2.5 Ring of polynomials

def: Polynomial

$$\begin{align*} &\sphericalangle \\ &R \in \mathcal R \\ &a_0, a_1, \ldots, a_n \in R \\ &a_n \ne 0 \\ &p: x \mapsto a_0+a_1x + \ldots + a_nx^n \\ \hline \\ & p \in R[x] \,\,\, (p- \text{polynomial}) \\ \end{align*}$$

In gerenal the notation $R[\alpha]$ means a minimal ring containing both $R$ and $\alpha$. For instance $\Z[\sqrt 2] = \{a + b \sqrt 2, a, b \in \Z\}$. Thus $R[x]$ is a minimal ring that contains both $R$ and indeterminate $x$. It can be easily seen that it's exactly the set of polynomials in one variable.

In particular, the set of all polynomial form a ring. We define the operations in the following manner. First, we need to make a convention that we pad polynomials of different degree with zeros, so they have the same number of coefficients. For example $f: x \mapsto 3x^2+2x+1$ and $g: x \mapsto x+2$, then we say that $g: x \mapsto 0x^2+x+2$. Then the operations are defined by:

$$f+g = (a_0+b_0)+(a_1+b_1)x + \ldots + (a_n+b_n)x^n \\ f\cdot g: x \mapsto c_0+c_1x+\ldots + c_{m+n}x^{m+n} \\ c_i=\sum_{j=0}^{i}a_jb_{i-j}$$

def: Polynomial degree

$$\begin{align*} &\sphericalangle \\ &R \in \mathcal R \\ &a_0, a_1, \ldots, a_n \in R \\ &a_n \ne 0 \\ &p: x \mapsto a_0+a_1x + \ldots + a_nx^n \\ \hline \\ & \deg p := n \,\,\,(p - \text{polynomial of degree } n) \\ & \deg (x \mapsto 0) := -\infty \\ \end{align*}$$

def: Leading term

$$\begin{align*} &\sphericalangle \\ &R \in \mathcal R \\ &a_0, a_1, \ldots, a_n \in R \\ &a_n \ne 0 \\ &p: x \mapsto a_0+a_1x + \ldots + a_nx^n \\ \hline \\ & LT(p):=a_nx^n - \text{leading term} \\ & LC(p):=a_n - \text{leading coefficient} \\ \end{align*}$$

def: Monic polynomial

$$\begin{align*} &\sphericalangle \\ &R \in \mathcal R \\ &p \in R[x] \\ &LC(p)=1 \\ \hline \\ &p- \text{monic polynomial} \end{align*}$$

def: Action on coefficents

$$\begin{align*} &\sphericalangle \\ &R \in \mathcal R \\ &p \in R[x] \\ &p(x)= a_0+a_1x + \ldots + a_nx^n \\ \hline \\ &p^{\phi}(x):= \phi(a_0)+\ldots + \phi(a_n)x^n \end{align*}$$

Polynomials over commutative rings

Proposition 2.5.1: Polynomial rings with isomorphic base rings are isomorphic

$$\begin{align*} &\sphericalangle \\ &R_1, R_2 \in \mathcal R^{\mathcal C} \\ &\phi: R_1 \rightsquigarrow_R R_2 \\ &\phi_x: R_1[x] \to R_2[x], p(x) \mapsto p^{\phi}(x) \\ \hline \\ &\begin{align*} & \phi_x: R_1[x] \rightsquigarrow_R R_2[x] \hspace{0.5cm} \tag{a}\\ &\phi: R_1 \cong_R R_2 \implies R_1[x] \overset{\phi_x}\cong R_2[x] \hspace{0.5cm} \tag{b}\\ \end{align*} \end{align*}$$

Proof

a.

Further we'll assume that the two polynomials are of the same degree. If not, we just pad one of them with zeros.

Let's prove $R_1[x] \overset{\phi_x}\rightsquigarrow R_2[x]$:

$$\phi_x(a_0+\ldots +a_nx^n + b_0+\ldots +b_nx^n)= \\ \phi(a_0+b_0)+\ldots+\phi(a_n+b_n)x^n=\\ \phi(a_0)+\ldots + \phi(a_n)x^n+\phi(b_0)+\ldots + \phi(b_n)x^n=\\ \phi_x(a_0+\ldots +a_nx^n)+\phi_x(b_0+\ldots +b_nx^n)$$ $$\phi_x((a_0+\ldots +a_nx^n) \cdot (b_0+\ldots +b_nx^n))=\\ \phi(a_0b_0)+\ldots+\phi(\sum_{i=0}^ka_ib_{k-i})x^k + \ldots + \phi(a_nb_n)x^{2n}=\\ \phi(a_0)\phi(b_0)+\ldots+\sum_{i=0}^k\phi(a_i)\phi(b_{k-i})x^k + \ldots + \phi(a_n)\phi(b_n)x^{2n}= \\ (\phi(a_0)+\ldots +\phi(a_n)x^n)\cdot (\phi(b_0)+\ldots +\phi(b_n)x^n) = \\ \phi_x(a_0+\ldots +a_nx^n) \cdot \phi_x(b_0+\ldots +b_nx^n)$$

b.

Obviously $\phi_x: R_1[x] \leftrightarrow R_2[x]$

$\square$

note

We'll use the notation $\phi_x$ to denote the homomorphism that maps coefficents of polynomials using $\phi$

Polynomials over integral domains

Proposition 2.5.2 Degree of a polynomials product

$$\begin{align*} &\sphericalangle \\ &R \in \mathcal R^{\mathcal {ID}} \\ \hline \\ &\forall f, g \in R[x] \setminus \{0\}: \deg fg=\deg f+\deg g \end{align*}$$

Proof

$$\forall f, g \in R[x] \ \setminus \{0\}: LT(f)=a_nx^n, LT(g)=b_mx^m: \\ LT(fg)=LT(f)\cdot LT(g) = a_nb_mx^{n+m} \overset{R \in R^{\mathcal {ID}}}{\implies} LT(fg) \ne 0 \implies \\ \deg fg=\deg f+\deg g \\$$

$\square$

Example: Degree of a product over non-integral domain

In $\Z_6$ we have $2x \cdot 3x = 0$, so obviously $\deg 2x + \deg 3x \ne \deg 2x \cdot 3x$.

Proposition 2.5.3 Units for polynomials

$$\begin{align*} &\sphericalangle \\ &R \in \mathcal R^{\mathcal {ID}} \\ \hline \\ &R[x]^* = R^* \end{align*}$$

Proof

$R^* \subseteq R[x]^*$ is obvious. Let's prove $R^* \supseteq R[x]^*$:

$$f \in R[x]^* \implies \exists g \in R[x]: fg = 1 \implies \\\deg f + \deg g = \deg 1 = 0 \overset{\forall h \ne 0: \deg h \ge 0}{\implies} \deg f=0, \deg g =0 \implies \\ f \in R, g \in R, fg = 1 \implies f \in R^*$$

$\square$

Proposition 2.5.4 Polynomials over integral domain is an integral domain

$$\begin{align*} &\sphericalangle \\ &R \in \mathcal R^{\mathcal {ID}} \\ \hline \\ &R[x] \in \mathcal R^{\mathcal {ID}} \\ \end{align*}$$

Proof

$$\forall f,g \in R[x] \setminus \{0\}: \deg fg \overset{(2.5.2)}{=} \deg f +\deg g \ge 0 \implies fg \ne 0$$

It's easy to see that $R[x] \in \mathcal G^{\mathcal A}_+$. The identity in this case is 0, the inverse for $\sum_ia_ix^i$ is $\sum_i(-a_i)x^i$. The abelian group properties follows from the abelian group properties of $R$.

Let's prove multiplicative associativity:

$$p := \sum_i^ma_ix^i, q := \sum_i^nb_ix^i, r:=\sum_i^pc_ix^i: \\ (pq)r = (\sum_i^na_ix^i\sum_i^mb_ix^i)\sum_i^pc_ix^i = \\ = \sum_i^{n+m}(\sum_j^{i}a_ib_{i-j})x^i\sum_i^pc_ix^i = \\ \sum_i^{n+m+p}[\sum_j^i(\sum_k^ja_kb_{j-k})c_{i-j}]x^i = \\ \sum_i^{n+m+p}[\sum_{j+k+l=i}a_kb_{j}c_{l}]x^i$$

By the symmetry of this expression, we can see that $p(qr)=\sum_i^{n+m+p}[\sum_{j+k+l=i}a_kb_{j}c_{l}]x^i$ as well. Thus, $(pq)r=p(qr)$.

Distributivity and commutativity are derived similarly.

$\square$

Polynomials over UFDs

Proposition 2.5.5 Factor ring for polynomials

$$\begin{align*} &\sphericalangle \\ &R \in \mathcal R^{\mathcal C} \\ &I \lhd_R R \\ \hline \\ &\begin{align*} &R[x] / I[x] \cong (R/I)[x] \tag{a}\\ &I \in \mathfrak P_I(R) \implies I[x] \in \mathfrak P_I(R[x]) \hspace{1cm} \tag{b}\\ \end{align*} \end{align*}$$

Proof

a.

Consider a map $\phi: R[x] \mapsto (R/I)[x]$ that maps coefficents to a factor ring. It's easy to show that it's a homomorphism. $\ker \phi = I[x]$, thus by $(2.3.9)$ $R[x] / I[x] \cong (R/I)[x]$.

b.

$$I \in \mathfrak P_I(R) \overset{(2.4.8)}{\implies} R/I \in \mathcal R^{\mathcal ID} \overset{(2.5.4)}{\implies} (R/I)[x] \in \mathcal R^{\mathcal ID} \overset{(a)}{\implies} \\ R[x]/I[x] \in \mathcal R^{\mathcal ID} \overset{(2.4.8)}{\implies} I[x] \in \mathfrak P_I(R[x])$$

$\square$

Proposition 2.5.6: Gauss' lemma

$$\begin{align*} &\sphericalangle \\ &R \in \mathcal R ^{\mathcal {UFD}} \\ &F:=\mathfrak F(R) \\ &p \in R[x] \\ \hline \\ &\begin{align*} & a(x), b(x) \in F[x] \setminus F, p(x) = a(x)b(x) \implies \\ &\exists r, s \in F, a'(x), b'(x) \in R[x] \setminus R: \\ & a(x)=ra'(x), b(x) = sb'(x), p(x)=a'(x)b'(x) \hspace{0.5cm} \tag{a}\\ & p(x) \in F[x]^+ \implies p(x) \in R[x]^+ \tag{b}\\ \end{align*} \end{align*}$$

Proof

a.

$$a(x), b(x) \in {F[x] \setminus F}: p(x)=a(x)b(x) \implies \\ \exists d \in R \setminus \{0\}: dp(x)=a'(x)b'(x), a'(x), b'(x) \in R[x] \setminus R$$

If $d \in R^*$, then $p(x)=d^{-1}a'(x)b'(x)$ and we're done. Otherwise using the fact that $R \in \mathcal R ^{\mathcal {UFD}}$ we can write $d = p_1\cdot \ldots \cdot p_n, p_i \in R^-$:

$$p_1 \in R^- \overset{(2.4.12)}{\implies} p_1 \in \mathfrak P(R) \overset{(2.4.1)}\implies \lang p \rang_I = p_1R \in \mathfrak P_I(R) \overset{(2.5.5.b)}\implies \\ (p_1R)[x] \in \mathfrak P_I(R[x]) \overset{(2.4.8)}{\implies} R[x]/(p_1R)[x] \in \mathcal R^{\mathcal {ID}} \overset{(2.5.5.a)}\implies \\ (R/p_1R)[x] \in \mathcal R^{\mathcal {ID}} \\$$

Below we'll use the notation (mod $p_1$) to denote that only its coefficients being taken modulo $p_1$. It's important to note that this doesn't change the polynomial's function in terms of $x$; it remains a part of $R[x]$, the ring of polynomials over $R$.

$$d \equiv 0 \pmod{p_1} \implies a'(x)b'(x)=dp(x) \equiv 0 \pmod {p_1} \overset{(R/p_1R)[x] \in \mathcal R^{\mathcal {ID}}}{\implies} \\ (a'(x) \equiv 0) \vee(b'(x) \equiv 0) \pmod {p_1}$$

Without loss of generality, assume $a'(x) \equiv 0 \pmod {p_1}$. Then it means that if $a'(x)=a_0 + \ldots + a_n x^n$, then $\forall i: p_1 \mid a_i$, and we can write $d_2p(x)=a_2(x)b(x), d_2 = p_2\cdot \ldots \cdot p_n$. Continuing this process for other $p_n$, we finish the proof.

b.

Follows from $(a)$

$\square$

Corrolary 2.5.7: Gauss' lemma for n factors

$$\begin{align*} &\sphericalangle \\ &R \in \mathcal R ^{\mathcal {UFD}} \\ &F:=\mathfrak F(R) \\ &p \in R[x] \\ &p(x) = p_1(x)\ldots p_k(x), p_i(x) \in F[x]\setminus F\\ \hline \\ &\exists q_1(x), \ldots, q_k(x) \in R[x], r_1, \ldots r_k \in F: \\ &\begin{align*} & p(x)=q_1(x)\ldots q_k(x), q_i(x)=r_ip_i(x) \hspace{1cm} \tag{a}\\ & p_i(x) \in F[x]^- \implies q_i(x) \in F[x]^- \tag{b}\\ & p_i(x) \in F[x]^+ \implies q_i(x) \in F[x]^+ \tag{c}\\ \end{align*} \end{align*}$$

Proof

a.

If $k=1$, then obviously $q_1(x)=r_1p_1(x)$ and we're done.

If $k=n$, consider $p_{12}(x):=p_1(x)p_2(x)$. By $(2.5.6)$ $\exists q_1(x)q_2(x) \in R[x]: p_{12}(x)=q_1(x)q_2(x)$. Thus we can write $p(x)=q_1(x)q_2(x)p_3(x)\ldots p_k(x)$. Now note that by $(2.5.6)$: $q_2(x) = rp_2(x), r \in F, p_2(x) \in F[x] \setminus F \implies q_2(x) \in F[x] \setminus F$. So we can continue with $p_{23}(x)=q_2(x)p_3(x)$.

Continuing this process, we finish the proof

b, c.

Follows trivially from $(a)$ because $q_i(x)=c_ip_i(x), c_i \in F$.

$\square$

Corrolary 2.5.8: Irreducibility for primary polynomials

$$\begin{align*} &\sphericalangle \\ &R \in \mathcal R ^{\mathcal {UFD}} \\ &F:=\mathfrak F(R) \\ &a(x) \in R[x] \setminus R, a(x) = \sum_i a_ix^i \\ &\gcd(a_0, \ldots, a_n) = 1 \\ \hline \\ &a(x) \in R[x]^- \iff a(x) \in F[x]^- \end{align*}$$

Proof

Remember, from the note to the irreducible element defintion, we have $F[x] = \{0\} \cup F^* \cup F[x]^- \cup F[x]^+, R[x]= \{0\} \cup R^* \cup R[x]^- \cup R[x]^+$ and all sets are disjoint.

Thus, the statement $a(x) \in R[x]^- \iff a(x) \in F[x]^-$ is equivalent to proving:

1. $a(x) \in R^* \cup \{0\} \iff a(x)\in F^* \cup \{0\}$
2. $a(x) \in R[x]^+ \iff a(x)\in F[x]^+$

We don't need to prove the first one, as the theorem preconditions state that we only consider $a(x) \in R[x] \setminus R$.

Thus, all we need to prove is $a(x) \in R[x]^+ \iff a(x)\in F[x]^+$

$\implies$

It's obvious that:

$\forall p(x) \in R[x] \setminus R \implies p(x) \in F[x] \setminus F$

Now we have:

$$a(x) \in R[x]^+ \implies \\ \exists b(x), c(x) \in R[x] \setminus (R^* \cup \{0\}): a(x) = b(x)c(x) \\$$

From our theorem preconditions $a(x) \in R[x] \setminus R$, as a result - $a(x) \in F[x] \setminus F$.

Next we want to prove that $b(x), c(x) \in R[x] \setminus R$. Assume it's not true for $b(x)$, then it will mean that $b(x)$ is constant and $b(x) = b \in R \setminus R^* \cup \{0\}$. But that means that $a(x) = b c(x), b \notin B^* \cup \{0\}$ which contradicts the precondition $\gcd(a_0, \ldots, a_n) = 1$.

Thus $b(x), c(x) \in R[x] \setminus R$ and so $b(x), c(x) \in F[x] \setminus F$.

Finally, taking into account that $F = F^* \cup \{0\}$ we can write:

$$a(x) \in R[x] \setminus R, \exists b(x), c(x) \in R[x] \setminus R: a(x) = b(x)c(x) \implies \\ a(x) \in F[x] \setminus F, \exists b(x), c(x) \in F[x] \setminus F: a(x) = b(x)c(x) \implies \\ a(x) \in F[x]^+$$

$\impliedby$

Follows from $(2.5.6.b)$

$\square$

Proposition 2.5.9 Polynomials over UFD is UFD

$$R \in \mathcal R^{\mathcal {UFD}} \implies R[x] \in \mathcal R^{\mathcal {UFD}}$$

Proof

note

Only for this proof, we'll adopt the notation $\gcd a(x)=\gcd(a_0, \ldots, a_n)$, where $a(x)=a_0+\ldots+a_nx^n$.

Consider $a(x) \in R[x] \setminus (R^* \cup \{0\})$ and define

$$b(x):= a(x) / \gcd(a(x))$$

Now note that since $R \in \mathcal R^{\mathcal {UFD}}$, $\gcd a(x)$ can be uniquely factored into irreducibles in $R$ and thus irreducibles in $R[x]$. Thus, all we need to prove that $b(x)$ with $\gcd b(x)=1$ can be uniquely factored into irreducibles. Then $a(x) = b(x)\gcd a(x)$ both of which are factored into irreducibles.

Existence

Let's make a notation $F:= \mathfrak F(R)$. It's clear that should the $b(x) \in R$, it's uniquely factorized because $R \in \mathcal R^{\mathcal {UFD}}$ and we're done. So we consider the case $b(x) \in R[x] \setminus R$. Since $b(x)$ is non-constant in $R[x]$, it's also non-constant in $F[x]$, in other words:

$$b(x) \in R[x] \setminus R \implies b(x) \in F[x] \setminus F \\$$

Next,

$$(2.5.12),(2.4.17), (2.4.18) \implies F[x] \in \mathcal R^{\mathcal {UFD}} \implies \\ \exists p_i(x) \in F[x]^-: b(x) = p_1(x)\ldots p_k(x) \overset{(2.5.7)}\implies \\ \exists q_1(x),\ldots,q_k(x) \in R[x] \cap F[x]^-: b(x) = q_1(x)\ldots q_k(x)$$

Note that $q_i(x) \in F[x]^-$ implies it's nonconstant. In particular, $q_i(x) \in R[x] \setminus R$. Next, it's obvious that $\forall i: \gcd q_i(x) \ne 1 \implies \gcd b(x) \ne 1$. So we have:

$$\gcd b(x) = 1 \implies \forall i: \gcd q_i(x) = 1 \overset{(2.5.8), q_i \in F[x]^-, q_i \in R[x]\setminus R}\implies \forall i: q_i(x) \in R[x]^-$$

Uniqueness

Assume $b(x) = p_1(x)\ldots p_r(x)=q_1(x)\ldots q_s(x), q_i, p_j \in R[x]^-$.

$$\gcd b(x)=1 \implies \gcd q_i = 1, \gcd p_j = 1 \overset{(2.5.8)}{\implies} q_i \in F[x]^-, p_j \in F[x]^- \overset{F[x] \in \mathcal R^{\mathcal {UFD}}}{\implies} \\ r = s, q_i(x)=c_ip_j(x), c_i \in F \implies a_iq_i(x) = b_ip_j(x)=m(x), a_i, b_i \in R$$

So we have $\gcd(m_0, \ldots, m_n)=a_i, \gcd(m_0, \ldots, m_n)=b_i$ for coefficents of a polynomial $m(x)$. By $(2.4.11)$, $a_i = ub_i, u \in R^* \implies p_j(x)=uq_i(x)$.

$\square$

Polynomials over Noetherian rings

Proposition 2.5.10 Polynomials over Noetherian ring is Noetherian ring (Hilbert's Basis Theorem)

$$\begin{align*} &\sphericalangle \\ & R \in \mathcal R^{\mathcal N} \\ \hline \\ &R[x] \in \mathcal R^{\mathcal N} \end{align*}$$

Proof

We'll use the following notation for this proof: $LC_I^{-1}(a)$ means arbitrary polynomial $f$ from set $I$ with leading coefficent $a$.

Consider some ideal $I \lhd_R R[x]$. If we prove that it's finitely generated then by $(2.4.9)$ $R[x] \in \mathcal R^{\mathcal N}$.

First, consider a set $L:=\{LC(f), f \in I\}$. We want to prove that $L \lhd_R R$:

$$0 \in I \implies 0 \in L \implies L \ne \empty \\ \forall a, b \in L, r \in R: \\ f(x):= LC^{-1}_I(a), g(x):= LC^{-1}_I(b), d:= \deg f, e:=\deg g,\\ h(x):=rx^ef(x)-x^dg(x) \implies \\ LT(h)=rx^eLT(f(x))-x^dLT(g(x)) \implies LC(h)=ra-b \\ \begin{cases} ra-b = 0 \overset{0 \in L}\implies ra-b \in L\\ ra-b \ne 0 \implies ra-b=LC(h), h \in I \implies ra-b \in L \end{cases}$$

Thus we proved that $L \lhd_R R$.

$$L \lhd_R R, R \in \mathcal R^{\mathcal N} \overset{(2.4.9)}\implies \exists a_1, \ldots, a_n \in L: \lang a_1, ..., a_n\rang_I = L \\ f_i :=LC_I^{-1}(a_i), e_i:=\deg f_i, N:=\max_i e_i \\ \forall d \le N-1: L_d:=\{LC(f), f \in I, \deg f = d\} \cup \{0\}$$

Similarly to $L \lhd_R R$ we can prove $L_d \lhd_R R$.

$$L_d \lhd_R R, R \in \mathcal R^{\mathcal N} \overset{(2.4.9)}\implies \exists b_{d,1}, \ldots, b_{d,n_d} \in L_d: \\ \lang b_{d,1}, \ldots, b_{d,n_d}\rang_I = L_d \\ f_{d, i} :=LC_I^{-1}(b_{d,i})$$

Consider $I_1:= \lang \{f_1, \ldots, f_n\} \cup \{f_{d,i}, 0\le d<N, 1 \le i \le n_d\}\rang_I$ and let's prove that $I = I_1$.

Since every polynomial in $I_1$ is by design from $I$, we have $I_1 \subseteq I$. Assume $I_1 \ne I$, define:

$$f:= \argmin_{\deg h} I \setminus I_1, d:= \deg f, a:=LC(f)$$

We want to build some polynomial $g$ with the following properties:

$$g \in I_1, \deg g = \deg f, LC(f) = LC(g)$$

Assume $d \ge N$:

$$f \in I \implies a \in L \overset{L \lhd_R R, R \in \mathcal R^{\mathcal N}}\implies \exists r_1, \ldots r_n \in R:a = r_1a_1 + \ldots + r_na_n \\ g:= r_1x^{d-e_1}f_1+\ldots+r_nx^{d-e_n}f_n \implies \\ g \in I_1, \deg g = \deg f = d, LC(f) = a = r_1a_1 + \ldots + r_na_n = LC(g)$$

Assume $d < N$:

$$\exists d: a \in L_d \implies \exists r_1, \ldots r_n \in R: a = r_1b_{d, 1}+\ldots+r_nb_{d, n} \implies \\ g:=r_1f_{d, 1} + \ldots + r_{n_d}f_{d, n_d} \implies \\ g \in I_1, \deg g = \deg f, LC(f) = LC(g)$$

Now that we constructed a polynomial $g$, let's see why it's existence is a contradiction to the assumption $I \ne I_1$.

If we had $f-g \in I_1$ then $g\in I_1 \implies f \in I_1$, but we know that $f \in I \setminus I_1$, so this is a contradiction. So we established that $f-g \notin I_1$.

On the other hand $f-g \in I$, so we have $f-g \in I \setminus I_1$. But since $LC(f)=LC(g)$, we have $\deg (f-g) < \deg f$ which is a contradiction to the definition of $f$.

Thus $I=I_1$ and $I_1$ is finitely generated so by $(2.4.9)$ we finish the proof.

$\square$

Polynomials over fields

Proposition 2.5.11 Division for polynomials over field

$$\begin{align*} &\sphericalangle \\ &F \in \mathcal F\\ &a(x), b(x) \in F[x] \\ \hline \\ &\exists! q(x), r(x) \in F[x]: \\ &a(x) = q(x)b(x)+r(x), (r(x) = 0) \vee (\deg r < \deg b) \end{align*}$$

Proof

Existence:

If $a(x)=0$, we take $q(x) =0, r(x)= 0$.

We'll make a proof inductively on $n: = \deg a(x)$. Assume $a(x)=a_0 + a_1x + \ldots + a_nx^n, b(x)=b_0 + b_1x + \ldots + b_mx^m$. If $n < m$ then $q(x)=0, r(x)=a(x)$ and we're done.

Now $n\ge m$, define $a'(x):=a(x)-a_n/b_mx^{n-m}b(x)$ Since $\deg a' < \deg a$, by induction:

$$a'(x)=q'(x)b(x) + r(x), (r(x) = 0) \vee (\deg r < \deg b) \implies \\ a(x)-a_n/b_mx^{n-m}b(x) = q'(x)b(x) + r(x) \implies \\ a(x) = (q'(x)+a_n/b_mx^{n-m})b(x)+r(x)$$

Uniqueness:

$$a(x)=q_1(x)b(x)+r_1(x)=q_2(x)b(x)+r_2(x) \implies \\ \deg (a(x)-q_1(x)b(x)) < \deg b, \\ \deg (a(x)-q_2(x)b(x)) < \deg b \implies \\ \deg b(x)(q_1(x) - q_2(x)) < \deg b \implies \\ \deg (q_1(x) - q_2(x)) < 0 \implies q_1(x) = q_2(x), r_1(x) = r_2(x)$$

$\square$

Proposition 2.5.12 Polynomials over field is an Euclidean domain

$$\begin{align*} &\sphericalangle \\ &F \in \mathcal F\\ \hline \\ &F[x] \in \mathcal R^{\mathcal E} \end{align*}$$

Proof

We define a norm for this euclidean domain as $\forall f \in F \setminus \{0\}: N(f) := \deg f, N(0):=0$. (2.5.11) completes the proof.

Multivariate polynomials

We can define polynomial in many variables in two ways. First, stating the form of polynomial explicitly:

$$f(x_1, ..., x_n)=\sum_i a_ix_1^{d_{i, 1}}x_2^{d_{i, 2}}\ldots x_n^{d_{i, n}}$$

The other way we can define the ring of polynomials in many variable inductively by $R[x_1, ..., x_n]=R[x_1, ..., x_{n-1}][x_n]$. The latter form meaning we're considering polynomials in a variable $x_n$, where the coefficients are in the ring $R[x_1, \ldots, x_{n-1}]$.

Using these defintions, we'll sum up all the propositions from the previous section in the following table:

Proposition 2.5.13: Relations of base fields and polynomial fields

$$\def\arraystretch{1.5} \begin{array}{c:c:c} R & R[x] & R[x_1, \ldots, x_n] \\ \hline \mathcal R^{\mathcal{ID}} & \mathcal R^{\mathcal{ID}} & \mathcal R^{\mathcal{ID}} \\ \hdashline \mathcal R^{\mathcal{UFD}} & \mathcal R^{\mathcal{UFD}} & \mathcal R^{\mathcal{UFD}} \\ \hdashline \mathcal R^{\mathcal{N}} & \mathcal R^{\mathcal{N}} & \mathcal R^{\mathcal{N}} \\ \hdashline \mathcal F & \mathcal R^{\mathcal E}, \mathcal R^{\mathcal {PID}} & R^{\mathcal{UFD}} \\ \end{array}$$