Polynomial Rings

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

Remark In this post, $A$ is always a commutative ring.

We defined the polynomial algebra $A[\x_i]_{i\in I}$ for an arbitrary (commutative) ring $A$ in §Algebras, ⁋Definition 7. This carries an $A$-algebra structure, but since the scalar multiplication of $A$ on $A[\x_i]_{i\in I}$ arises from the inclusion $A\hookrightarrow A[\x_i]_{i\in I}$ when we regard $A[\x_i]_{i\in I}$ as a ring, it suffices to consider $A[\x_i]_{i\in I}$ merely as a ring when studying its properties.

Degree of Polynomials

Before treating polynomials in earnest, let us first define the tools needed to handle them. By a polynomial defined over $A$ we mean an element of the polynomial ring $P=A[\x_i]_{i\in I}$. Let $\mathbb{N}^{(I)}$ denote the set of finitely supported functions from $I$ to $\mathbb{N}$:

\[\mathbb{N}^{(I)}=\{\nu:I \rightarrow \mathbb{N}\mid\text{$f(i)=0$ for all but finitely many $i\in I$}\}\]

For any $\nu\in \mathbb{N}^{(I)}$, setting

\[\x^\nu=\prod_{i\in I} \x_i^{\nu_i}\]

makes $\x^\nu$ an element of $P$. We call elements of the form

\[a\x^\nu\]

monomials. Then any polynomial $u$ can be expressed as a finite sum of monomials

\[u(\x)=\sum_{\nu\in \mathbb{N}^{(I)}} a_\nu \x^\nu,\qquad\text{$a_\nu=0$ for all but finitely many $\nu$}\]

On the other hand, for any $\nu\in \mathbb{N}^{(I)}$, defining

\[\lvert\nu\rvert=\sum_{i\in I} \nu_i\]

allows us to view $P=A[\x_i]_{i\in I}$ as an $\mathbb{N}$-graded ring

\[P=\bigoplus_{n\in \mathbb{N}}\bigoplus_{\lvert\nu\rvert=n}(A[\x_i]_{i\in I})_\nu=\bigoplus_{n\in \mathbb{N}} P_n\]

For each $n$, we call the elements of $P_n$ homogeneous polynomials of degree $n$. Also, for any polynomial $u\in P$, we write the component of $u$ of degree $n$ in this homogeneous decomposition as $u_n$.

Definition 1 Let an arbitrary element of the polynomial ring $P=A[\x]_{i\in I}$

\[u(\x)=\sum_{\nu\in \mathbb{N}^{(I)}} a_\nu \x^\nu,\qquad\text{$a_\nu=0$ for all but finitely many $\nu$}\]

be given. Then we call the largest $n$ satisfying $u_n\neq 0$ the degree of $u$, and denote it by $\deg(u)$. By definition, the degree of a polynomial having only a constant term is $0$, but in particular for the additive identity $0$ of $P$, we define $\deg(0)=-\infty$.

Then the following proposition holds trivially.

Proposition 2 For two polynomials $u,v$, the following hold.

If $\deg (u)\neq\deg(v)$, then $u+v\neq 0$ and the equality
\[\deg(u+v)=\sup(\deg(u), \deg(v))\]
holds. If $\deg(u)=\deg(v)$, then the inequality $\deg(u+v)\leq\deg(u)$ holds.
The inequality $\deg(uv)\leq\deg(u)+\deg(v)$ holds.

The reason the second condition of this proposition is not an equality lies in $A$. To examine this more closely, consider the polynomial ring $A[\x]$ in one variable. Then any polynomial of $A[\x]$ can be written in the form

\[u(\x)=\sum_{i=0}^n a_i\x^i\qquad\text{($a_n\neq 0$)}\]

and we call $a_n\x^n$ the leading term of the polynomial $p$, and its coefficient $a_n$ the leading coefficient of the polynomial $p$. If the leading coefficient of $p$ is $1$, we call $p$ a monic polynomial.

Now for any two (univariate) polynomials

\[u(\x)=\sum_{i=0}^n a_i\x^i,\qquad v(\x)=\sum_{j=0}^m b_j \x^j\]

their product is explicitly given by the formula

\[u(\x)v(\x)=\sum_{k=0}^{n+m}\left(\sum_{i=0}^k a_ib_{k-i}\right)\x^k\]

and therefore the coefficient of the highest degree term is $a_nb_m$. However, if $A$ is not an integral domain, the product of two nonzero elements $a_n$, $b_m$ could be $0$, so $uv$ might not be a polynomial of degree $m+n$. Using this discussion, we obtain the following.

Lemma 3 For an integral domain $A$, the following hold.

For any $u,v\in A[\x]$, we have $\deg(uv)=\deg(u)+\deg(v)$.
The units of $A[\x]$ are exactly the units of $A$.
$A[\x]$ is an integral domain.

Now let us return to the general case. If any two polynomials $u,v\in A[\x_i]_{i\in I}$ are given, since only finitely many indeterminates appear in these polynomials, when computing $uv$ it suffices to consider $A[\x_j]_{j\in J}$ for some finite set $J\subset I$ instead of $A[\x_i]_{i\in I}$. Then the fact that $A[\x_j]_{j\in J}$ is an integral domain follows from Lemma 3 and the isomorphism

\[A[\x_1,\x_2]\cong (A[\x_1])[\x_2]\]

That is, the following holds.

Proposition 4 For an integral domain $A$ and a polynomial ring $P=A[\x_i]_{i\in I}$, the following hold.

For any $u,v\in P$, we have $\deg(uv)=\deg(u)+\deg(v)$.
The units of $P$ are exactly the units of $A$.
$P$ is an integral domain.

Univariate Polynomials

We now examine univariate polynomials in somewhat more detail.

Proposition 5 Let a polynomial $u(\x)$ of degree $m$ and a polynomial $v(\x)$ of degree $n$ in $A[\x]$ be given, and let the leading coefficient of $v$ be $b_n$. Setting $k=\sup(m-n+1,0)$, there exist $q,r\in A[\x]$ satisfying the formula

\[b_n^k u=qv+r,\qquad \deg r < n\]

Proof

If $n>m$, then $k=0$, and in this case setting $q=0$, $r=u$ satisfies the given formula, so there is nothing to prove in this case. Therefore, assume $n\leq m$.

We use induction on $m$. Let the leading coefficient of $u$ be $a_m$. If

\[v(\x)=\sum_{j=0}^n b_j\x^j\]

then there exists a polynomial $u_1\in A[\x]$ of degree less than $m$ such that

\[b_n^k u(\x)=b_n^{k-1}a_m\x^{m-n}v(\x)+b_n^{k-1}u_1(\x)\]

This formula is obtained by multiplying $v(\x)$ by $a_m\x^{m-n}$ to match the leading term with that of $b_n u$, obtaining

\[b_nu(\x)=a_m\x^{m-n}v(\x)+u_1(\x)\]

and then multiplying both sides by $b_n^{k-1}$. Now by the inductive hypothesis, there exist suitable polynomials $q_1,r\in A[\x]$ such that the formula

\[b_n^{k-1}u_1(\x)=q_1(\x)v(\x)+r(\x),\qquad \deg(r) < n\]

holds. Substituting this back into the previous formula, we obtain

\[b_n^k u(\x)=(b_n^{k-1}a_m\x^{m-n}+q_1(\x))v(\x)+r(\x)\]

The coefficient $b_n^k$ appearing here arises when we raise the degree one by one from $v$ to match the leading term with $u$, and if $b_n$ is invertible, we can verify that the $q$ and $r$ satisfying the above formula are uniquely determined. In particular, if every nonzero element of $A$ is invertible, that is, if $A=\mathbb{K}$, then in the above situation we can uniquely determine $q,r$ satisfying

\[u=qv+r,\qquad \deg r < n\]

Moreover, in this case the polynomial ring $\mathbb{K}[\x]$ is an integral domain by Lemma 3, and thus defining $N:\mathbb{K}[\x] \rightarrow \mathbb{Z}^{\geq0}$ by

\[N: u\mapsto \deg(u)\qquad \text{where $N(0)=0$}\]

we see that $\mathbb{K}[\x]$ is a Euclidean domain.

Proposition 6 For any field $\mathbb{K}$, $\mathbb{K}[\x]$ is a Euclidean domain.

In particular, the notion of greatest common divisor is well-defined over a Euclidean domain, and related to this, Bézout’s lemma also holds. (§Integral Domains, ⁋Theorem 7)

We have previously seen that the images of arbitrary (nonzero) elements of $\mathbb{K}$ in $\mathbb{K}[\x]$ are units of $\mathbb{K}[\x]$. (Lemma 3) On the other hand, in a Euclidean domain whether one element divides another can be determined by running the Euclidean algorithm, so an arbitrary non-constant $u\in \mathbb{K}[\x]$ being irreducible is equivalent to $u$ not being divisible by any $v\in \mathbb{K}[\x]$ satisfying $\deg(v)<\deg(u)$.

On the other hand, $\mathbb{K}[\x]$ is a UFD, and thus we can define irreducible elements of $\mathbb{K}[\x]$. Since the units of $\mathbb{K}[\x]$ are exactly the units of $\mathbb{K}$, any irreducible polynomial $u$ satisfies $\deg(u)\geq 1$ by definition, and since $u$ is irreducible, if $v\mid u$ then $v$ is either a constant polynomial or a constant multiple of $u$. In particular, any two irreducible polynomials must be constant multiples of each other, so two distinct monic irreducible polynomials are coprime. In this way, any polynomial in $A[\x]$ can be uniquely expressed as a product of its leading coefficient and monic irreducible polynomials.

Proposition 7 For any polynomial $u\in A[\x]$ and any $a\in A$, the remainder when $u(\x)$ is divided by $\x-a$ is $u(a)$. Therefore, $u$ having $a$ as a root is equivalent to $\x-a$ being a divisor of $u$ in $A[\x]$.

The proof of this is of course to run the Euclidean algorithm, and this is in fact a familiar result from middle school. As another result, if $u$ has $a$ as a root, then $u$ must necessarily be written in the form

\[u(\x)=(\x-a)^p v(\x),\qquad v(a)\neq 0\]

In this case, we call $p$ the multiplicity of the root $a$. Then the following holds.

Proposition 8 Suppose any two polynomials $u,v\in A[\x]$ have a common root $a$, and let the multiplicity of $a$ in $u$ and $v$ be $p,q$ respectively. Then the following hold.

The multiplicity of $a$ in the polynomial $u+v$ is at least $\inf(p,q)$, and equality holds when $p\neq q$.
The multiplicity of $a$ in the polynomial $uv$ is at least $p+q$, and equality holds when $A$ is an integral domain.

Proof

Suppose $u(\x) = (\x-a)^p u_1(\x)$, $v(\x) = (\x-a)^q v_1(\x)$ with $u_1(a) \neq 0$, $v_1(a) \neq 0$. Without loss of generality, assume $p \leq q$; then we obtain the formula

\[u(\x) + v(\x) = (\x-a)^p (u_1(\x) + (\x-a)^{q-p}v_1(\x))\]

and thus obtain the desired inequality. If here $p < q$, then $a$ is not a root of $u_1(\x) + (\x-a)^{q-p}v_1(\x)$, so we obtain the desired result.

For the second result, under the same assumption,

\[u(\x)v(\x) = (\x-a)^{p+q}u_1(\x)v_1(\x)\]

If $A$ is an integral domain, then $u_1(a)v_1(a) \neq 0$, from which we obtain the second result.

Furthermore, if $A$ is an integral domain, we can show the following inductively.

Proposition 9 Fix an integral domain $A$. Let a nonzero element $u$ of $A[\x]$ have roots $a_1,\ldots, a_r$, and let their multiplicities be $p_1,\ldots, p_r$ respectively. Then there exists $v\in A[\x]$ such that

\[u(\x)=(\x-a_1)^{p_1}\cdots(\x-a_r)^{p_r}v(\x),\qquad v(a_1),\ldots, v(a_r)\neq0\]

Proof

We proceed by induction on $r$. The case $p=1$ is trivial, so suppose $r$ roots $a_1,\ldots, a_r$ of $u$ are given, and apply the inductive hypothesis to the first $r-1$ roots to write

\[u(\x)=u_1(\x)u_2(\x)=(\x-a_1)^{p_1}\cdots(\x-a_{r-1})^{p_{r-1}}u_2(\x)\]

Then since $A$ is an integral domain, we know $u_1(a_r)\neq 0$, so necessarily $u_2(a_r)=0$ and the multiplicity of $a_r$ in $u_2$ must be $p_r$. From this we obtain the desired claim.

In particular, for any integral domain $A$, any polynomial of degree $n$ in $A[\x]$ has at most $n$ roots (counted with multiplicity). Therefore, if two polynomials $f,g\in A[\x]$ of degree at most $n$ are given, and if $f(a_i)=g(a_i)$ holds for $n+1$ distinct elements $a_1,\ldots, a_{n+1}$, then necessarily $f=g$. From this we obtain the following result.

Proposition 10 Fix $n$ distinct elements $a_1,\ldots, a_n$ of a field $\mathbb{K}$. For arbitrary elements $b_1,\ldots, b_n$ of $\mathbb{K}$, define for each $i$

\[u_i(\x)=\prod_{j\neq i}\frac{\x-a_j}{a_i-a_j}\]

Then

\[u=b_1 u_1 + \cdots + b_n u_n\]

is the unique polynomial of degree less than $n$ satisfying $u(a_i)=b_i$ for each $i$.

Proof

Uniqueness is trivial by the above argument, and it remains only to verify that substituting each $a_i$ into $u$ yields the value $b_i$.

On the other hand, one useful method for finding multiple roots is to differentiate the given polynomial. We can define algebraically what differentiation is (§Differentiation), but in this post we define $D: A[\x] \rightarrow A[\x]$ by the formula

\[D:\left(u(\x)=\sum_{i=0}^n a_i\x^i\right)\mapsto \left((Du)(\x)=i.a_i\x^{i-1}\right)\tag{$\ast$}\]

Here, $i.a_i$ is the element of $A$ defined by

\[i.a_i=\underbrace{a_i+\cdots+a_i}_\text{\scriptsize$i$ times}\]

The only property we will use in the remaining discussion is the Leibniz rule

\[D(uv)=(Du)v+u(Dv)\]

which in §Differentiation is the definition of a derivation, but if we accept the formula ($\ast$) as a definition, it can be verified by direct computation.

Proposition 11 For any polynomial $u \in A[\x]$, a necessary and sufficient condition for a root $a \in A$ of $u$ to be a simple root is that $a$ is not a root of $Du$.

Proof

By the assumption that $a$ is a root of $u$, there exists $v \in A[\x]$ such that $u = (\x - a)v$, and in this case the necessary and sufficient condition for $a$ to be a simple root of $u$ is $v(a) \neq 0$. Now since $Du = v + (\x - a)Dv$, we have $(Du)(a) = v(a)$.

More generally, using induction we can show the following.

Proposition 12 For any polynomial $u \in A[\x]$, if a root $a \in A$ of $u$ has multiplicity $k$, then $a$ is a root of $Du$ of multiplicity $\geq k-1$. If $k \cdot 1$ is cancellable in $A$, then $a$ has order $k-1$ in $Du$.

Proof

Similarly to the above proposition, set $u=(\x-a)^k v$ with $v(a)\neq 0$; then

\[Du=k(\x-a)^{k-1}v+(\x-a)^k Dv=(\x - a)^{k-1}(kv + (\x - a)Dv)\]

from which we obtain the first claim. If $k\cdot 1$ is cancellable in $A$, then substituting $\x=a$ into $kv+(\x-a)Dv$ gives $k.v(a)$, which is not $0$, so the latter claim holds.

Proposition 13 Let an integral domain $A$ and a set $I$ be given, and let $(H_i)_{i\in I}$ be a family of infinite subsets of $A$ indexed by $I$. Also, let $H=\prod_{i\in I} H_i\subseteq A^I$. Then if $u$ is a nonzero element of $A[\x_i]_{i\in I}$, the set

\[H_u=\{(\x_i)\in H\mid u(\x_i)\neq 0\}\]

has the same cardinality as $H$.

Proof

The case where $I$ is a finite set can be shown by induction. Let $\lvert I\rvert=n$, and prove by induction on $n$. First, the case $n=0$ is trivial. To simplify notation for the induction, let $I=\{1,\ldots, n\}$, $J=\{1,\ldots, n-1\}$, and $B=A[\x_i]_{i\in J}$. Then since $u\neq 0$, there exist $v_k\in B$ such that

\[u=\sum_{k=0}^m v_k\x_n^k\]

Here, $v_m\neq 0$. Now by the inductive hypothesis, the set

\[H_{v_m}=\{(\x_i)\in H\mid v_m(\x_i)\neq 0\}\]

has the same cardinality as $\prod_{i\in J} H_i$. Now for any element $(x_i)_{i\in J}$ of $\prod_{i\in J} H_i$,

\[w(\x)=\sum_{k=0}^mv_k(x_1,\ldots, x_{n-1})\x_n^k\]

is a nonzero polynomial. On the other hand, the set

\[\{a\in H_n\mid w(a)\neq 0\}\]

has, by the fact that $H_n$ is infinite and the argument after Proposition 9, and by §Natural Numbers and Infinite Sets, ⁋Proposition 13,

\[\lvert H\rvert\geq \lvert H_u\rvert\geq \lvert H_{v_m}\rvert\lvert H_n\rvert=\lvert H\rvert\]

The case where $I$ is an infinite set can be reduced to the finite case by considering the polynomial ring containing only the indeterminates appearing in $u$.

In particular, if $I$ is nonempty, then $H_u$ is an infinite set.

Factorization

If an arbitrary ring homomorphism $\phi:A \rightarrow B$ is given, any $u\in A[\x_i]_{i\in I}$ can be viewed as an element of $B[\x_i]_{i\in I}$ via $\phi$.

In particular, consider the case where $A$ is an integral domain and $\phi$ is the canonical inclusion $A \hookrightarrow \Frac(A)$. Then since $\Frac(A)$ is a field, $(\Frac A)[\x]$ is a Euclidean domain by Proposition 6, and thus if we view any $u\in A[\x]$ in $(\Frac A)[\x]$, then $u$ can be factored at least in $(\Frac A)[\x]$. It is then natural to examine how this factorization is reflected in $A[\x]$.

Proposition 14 (Gauss) Consider a UFD $A$, its field of fractions $\Frac(A)$, and an element $u$ of $A[\x]$. If $u$ is reducible in $(\Frac A)[\x]$, then $u$ is also reducible in $A[\x]$.

Proof

Suppose a polynomial $u$ in $(\Frac A)[\x]$ is expressed as a product of two polynomials

\[u(\x)=\tilde{v}_1(\x)\tilde{v}_2(\x),\qquad \tilde{v}_i\in (\Frac A)[\x]\]

Then multiplying both sides by the least common multiple of the coefficients of $\tilde{v}_1$ and $\tilde{v}_2$, there exist $v_i\in A[\x]$ such that for some appropriate $a\in A$,

\[a u(\x)=v_1(\x)v_2(\x)\]

If $a$ is a unit, there is nothing more to prove, so assume $a$ is not a unit. Then there exist irreducible elements $p_i\in A$ such that $a=p_1\cdots p_r$. In this case, by §Integral Domains, ⁋Proposition 17, $(p_i)$ is a prime ideal of $A$, and thus

\[(A/p_iA)[\x]\cong A[\x]/P_i\]

is an integral domain by Proposition 4. Therefore, reducing the equation $au=v_1v_2$ modulo $P_i$, we see that either $v_1$ or $v_2$ must become $0$ in $(A/p_iA)[x]$. That is, the coefficients of one of $v_1$ or $v_2$ are all multiples of $p_i$, and thus we can cancel this $p_i$ to obtain another polynomial in $A[\x]$. Applying this to all $p_i$ gives the desired result.

From this we obtain the following.

Corollary 15 Consider a UFD $A$ and its field of fractions $\Frac A$, and a polynomial $u(\x) \in A[\x]$, and let the greatest common divisor of the coefficients of $u(\x)$ be $1$. Then $u(\x)$ being irreducible in $A[\x]$ is equivalent to $u(\x)$ being irreducible in $(\Frac A)[\x]$.

Proof

By Proposition 14 (Gauss), if $u(\x)$ is reducible in $(\Frac A)[\x]$, then it is also reducible in $A[\x]$.

Conversely, suppose the greatest common divisor of the coefficients of $u(\x)$ is $1$, and assume $u(\x)$ is reducible in $A[\x]$. That is, we can write

\[u(\x) = v_1(\x) v_2(\x)\]

and since the greatest common divisor of the coefficients of $v_1(\x)$ and $v_2(\x)$ must also be $1$, in particular neither $v_1$ nor $v_2$ is constant. From this, $u=v_1v_2$ means that $u$ is also reducible when viewed in $(\Frac A)[\x]$.

Then one of the key results related to factorization is the following theorem.

Theorem 16 A necessary and sufficient condition for $A$ to be a UFD is that $A[\x]$ is a UFD.

Proof

That $A[\x]$ being a UFD implies $A$ is also a UFD is trivial, so it suffices to show the converse.

For a UFD $A$, let a nonzero element of $A[\x]$ be $u(\x)$, and if the greatest common divisor of the coefficients of $u(\x)$ is $d$, then we can write $u(\x)=du_0(\x)$ to obtain a polynomial $u_0\in A[\x]$ whose coefficients have greatest common divisor $1$; thus we may assume without loss of generality that the greatest common divisor of the coefficients of $u$ is $1$.

By Proposition 6, $u$ can be uniquely factored in $(\Frac A)[\x]$, and applying Proposition 14 (Gauss) to this factorization, we can factor $u$ in $A[\x]$. On the other hand, since the greatest common divisor of the coefficients of $u$ is $1$, the greatest common divisor of the coefficients of the factors of $u$ thus obtained is also $1$, and therefore by Corollary 15 they are irreducible in $A[\x]$. From this we obtain a factorization of $u$ in $A[\x]$, and uniqueness is trivial using the fact that each of these components is a $\Frac A$-multiple of the corresponding factor in $(\Frac A)[\x]$.

Field of Rational Functions

We now define rational function fields and power series rings, which are variants of polynomial rings. Earlier in Proposition 4, we proved that for any field $\mathbb{K}$, $\mathbb{K}[\x_i]_{i\in I}$ is an integral domain. Therefore, the field of fractions of $\mathbb{K}[\x_i]_{i\in I}$ is well-defined.

Definition 17 We call the field of fractions of the polynomial ring $\mathbb{K}[\x_i]_{i\in I}$ defined over a field $\mathbb{K}$ the field of rational functions, and denote it by $\mathbb{K}(\x_i)_{i\in I}$.

Earlier, we saw that a polynomial ring can be viewed as an $\mathbb{N}$-graded ring using multidegree. In a similar way, let us define a degree on $\mathbb{K}(\x_i)_{i\in I}$ as well. The natural choice is to define for any element $u/v$ of $\mathbb{K}(\x_i)_{i\in I}$

\[\deg(u/v)=\deg(u)-\deg(v)\]

and we can verify that this is well-defined. As with polynomials, we define $\deg(0)=-\infty$.

Then the following proposition is the analogue of Proposition 2.

Proposition 18 For two rational fractions $r, s$, the following hold.

If $\deg r \ne \deg s$, then
\[r + s \ne 0 \quad \text{and} \quad \deg(r + s) = \sup(\deg r, \deg s)\]
holds. If $\deg r = \deg s$, then $\deg(r + s) \leq \deg r$.
$\deg(rs) = \deg r + \deg s$.

Proof

For both claims, it suffices to consider only the case where $r, s \ne 0$. Therefore, let $r =u/v$, $s = w/z$ with $u, v, w, z$ all nonzero polynomials. In any case, the idea is to first compute the given expression and then apply Proposition 2 and Lemma 3.

$r + s = (uz+vw)/(vz)$. First, suppose $\deg r \ne \deg s$, that is, $\deg u + \deg z \ne \deg w + \deg v$. Then $uz + vw \ne 0$, and
\[\begin{aligned}\deg(r + s) &= \deg(uz + vw) - \deg(vz) \\ &= \sup(\deg(uz), \deg(vw)) - \deg(vz) \\ &= \sup(\deg(uz) - \deg(vz), \deg(vw) - \deg(vz)) \\ &= \sup(\deg r, \deg s).\end{aligned}\]
Now suppose $\deg r = \deg s$, that is, $\deg u + \deg z = \deg w + \deg v$, and if $r + s \ne 0$, then
\[\deg(r + s) = \deg(uz + vw) - \deg(vz)\leq \deg(uz) - \deg(vz) = \deg r\]
so the claim holds.
$rs = (uw)/(vz)$, and therefore
\[\deg(rs) = \deg(uw) - \deg(vz) = \deg u - \deg v + \deg w - \deg z = \deg r + \deg s.\]

Power Series Ring

A power series ring is another variant of a polynomial ring, and is now the set of infinite sums of monomials

\[u(\x)=\sum_{\nu\in \mathbb{N}^{(I)}} a_\nu \x^\nu,\qquad\text{$a_\nu$ need not satisfy finiteness condition}\]

We write this as $A[[\x_i]]_{i\in I}$ and call it the ring of formal power series.

Formal power series can be defined with concepts similar to polynomials, except for the fact that they are infinite sums of monomials. For example, we can define the degree $p$ component $u_p$ of a formal power series $u$. However, for the (total) degree of $u$, since the degree of elements other than polynomials would be $+\infty$, there is no reason to use this in $A[[\x_i]]_{i\in I}$. Instead, for any formal power series $u$, we define the order $\omega(u)$ of $u$ as the smallest $p$ such that $u_p\neq 0$. Similarly, setting $\omega(0)=\infty$, the formulas

\[\omega(u+v)\geq \inf(\omega(u),\omega(v)),\quad\text{equality if $\omega(u)\neq\omega(v)$}\]

and

\[\omega(uv)\geq \omega(u)+\omega(v)\]

hold.

For degree, the above inequality would have held if $A$ were an integral domain. Similarly, the following holds.

Proposition 19 For an integral domain $A$, the following hold.

$A[[x_i]]_{i\in I}$ is an integral domain.
For two nonzero elements $u,v\in A[[x_i]]_{i\in I}$, we have $\omega(uv)=\omega(u)+\omega(v)$.

However, one must be somewhat careful: unlike Proposition 4, the units of $A[[x_i]]_{i\in I}$ are larger than those of $A$. For example, consider the formula

\[(1-\x)\left( \sum_{n=0}^\infty \x^n\right)=1\]

More generally, the following holds.

Proposition 20 For any $u\in A[[x_i]]_{i\in I}$, $u$ being invertible in $A[[x_i]]_{i\in I}$ is equivalent to the constant term of $u$ being invertible in $A$.

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Degree of Polynomials

Univariate Polynomials

Factorization

Field of Rational Functions

Power Series Ring

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