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).

• 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.