Theorem 1.4.1 (Division Algorithm) If $K$ is a field and $f(x),g(x)\in K[x]$ with $f\ne 0$, then there are unique polynomials $q(x),r(X)\in K[x]$ with
$$g(x)=q(x)f(x)+r(x),
$$
where either $r=0$ or $\deg r(x)<\deg f(x)$.
Definition 1.4.1 If $f(x)$ and $g(x)$ are polynomials in $K[x]$, where $K$ is a field, then the polynomials $q(x)$ and $r(x)$ occurring in the Division Algorithm are called the quotient and the remainder after dividing $g$ by $f$.
Corollary 1.4.1 Let $R$ be a commutative ring, and let $f(x)\in R[x]$ be a monic polynomial. If $g(x)\in R[x]$, then there exist $q(x),r(x)\in R[x]$ with
$$g(x)=q(x)f(x)+r(x),
$$
where either $r=0$ or $\deg r(x)<\deg f(x)$.
Theorem 1.4.2 If $K$ is a field, then every ideal $I$ in $K[x]$ is a principal ideal; that is, there is $d\in I$ with $I=(d)$, then $d$ can be chosen to be a monic polynomial. Moreover, if $I\ne (0)$, then $d$ can be chosen to be a monic polynomial.
Definition 1.4.2 If $f(x)\in R[x]$, where $R$ is a ring, then a root of $f(x)$ in $R$ is an element $a\in R$ with $f(a)=0$.
Lemma 1.4.1 Let $f(x)\in K[x]$, where $K$ is a field, and let $u\in K$. Then there is $q(x)\in K[x]$ with
$$f(x)=q(x)(x-u)+f(u).
$$
Proposition 1.4.1 If $f(x)\in K[x]$, where $K$ is a field, then $a$ is a root of $f(x)$ in $K$ if and only if $x-a\mid f(x)$.
Proposition 1.4.2 Let $K$ be a field and let $f(x)\in K[x]$. If $f$ has degree $n$, then $f$ has at most $n$ roots in $K$.
Corollary 1.4.2 Every $n$th root of unity in $\mathbb C$ is equal to
$$e^{2\pi ik/n}=\cos(2\pi k/n)+i\sin(2\pi k/n),
$$
where $k=0,1,2,\dots,n-1$.
Corollary 1.4.3 Let $K$ be an infinite field and let $f(x)$ and $g(x)$ be polynomials in $K[x]$. If $f$ and $g$ determine the same polynomial function (that is, $f(a)=g(a)$ for all $a\in K$), then $f=g$.
Proof.
Let $h(x)=f(x)-g(x)$. Then if $h\ne 0$, $h$ has some degree.
However, the roots of $h$ are too much, which contradicts the Proposition 1.4.2. $\square$
Corollary 1.4.4 Let $K$ be a (possibly finite) field, let $f(x),g(x)\in K[x]$, and let $\deg f(x)\le\deg g(x)=n$. If $f(a)=g(a)$ for $n+1$ elements $a\in K$, then $f=g$.
Proof.
Let $h(x)=f(x)-g(x)$. Then if $h\ne 0$, $\deg h(x)\le n$.
That $h(x)$ have $n+1$ roots contradicts the Proposition 1.4.2. $\square$
Proposition 1.4.3 Let $f(X),g(X)\in K[X]=K[x_1,\dots,x_n]$, where $K$ is an infinite field.
- If $f(X)$ is nonzero, then there are $a_1,\dots,a_n\in K$ with $f(a_1,\dots,a_n)\ne 0$.
- If $f(a_1,\dots,a_n)=g(a_1,\dots,a_n)$ for all $(a_1,\dots,a_n)\in K^n$, then $f=g$.
Theorem 1.4.3 Let $K$ be a field. If $G$ is a finite subgroup of the multiplicative group $K^\times$, then $G$ is cyclic. In particular, if $K$ is finite, then $K^\times$ is cyclic.
Proof.
$G$ is an abelian group and $x^m-1$ have less than $m$ roots in $K$ (also in $G$).
For 命题1.1.7 in 抽象代数笔记, $G$ is cyclic. $\square$
Definition 1.4.3 If $K$ is a finite field, a generator of the cyclic group $K^\times$ is called a primitive element of $K$.
Definition 1.4.4 If $f(x)$ and $g(x)$ are polynomials in $K[x]$, where $K$ is a field, then a common divisor is a polynomial $c(x)\in K[x]$ with $c\mid f$ and $c\mid g$. If $f$ and $g$ in $K[x]$ are not both $0$, define their greatest common divisor abbreviated $\gcd$, to be the monic common divisor having largest degree. If $f=0=g$, define $\gcd(f,g)=0$.
Theorem 1.4.4 If $K$ is a field and $f(x),g(x)\in K[x]$, then their gcd $d(x)$ is a linear combination of $f$ and $g$; that is there are $s(x),t(x)\in K[x]$ with
$$d=sf+tg.
$$
Proof.
The set $(f,g)$ of all linear combinations of $f$ and $g$ is an ideal in $K[x]$.
Since every ideal in $K[x]$ is a principle ideal, there is $d(x)\in K[x]$ such that $(d)=(f,g)$.
Then there are $s(x),t(x)\in K[x]$ with $d=sf+tg$.
Because $f(x),g(x)\in (d)$, $d$ is a common divisor of $f$ and $g$.
If $q(x)$ is a common divisor of $f$ and $g$, then there are $f_1(x),g_1(x)\in K[x]$ with $f=f_1q,g=g_1q$.
$d=sf+tg=(sf_1+tg_1)q$ shows that $q$ is a divisor of $d$, and $\deg q(x)\le\deg d(x)$.
Therefore, $d$ is the gcd of $f$ and $g$. $\square$
Corollary 1.4.5 Let $K$ be a field and let $f(x),g(x)\in K[x]$.
- A monic common divisor $d(x)$ is the gcd if and only if $d$ is divisible by every common divisor; that is, if $h(x)$ is a common divisor, then $h\mid d$.
- $f$ and $g$ have a unique gcd.
Definition 1.4.5 An element $p$ in a domain $R$ is irreducible if $p$ is neither $0$ nor a unit and, in every factorization $p=uv$ in $R$, either $u$ or $v$ is a unit.
For example, a prime $p\in \mathbb Z$ is irreducible element, as is $-p$.
Proposition 1.4.4 If $K$ is a field, then a polynomial $p(x)\in K[x]$ is irreducible if and only if $\deg(p)=n\ge 1$ and there is no factorization in $K[x]$ for the form $p(x)=g(x)h(x)$ in which both factors have degree smaller than $n$.
Corollary 1.4.6 Let $K$ be a field and let $f(x)\in K[x]$ be a quadratic or cubic polynomial. Then $f$ is irreducible in $K[x]$ if and only if $f$ has no roots in $K$.
Theorem 1.4.5 (Gauss's Lemma) Let $f(x)\in \mathbb Z[x]$. If $f(x)=G(x)H(x)$ in $\mathbb Q[x]$, where $\deg(G),\deg(H)<\deg(f)$, then $f(x)=g(x)h(h)$ in $\mathbb Z[x]$, where $\deg(g)=\deg(G)$ and $\deg(h)=\deg(H)$.
Proof.
Clearing denominators, there are positive integers $n',n''$ such that $g(x)=n'G(x)$ and $h(x)=n''H(x)$.
Setting $n=n'n''$, we have
$$nf(x)=n'G(x)n''H(x)=g(x)h(x)\text{ in }\mathbb Z[x].
$$
If $p$ is a prime divisor of $n$, consider the map $\mathbb Z[x]\to \mathbb Z_p[x]$, denoted by $g\to \overline g$, which reduces all coefficients mod $p$. The equation becomes
$$0=\overline g(x)\overline h(x).
$$
But $\mathbb Z_p[x]$ is a domain, and so at least one of these factors is $0$.
Let's assume that $\overline g(x)=0$; that is all the coefficients of $g(x)$ are multiples of $p$.
Therefore, we may write $g(x)=pg'(x)$, where all the coefficients of $g'(x)$ lie in $\mathbb Z$. If $n=pm$, then
$$pmf(x)=pg'(x)h(x)\text{ in } \mathbb Z.
$$
Cancel $p$, and continue canceling primes until we reach a factorization $f(x)=g^*(x)h^*(x)$ in $\mathbb Z[x]$. $\square$
The contrapositive version of Gauss's Lemma is more convenient to use. If $f(x)\in \mathbb Z[x]$ has no factorization in $\mathbb Z[x]$ as a product of two polynomials, each having degree smaller than $\deg(f)$, then $f$ is irreducible in $\mathbb Q[x]$.
Lemma 1.4.2 Let $K$ be a field, let $p(x),f(x)\in K[x]$, and let $d(x)=\gcd(p,f)$. If $p$ is a monic irreducible polynomial, then
$$d(x)=\begin{cases}
1,&p\nmid f,\\
p(x),&p\mid f.
\end{cases}
$$
Proof. Since $d\mid p$, we have $d=1$ or $d=p$. $\square$
Theorem 1.4.6 (Euclid's Lemma) Let $K$ be a field and let $f(x),g(x)\in K[x]$. If $p(x)$ is an irreducible polynomial in $K[x]$, and $p\mid fg$, then either
$$p\mid f\text{ or }p\mid g.
$$
More generally, if $p\mid f_1(x)\cdots f_n(x)$, then $p\mid f_i$ for some $i$.
Proof.
Assume that $p\mid fg$ but that $p\nmid f$.
Since $p$ is irreducible, $\gcd(p,f)=1$, and so $1=sp+sf$ for some polynomials $s$ and $t$.
Therefore $g=spg+sfg$. Because $p\mid fg$, $fg=hp$ for some polynomial $h$.
Then $g=spg+shp=(sg+sh)p$, so $p\mid g$. $\square$
Definition 1.4.6 Two polynomials $f(x),g(x)\in K[x]$, where $K$ is a field, are called relatively prime if their gcd is $1$.
Corollary 1.4.7 Let $f(x),g(x),h(x)\in K[x]$, where $K$ is a field, and let $h$ and $f$ be relatively prime. If $h\mid fg$, then $h\mid g$.
Definition 1.4.7 If $K$ is a field, then a retaional function $f(x)/g(x)\in K(x)$ is in lowest terms if $f(x)$ and $g(x)$ are relatively prime.
Theorem 1.4.7 (Euclidean Algorithms) If $K$ is a field $f(x),g(x)\in K[x]$, then there are algorithms for computing $\gcd(f,g)$, as well as for finding a pair of polynomials $s(x)$ and $t(x)$ with
$$\gcd(f,g)=sf+tg.
$$
Corollary 1.4.8 Let $k$ be a subfield of a field $K$, so that $k[x]$ is a subring of $K[x]$. If $f(x),g(x)\in k[x]$, then their gcd in $k[x]$ is equal to their gcd in $K[x]$.
Corollary 1.4.9 If $f(x),g(x)\in \mathbb R[x]$ have no common root in $\mathbb C$, then $f,g$ are relatively prime in $\mathbb R[x]$.
Theorem 1.4.8 (Unique Factorization) If $K$ is a field, then every polynomial $f(x)\in K[x]$ of degree $\ge1$ is a product of a nonzero constant and monic irreducibles. Moreover, if $f(x)$ has two such factorizations,
$$f(x)=ap_1(x)\cdots p_m(x) \text{ and } f(x)=bq_1(x)\cdots q_n(x)
$$
that is, $a$ and $b$ are nonzero constants and the $p$'s and $q$'s are monic irreducibles, then $a=b,m=n$ and the $q$'s may be reindexed so that $q_i=p_i$ for all $i$.
Definition 1.4.8 Let $f(x)\in K[x]$, where $K$ is a field. A prime factorization of $f(x)$ is
$$f(x)=ap_1(x)^{e_1}\cdots p_m(x)^{e_m},
$$
where $a$ is a nonzero constant, the $p_i$ are distinct monic irreducible polynomials, and $e_i\ge 0$ for all $i$.
Let $K$ be a field, and assume that there are $a,r_1,\dots,r_n\in K$ with
$$f(x)=a\prod_{i=1}^n(x-r_i);
$$
we say that $f$ splits over $k$. If $r_1,\dots,r_s$, where $s\le n$, are the distinct roots of $f(x)$, then a prime factorization of $f(x)$ is
$$f(x)=a(x-r_1)^{e_1}(x-r_2)^{e_2}\cdots(x-r_s)^{e_s}.
$$
We call $e_j$ the multiplicity of the root $r_j$.
Definition 1.4.9 If $f$ and $g$ are elements in a commutative ring $R$, then a common multiple is an element $m\in R$ with $f\mid m$ and $g\mid m$. If $f$ and $g$ in $R$ are not both $0$, define their last common multiple, abbreviated $\operatorname{lcm}(f,g)$, to be a common mutiple $c$ of them with $c\mid m$ for every common multiple $m$. If $f=0=g$, define their $\operatorname{lcm}(f,g)=0$, If $R=K[x]$, we require lcm's to be monic.
Proposition 1.4.5 If $K$ is a field and $f(x),g(x)\in K[x]$ have prime factorizations $f(x)=a_0p_1^{a_1}\cdots p_n^{a_n}$ and $g(x)=b_0p_1^{b_1}\cdots p_n^{b_n}$ in $K[x]$, then
-
$f\mid g$ if and only if $a_i\le b_i,\forall i$.
-
If $m_i=\min\{a_i,b_i\}$ and $M_i=\max\{a_i,b_i\}$, then
$$
\gcd(f,g)=p_1^{m_1}\cdots p_n^{m_n}\text{ and }\operatorname{lcm}(f,g)=p_1^{M_1}\cdots p_n^{M_n}.
$$
Corollary 1.4.10 If $K$ is a field and $f(x),g(x)\in K[x]$ are monic polynomials, then
$$\gcd(f,g)\operatorname{lcm}(f,g)=fg.
$$