Rings and Modules HT24, Eisenstein's criterion

Suppose:

• $f \in \mathbb Z[t]$, primitive (i.e. content is $1$)
• $f(t) = a _ n t^n + a _ {n-1} t^{n-1} + \cdots + a _ 0$
• $p \in \mathbb Z$ prime
• $p \mid a _ i$, $\forall i, 0 \le i \le n-1$
• $p \not\mid a _ n$
• $p^2 \not\mid a _ 0$

Then:

• $f$ is irreducible in $\mathbb Z[t]$ and $\mathbb Q[t]$

Since $f$ is primitive (i.e. has content $1$), irreducibility in $\mathbb Z[t]$ and $\mathbb Q[t]$ are equivalent. Hence it suffices just to consider the $\mathbb Z[t]$ case.

Suppose for a contradiction that $f$ were reducible in $\mathbb Z[t]$, i.e. $\exists g, h \in \mathbb Z[t]$ such that $f = gh$ and they are not units. Consider the quotient map $\phi _ p : \mathbb Z[t] \to \mathbb F _ p [t]$ applied to this factorisation:

\[\phi_p(f) = \phi_p(g) \phi_p(h)\]

By assumption, $\phi _ p(f) = \bar a _ n t^n$ and since $\mathbb F _ p[t]$ is an integral domain we must have

\[\phi_p(g) = \overline b_k t^k, \phi_p (h) = \overline c_{n-k}t^{n -k}\]

where $k = \deg g$ and $\overline{b} _ k, \overline{c} _ {n-k}$ are some coefficients. But then the constant terms of $\phi _ p(g)$ and $\phi _ p(h)$ must both be divisible by $p$, and hence $a _ 0$ must be divisible by $p^2$, which contradicts our assumption, so no such factorisation exists.

(A similar proof works for any $R$ as a principal ideal domain and $p \in R$ a prime)

Suppose for a contradiction that $2^{1/3} \in \mathbb Q$. Then the polynomial $t^3 - 2$ would be reducible, but applying Eisenstein’s criterion with $p = 2$, we see that $t^3 - 2$ is not irreducible.

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For reference, Eisenstein’s condition states that if:

• $f \in \mathbb Z[t]$, primitive (i.e. content is $1$)
• $f(t) = a _ n t^n + a _ {n-1} t^{n-1} + \cdots + a _ 0$
• $p \in \mathbb Z$ prime
• $p \mid a _ i$, $\forall i, 0 \le i \le n-1$
• $p \not\mid a _ n$
• $p^2 \not\mid a _ 0$

then:

• $f$ is irreducible in $\mathbb Z[t]$ and $\mathbb Q[t]$

Here $a _ 3 = 1$, $a _ 2 = 0$, $a _ 1 = 0$ and $a _ 0 = 2$. Since $2 \mid a _ i$ for each $0 \le i < 3$, $2 \not\mid a _ 3$ and $4 \not\mid 2$, Eisenstein’s condition is satisfied.

Consider

\[q(x) = p(x + 3) = x^2 + 7x + 14\]

Then this polynomial satisfies Eisenstein’s criterion for the prime $7$, and since any factorisation of $q(x)$ would imply one for $p(x)$ (by undoing the substitution), it implies that $p(x)$ is irreducible.

Note that

\[1 + x + x^2 + \cdots + x^{p-1} = \frac{x^p - 1}{x-1}\]

Then applying the substitution $x \mapsto x + 1$ gives

\[\frac{(x + 1)^p - 1}{x} = {p \choose 1} + {p \choose 2} x + \cdots + {p \choose{p - 1} } x^{p-2} + x^{p-1}\]

which satisfies Eisenstein’s criterion using $p$ as the prime.