📂Abstract Algebra

Eisenstein's Criterion

Theorem¹

If $f(x) = a_{n} x^{n} + \cdots + a_{0 } \in \mathbb{Z} [ x ]$ satisfies the following conditions for a prime $p \in \mathbb{Z}$ and $k = 0,1,2, \cdots , n-1$, then $f(x)$ is an irreducible function over $\mathbb{Q}$.

• (i): $a_{n} \not\equiv 0 \pmod{p}$
• (ii): $a_{k} \equiv 0 \pmod{p} $
• (iii): $a_{0} \not\equiv 0 \pmod{p^2}$

Explanation

It is meaningful as a very easy criterion for integer polynomials of the form $f(x) = ax^{n} + b$. Being a criterion over $\mathbb{Q}$, it can be usefully employed in discussions related to algebraic numbers.

Example

Show that $f(x) = 25 x^{5} - 9 x^4 - 3 x^2 - 12$ is an irreducible function over $\mathbb{Q}$.

Solution

• (i) $25 \not\equiv 0 \pmod{3}$
• (ii) $-9 \equiv -3 \equiv 0 \pmod{3}$
• (iii) $12 \not\equiv 0 \pmod{9}$

Applying Eisenstein’s criterion with $p=3$ shows that $f(x)$ is an irreducible function.

Proof

If we suppose that $f(x) \in \mathbb{Z} [ x ]$ satisfies $f(x) = R(x) S(x)$ for some $R(x)$ of degree $r<n$ and $S(x)$ of degree $s<n$, then we can see that $$ R(x) , S(x) \in \mathbb{Q} [ x ] \iff R(x) , S(x) \in \mathbb{Z} [ x ] $$ Assume that, in $Z[x]$, $$ f(x) = (b_{r} x^{r} + \cdots + b_{0}) (c_{s} x^{s} + \cdots + c_{0}) $$ satisfies the three conditions (i), (ii), and (iii). Then by condition (iii), $$ b_{0} c_{0 } = a_{0} \not\equiv 0 \pmod{p^2} $$ so $b_{0}$ and $c_{0}$ cannot simultaneously satisfy $$ b_{0} \equiv c_{0} \equiv p \equiv 0 \pmod{p} $$ Instead, considering the case where only one of them is congruent, namely $$\begin{cases} b_{0} \not\equiv 0 \pmod{p} \\ c_{0} \equiv 0 \pmod{p} \end{cases}$$ by condition (i), $$ b_{r} c_{s} = a_{n} \not\equiv 0 \pmod{p} $$ and therefore we must have $$ \begin{cases} b_{r} \not\equiv 0 \pmod{p} \qquad \cdots (\star) \\ c_{s} \not\equiv 0 \pmod{p} \end{cases} $$

Now, let $m$ be the smallest value among the $k$ satisfying $c_{k} \not\equiv 0 \pmod{p}$. Then $$ a_{m} = b_{0 } c_{m} + b_{1} c_{m-1} + \cdots + \begin{cases} b_{m} c_{0} & , r \ge m \\ b_{r} c_{m-r} & , r<m \end{cases} $$ By $(\star)$, $b_{0} \not\equiv 0 \pmod{p}$, and by the definition of $m$, $c_{m} \not\equiv 0 \pmod{p}$, so $$ c_{m-1} \equiv \cdots \equiv c_{0} \equiv 0 \pmod{p} $$ Therefore $$ a_{m} \not\equiv 0 \pmod{p} $$ and, according to conditions (i) and (ii), we must have $m=n$.

In the end, $s \ge m = n$, which contradicts the premise that $s < n$.

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