Notes - Rings and Modules HT24, Factorisation in polynomial rings

\[c(f) = \text{hcf}(a_0, \cdots, a_d) > 0\]

It has content $1$.

Suppose:

• $f(t), g(t) \in \mathbb Q[t]$

Then:

\[c(fg) = c(f) \cdot c(g)\]

Wlog, we can assume $f$ and $g$ are primitive, i.e. that

\[c(f) = c(g) = 1\]

since otherwise we could just take out a factor from each. Hence we want to show that if $f$ and $g$ are primitive, then $fg$ is also primitive.

Suppose:

• $f(t) = a _ 0 + a _ 1 t + \cdots + a _ n t^n$
• $g(t) = b _ 0 + b _ 1 t + \cdots + b _ m t^m$

Then

\[f(t)g(t) = c_0 + c_1 t + \cdots + c_{n + m} t^{n+m}\]

where

\[c_k = \sum_{i + j = k} a_i b_j\]

Suppose for a contradiction that $c(fg) \ne 1$, i.e. $\exists p$ prime such that $\forall i$, $p \mid c _ i$. By assumption $p$ does not divide all $a _ i$ or all $b _ j$ (inclusive).

Let $\ell$ be the smallest index with $p \not\mid a _ \ell$ and $s$ the corresponding smallest index with $p \not\mid b _ s$.

Then

\[c_{\ell + s} = a_\ell b_s + \sum_{i + j = \ell + s, (i, j) \ne (\ell, s)} a_i b_j\]

(here we have just removed the $a _ \ell b _ s$ term from the sum). For each term in the sum, we must have either $i < \ell$ or $j < s$ (exclusive) since the indices have to sum up to $\ell + s$. But this is a contradiction:

• Each $a _ i b _ j$ is divisible by $p$, since $i < \ell$ or $j < \ell$
• $c _ {\ell + s}$ is divisible by $p$ by assumption
• But $a _ \ell b _ s$ is not, which cannot happen

Hence there exists no such $p$.

The unique $\alpha$ such that

\[f = \alpha f_1\]

and $f _ 1$ is primitive in $\mathbb Z[t]$.

Suppose:

• $f(t) \in \mathbb Z[t] \setminus \{0\}$
• $f(t)$ factors in $\mathbb Q[t]$.

Then $f(t)$ factors in $\mathbb Z[t]$ also.

Suppose:

\[f(t) = g(t) \cdot h(t)\]

where $g(t), h(t) \in \mathbb Q[t]$. Then write

\[g = c(g) g_1\] \[h = c(h) h_1\]

where $c(g), c(h) \in \mathbb Q$ and $g _ 1, h _ 1$ are irreducible polynomials in $\mathbb Z[t]$.

Then

\[\begin{aligned} f &= g h \\\\ &= c(g) c(h) g_1 h_1 \\\\ &= c(gh) g_1 h_1 \\\\ &= c(f) g_1 h_1 \\\\ &= (c(f) g_1) h_1 \end{aligned}\]

gives a factorisation into elements of $\mathbb Z[t]$ as required (since $c(f) \in \mathbb Z$).

\[f(t) \text{ irreducible in } \mathbb Z[t] \iff f(t) \text{ irreducible in } \mathbb Q[t] \text{ and } f(t) \text{ is primitive}\]

(primitive means $c(f) = 1$)

Suppose:

• $f(t) \in \mathbb Z[t]$
• $f(t) = g(t) h(t)$, factorisation in $\mathbb Z[t]$.
• $p$ prime such that $0 \not\equiv g(t) \mod p$ and $0 \not\equiv h(t) \mod p$

Then

\[\bar f(t) = \bar g(t) \bar h(t)\]

gives a factorisation in $\mathbb F _ p[t]$.

This says if a polynomial is reducible in $\mathbb Z[t]$ and the nonzero condition is met, it is reducible in $\mathbb F _ p[t]$. The contrapositive then gives a useful criterion for irreducibility: if a polynomial is irreducible in $\mathbb F _ p[t]$, then either it is irreducible in $\mathbb Z[t]$, the nonzero condition is not met (or both).

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)

• If $f \in \mathbb Q[t]$ has $c(f) = 1$, then $f$ must lie in $\mathbb Z[t]$ (by definition).
• We want to show that if $g, h \in \mathbb Z[t]$, then if $f \mid gh$, $f \mid g$ or $f \mid h$ (where divisibility is considered in $\mathbb Z[t]$).
• If $f \mid gh$ in $\mathbb Z[t]$, then it does so in $\mathbb Q[t]$. Since $\mathbb Q[t]$ is a PID, the notion of irreducible is the same as being prime, so either $f \mid g$ or $f \mid h$ in $\mathbb Q[t]$. Wlog assume $f \mid g$ in $\mathbb Q[t]$.
• Then $\exists k \in \mathbb Q[t]$ such that $g = fk$. Writing $k = c(k) \cdot k’$ where $k’ \in \mathbb Z[t]$, we have by Gauss’ lemma that $c(g) = c(f) c(k) = c(k)$ since $c(f) = 1$.
• But $g \in \mathbb Z[t]$ (by assumption), hence $c(g) = c(k) \in \mathbb Z$. Since $c(k) \in \mathbb Z$, then $k \in \mathbb Z[t]$, so $f \mid g$ in $\mathbb Z[t]$.

\[f(t) \text{ irreducible in } \mathbb Z[t] \iff f(t) \text{ irreducible in } \mathbb Q[t] \text{ and } f(t) \text{ is primitive}\]

$\mathbb Z[t]$ is an integral domain (since $\mathbb Z$ is), so using the lemma that says “$R$ is a UFD $\iff$ $R$ is an integral domain and every non-zero non-unit $a \in R$ can be written as a product of prime elements”, we need to show every element $f \in \mathbb Z[t]$ is a product of primes.

If $f \in \mathbb Z[t]$, then we can write $f = a \cdot f’$ where $c(f’) = 1$ and $a \in \mathbb Z$. Since $\mathbb Z$ is a UFD, this can be factorised into a product of primes.

Hence we can assume that $c(f) = 1$ (now instead working with $f’$). Viewing $f’$ as an element of $\mathbb Q[t]$ (which we know is a PID as $\mathbb Q$ is a field, and hence a UFD), we have a factorisation in $\mathbb Q[t]$ as $f = p _ 1 \cdots p _ k$, where each $p _ i$ is prime in $\mathbb Q[t]$.

Then each $p _ i$ can be written $p _ i = a _ i q _ i$ where $a _ i \in \mathbb Q$ and $q _ i \in \mathbb Z[t]$, and $c(q _ i) = 1$.

Then $q _ i$ is a prime element of $\mathbb Z[t]$ (there is a lemma that says $f$ is irreducible in $\mathbb Q[t]$ and $c(f) = 1$, then $f$ is not just irreducible in $\mathbb Z[t]$, but also prime in $\mathbb Z[t]$).

Hence $f = (a _ 1 \cdots a _ k) q _ 1 \cdots q _ k$, and since $c(f) = 1$ and $c(q _ i) = 1$, we can conclude $a _ 1 \cdots a _ k = 1$ (since content is multiplicative), so we are done.

Note that it has no roots in $\mathbb Z _ 2$.

If $f$ has no roots in $\mathbb F$, then it is irreducible. This is because if it was reducible, say into a quadratic term and a linear term, the linear term would imply the existence of a root.

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.