Notes - Rings and Modules HT24, Polynomial rings

\[x \mapsto q(x)\]

and for all other $f(x)$,

\[f(x) \mapsto f(q(x))\]

(this is because we require $a \mapsto a$, and then we can map $x$ to any polynomial in $\mathbb Z[x]$, and since $\mathbb Z[x] = \langle 1, x \rangle$, we’ve fully described the homomorphism on a generating set).

Suppose:

• $R$ is a ring
• $f(x) = \sum^n _ {i = 0} a _ i x^i \in R[x]$
• $a _ n \in R^\times$
• $g(x) \in R[x]$

Then $\exists q(x), r(x) \in R[x]$ such that:

• $f(x) = q(x) g(x) + r(x)$
• $r = 0$ or $\deg r < \deg g$

(so in particular, if $R$ is a field, then the division algorithm works for all polynomials)

Overall idea: If $\deg f < \deg g$ then just take $r = f$. Otherwise, multiply $g$ by a polynomial that means it has the same leading term as $f$, and then subtract this from $f$ to get a polynomial with smaller degree.

We proceed by induction on $\deg f$. Since $a _ n \in R^\times$, if $h \in R[x] \setminus \{0\}$, it follows that

\[\deg g(x) h(x) = \deg g(x) + \deg h(x)\]

since if this were not true, the leading terms would multiply to $0$, but this contradicts the fact $a _ n \in R^\times$.

Thus if $\deg f < \deg g$, we must take $q = 0$ and thus $r = f$.

Now suppose $\deg f = m \ge \deg g = n$. Then

\[f(x) = \sum^m_{j = 0} b_j x^j\]

where $b _ m \ne 0$. Then

\[a_n^{-1} b_m t^{m - n} g\]

has leading term $b _ m t^m$, the polynomial

\[h = f - a_n^{-1} b_m t^{m - n} g\]

has $\deg h < \deg f$. Then it follows by induction that there are unique $q’, r’$ with $h = q’ \cdot g + r’$. Setting $q = a _ n^{-1} b _ m t^{m - n} + q’$ and $r = r’$, it follows

\[f = qg +r\]

Since $q$ and $r$ are uniquely determined by $q’$ and $r’$, they are also unique as required.

Alternative shorter but non-constructive proof for fields:

Suppose that $f, g \in \mathbb F[x]$ and $g \ne 0$. Then if $g \vert f$, take $r = 0$ and let $q$ be such that $f = qg$.

If not, consider

\[P := \\{f - gq \mid q \in \mathbb F[x]\\}\]

Then by assumption $P$ does not contain $0$. Take $q$ such that $r := f - gq$ has minimal degree among polynomials in $P$. Suppose that $\deg r \ge \deg g$. Write $\lambda$ for the coefficient of $x^{\deg r}$ in $r$, and note that

\[r' := r - g\lambda x^{\deg r - \deg g}\]

has $r’ \in P$ and $\deg r’ < \deg r$, a contradiction.

Take $N(f) := \deg f$, and the fact this makes $\mathbb F[x]$ into a Euclidean domain follows from the division algorithm.

Short but non-constructive proof for division algorithm: Suppose that $f, g \in \mathbb F[x]$ and $g \ne 0$. Then if $g \vert f$, take $r = 0$ and let $q$ be such that $a = bq$.

If not, consider

\[P := \\{f - gq \mid q \in \mathbb F[x]\\}\]

Then by assumption $P$ does not contain $0$. Take $q$ such that $r := f - gq$ is minimal for polynomials in $P$. Suppose that $\deg r \ge \deg g$. Write $\lambda$ for the coefficient of $x^{\deg r}$ in $r$, and note that

\[r' := r - g\lambda x^{\deg r - \deg g}\]

has $r’ \in P$ and $\deg r’ < \deg r$, a contradiction.

It is a field.

(A quick proof: suppose $a \in R$, $a \ne 0$. Then $\langle a, x \rangle = \langle f \rangle$ for some $f$. Then $a = fg$, so $f$ is a constant. Then $x = fh$ implies $f$ is a unit. So $\langle f \rangle = R[x]$. Then $1 = ap + xq$ for some $p, q$. Evaluating both sides at $x = 0$ gives $a p = 1$, so $a$ must be a unit).

(The contrapositive is also interesting, if $R$ is not a field, then $R[X]$ is not a PID, so e.g. this allows us to immediately deduce that $\mathbb Z[x]$ is not a PID).