 c_0 \equiv_p 0$, or $b_1 \equiv_p 0$ [since $c_0$ is a unit modulo $p$]. - Thus, we have now "domino"'d to show that $p$ divides
*both * $b_0, b_1$. - For induction, suppose $p$ divides everything $b_0, b_1, \dots, b_r$. We must show that $p$ divides $b_{r+1}$.
- Consider the coefficient of the term $xri$, ie $a_r$. This is divisible by $p$, and we have that $a_r = b_0 c_r + b_1 c_{r-1} + \dots + b_r c_0$. Modulo $p$, the left hand side vanishes (as $a_r$ is divisible by $p$), and every term $b_0, b_1, \dots, b_{r-1}$ vanishes, leaving behind $0 \equiv_p b_r c_0$. Since $c_0$ is a unit, we get $b_r \equiv_p 0$.
- Thus, every term $\{ b_i \}$ is divisible by $p$, implying $a_n$ is divisible by $p$, leading to contradiction.
- Again, the key idea: (1) $b_0$ is divisible by $p$ while $c_0$ is not. (This uses $p | a_0$ and $p^2 \not | a_0$). (2) This allows us to "domino" and show that all $b_i$ are divisible by $p$ (This uses $p | a_i$). (3) This show that $a_n$ is divisible by $p$, a contradiction. (This uses $p \not | a_n$).