Definition 1.9.1 A domain $R$ is a unique factorization domain (UFD) or factorial ring if

1. every $r\in R$, neither $0$ or a unit, is a product of irreducibles;
2. if $p_1\cdots p_m=q_1\cdots q_n$ where all $p_i$ and $q_j$ are irreducible, then $m=n$ and there is a permutation $\sigma \in S_n$ with $p_i$ and $q_{\sigma (i)}$ associates for all $i$.

Proposition 1.9.1 Let $R$ be a domain in which every $r\in R$, neither $0$ nor a unit, is a product of irreducibles. Then $R$ is a UFD if and only if $(p)$ is a prime ideal in $R$ for every irreducible element $p\in R$.

Proof.

• ($\Rightarrow$).

  Let $p\in R$ be an irreducible element.

  $\mathrm{\forall }ab\in (p)$, assume that $a\notin (p)$.

  For $ab\in (p)$, $\mathrm{\exists }r\in R$ with $ab=rp$.

  Then let $a=p_1\cdots p_n,b=q_1\cdots q_m,$ and $r=r_1\cdots r_t$ with $p$’s, $q$’s, and $r$’s are irreducible.

  $p_1\cdots p_n{q}_1\cdots q_m=r_1\cdots r_{t}p$, then $\mathrm{\exists }j$ such that $q_j$ and $p$ associates.

  Therefore $b\in (p)$, and $(p)$ is a prime ideal.

• ($\Leftarrow$).

  Let $a\in R$ with $a=p_1\cdots p_n=q_1\cdots q_m$ in which $p$’s and $q$’s are irreducible.

  By induction on $M=max\{m,n\}$.

  If $M=1$, we have $p_1=q_1$.

  For inductive step, because $q_1\cdots q_m\in (p_n)$, $\mathrm{\exists }j,q_j\in (p_n)$; that is $q_j$ and $p_n$ are associates.

  By reindexing, let $q_m=u{p}_n$, where $u$ is a unit.

  Then by cancelling,we have $p_1\cdots p_{n-1}=q_1\cdots (u{q}_{m-1})$.

  Since $q_{m-1}u$ is irreducible, the inductive hypothesis gives $n-1=m-1$ (hence, $m=k$) and, after reindexing, $p_i$ and $q_i$ are associates for all $i$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

Lemma 1.9.1

1. If $R$ is a commutative ring and
   $$I_1\subset I_2\subset \cdots \subset I_n\subset I_{n+1}\subset \cdots$$
   is a ascending chain of ideals in $R$, then $J=\bigcup _{n\ge 1}{I}_n$ is an ideal in $R$.

2. If $R$ is a PID, then it has no infinite strictly ascending chain of ideals
   $$I_1⊊I_2⊊\cdots ⊊I_n⊊I_{n+1}⊊\cdots .$$

3. If $R$ is a PID and $r\in R$ is neither $0$ or a unit, then $r$ is a product of irreducibles.

Proof.

1. • $\mathrm{\forall }x,y\in \bigcup _{n\ge 1}{I}_n$, $\mathrm{\exists }m,k\in {\mathbb{N}}_{+}$ with $x\in I_m,y\in I_k$.

     Let $M=max\{m,k\}$. Then $x,y\in I_M$.

     We have $x-y\in I_M\subset \bigcup _{n\ge 1}{I}_n$.

   • $\mathrm{\forall }x\in \bigcup _{n\ge 1}{I}_n,\mathrm{\exists }m\in {\mathbb{N}}_{+}$ with $x\in I_m$.

     $\mathrm{\forall }r\in R,rx\in I_m\subset \bigcup _{n\ge 1}{I}_n$ for $I_m$ is an ideal.

   Therefore $\bigcup _{n\ge 1}{I}_n$ is an ideal.

2. Assume that $I_1\subset I_2\subset \cdots \subset I_n\subset I_{n+1}\subset \cdots$ is a ascending chain of ideals in $R$.

   For $R$ is a PID, there is $p\in R$ with $\bigcup _{n\ge 1}{I}_n=(p)$.

   Then we have $p\in \bigcup _{n\ge 1}{I}_n$; that is $\mathrm{\exists }m$ with $p\in I_m$.

   Therefore $(p)\subset I_m\subset (p)\Rightarrow I_m=(p)$, so does $I_{m+1}$.

3. A divisor $r$ of an element $a$ is called a proper divisor of $a$ if $r$ is neither a unit nor an associate of $a$.

   If $r$ is a divisor of $a$, then $(a)\subset (r)$; if $r$ is a proper divisor, then $(a)⊊(r)$, for f the inequality if not strict, then $(a)=(r)$, and this force $a$ and $r$ to be associates.

   Call a nonzero non-unit $a\in R$ good, if it is a product of irreducibles; call it bad otherwise.

   If $a$ is bad, it is not irreducible, and so $a=rs$, where $r$ and $s$ are proper divisors. But the product of good elements is good, and so at least one of the factors, say $r$, is bad. We can get $(a)⊊(r)$. It follows, by induction, that there exists a sequence $a_1=1,a_2=r,a_3,\dots ,a_n,\dots$ of bad elements with each $a_{n+1}$ a proper divisor of $a_n$, and this sequence yields a strictly ascending chain
   $$(a_1)⊊(a_2)⊊\cdots ⊊(a_n)⊊(a_{n+1})⊊\cdots ,$$
   contradicting (2). $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

Theorem 1.9.1 Every PID is a UFD.

Proposition 1.9.2 If $R$ is a UFD, then a $gcd(a_1,\dots ,a_n)$ of any finite set of elements $a_1,\dots ,a_n$ in $R$ exists.

Proof.

We prove first that a gcd of two elements $a$ and $b$ exists. There are distinct irreducibles $p_1,\dots ,p_t$ with
$$a={p_1}^{e_1}{p}_2^{e_2}\cdots p_t^{e_t}\text{ and }b=p_1^{f_1}{p}_2^{f_2}\cdots p_t^{f_t},$$
where $e_i\ge 0$ and $f_i\ge 0$ for all $i$. It is easy to see that if $c\mid a$, then the factorization of $c$ into irreducibles is $c=w{p}_1^{g_1}\cdots p_t^{g_t}$, where $0\le g_i\le e_i$ for all $i$ and $w$ is a unit. Thus, $c$ is a common divisor of $a$ and $b$ if $g_i<m_i$ for all $i$, where $m_i=min\{e_i,f_i\}$.

It is now clear that $p_1^{m_1}{p}_2^{m_2}\cdots p_t^{m_t}$ is a gcd of $a$ and $b$.

More generally, if $a_i=u_i{p}_1^{e_{i1}}\dots p_t^{e_{it}}$, where $e_{ij}\ge 0$ and $i=1,\dots ,n$ and $u_i$ are units, then
$$d=p_1^{{\mu }_1}\cdots p_t^{{\mu }_t}$$
is a gcd of $a_1,\dots ,a_n$, where ${\mu }_j=min\{e_{1j},\dots ,e_{nj}\}$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

Proposition 1.9.3 Let $R$ be a UFD, and let $a_1,\dots ,a_n$ in $R$. An lcm of $a_1,\dots ,a_n$ exists, and
$$a_1\cdots a_n=gcd(a_1,\dots ,a_n)\mathrm{lcm}(a_1,\dots ,a_n).$$
Definition 1.9.2 A polynomial $f(x)=a_n{x}^n+\cdots +a_{1}x+a_0\in R[x]$, where $R$ is a UFD, is called primitive if its coefficients are relatively prime; that is, the only common divisors of $a_n,\dots ,a_1,a_0$ are units.

We can see that every monic polynomial is primitive. Observe that if $f(x)$ is not primitive, then there exists an irreducible $q\in R$ that divides each of its coefficients: if the gcd is a non-unit $d$, then take for $q$ any irreducible factor of $d$.

Lemma 1.9.2 (Gauss) If $R$ is a UFD and $f(x),g(x)\in R[x]$ are both primitive, then their product $fg$ is also primitive.

Proof.

If $fg$ is not primitive, there is an irreducible $p\in R$ which divides all its coefficients.

Let $P=(p)$ and let $\pi :R\to R/P$ be the natural map $a\mapsto a+P$.

By replacing each coefficient $c$ of a polynomial by $\pi (c)$, we have the homomorphism $\tilde{\pi }:R[x]\to (R/P)[x]$.

Now $\tilde{\pi }(fg)=0$ in $(R/P)[x]$. Since $P$ is a prime ideal, both $R/P$ and $(R/P)[x]$ are domains.

But neither $\tilde{\pi }(f)$ nor $\tilde{\pi }(g)$ is $0$ in $(R/P)[x]$, because $f$ and $g$ are primitive, and this contradicts $(R/P)[x]$ being a domain. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

Proposition 1.9.4 Let $R$ be a UFD and let $Q=\mathrm{Frac}(R)$ be its fraction field. Each nonzero $a/b\in Q$ has an expression in lowest terms; that is, $a$ and $b$ are relatively prime.

Proof.

Since $R$ is a UFD, there is a gcd of $a$ and $b$ in $R$. Let $d$ be a gcd of $a$ and $b$.

Then there are $u,v\in R$ with $a=ud,b=vd$ and $u,v$ are relatively prime.

Therefore $u/v\in Q$ is the expression in lowest term of $a/b$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

Proposition 1.9.5 Let $R$ be a UFD. If $a,b,c\in R$ and $a$ and $b$ are relatively prime. Then $a\mid bc$ implies $a\mid c$.

Proof.

Since $R$ is a UFD, there are irreducibles $p_1,\dots ,p_t$ with $a=p_1\cdots p_t$.

We will show $c=u{p}_1\cdots p_t$ for some $u\in R$.

Let’s prove by induction on $t$.

When $t=1$, since $a=p_1$ is an irreducible and $p_1\nmid b$, $p_1\mid c$.

Otherwise, $(p_t)$ is a prime ideal in $R$, then $c\in (p_t)$ for $bc\in (p_t)$ and $b\notin (p_t)$.

Therefore, $c=c^{\prime }{p}_t$ with some $c^{\prime }\in R$. By cancelling, we have $p_1\cdots p_{t-1}\mid bc$.

By inductive suppose we have $c^{\prime }=u{p}_1\cdots p_{t-1}$. Therefore $c=c^{\prime }{p}_t=u{p}_1\cdots p_t=ua$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

Lemma 1.9.3 Let $R$ be a UFD, let $Q=\mathrm{Frac}(R)$, and let $f(x)\in Q[x]$ be nonzero.

1. There is a factorization
   $$f(x)=c(f)f^{\ast }(x),$$
   where $c(f)\in Q$ and $f^{\ast }\in R[x]$ is primitive. This factorization is unique in the sense that if $f(x)=q{g}^{\ast }(x)$, where $q\in Q$ and $g^{\ast }\in R[x]$ is primitive, then there is a unit $w\in R$ with $q=wc(f)$ and $f^{\ast }=w{g}^{\ast }$.

2. If $f(x),g(x)\in R[x]$, then $c(fg)$ and $c(f)c(g)$ are associates in $R$ and $(fg{)}^{\ast }$ and $f^{\ast }{g}^{\ast }$ are associates in $R[x]$.

3. Let $f(x)\in Q[x]$ have a factorization $f=q{g}^{\ast }$, where $q\in Q$ and $g^{\ast }\in R[x]$ is primitive. Then $f\in R[x]$ if and only if $q\in R$.

4. Let $g^{\ast },f\in R[x]$. If $g^{\ast }$ is primitive and $g^{\ast }\mid bf$, where $b\in R$ and $b\ne 0$, then $g^{\ast }\mid f$.

Proof.

1. Clearing denominators, there is $b\in R$ with $bf\in R[x]$. If $d$ is the gcd of the coefficients of $bf$, then $f^{\ast }(x)=(b/d)f\in R[x]$ is a primitive polynomial. If we define $c(f)=d/b$, then $f=c(f)f^{\ast }$.

   To prove uniqueness, suppose that $c(f)f^{\ast }=f=q{g}^{\ast }$, where $c(f),q\in Q$ and $f^{\ast }(x),g^{\ast }(x)\in R[x]$ are primitive. Proposition 1.9.4 allows us to write $q/c(f)$ in lowest terms: $q/c(f)=u/v$, where $u$ and $v$ are relatively prime elements of $R$.

   The equation $v{f}^{\ast }(x)=u{g}^{\ast }(x)$ holds in $R[x]$. Since $u$ and $v$ are relatively prime, Proposition 1.9.5 says that $v$ is a common divisor of all the coefficients of $g^{\ast }$. But $g$ is primitive, and so $v$ is a unit.

   A similar argument shows that $u$ is a unit. Therefore, $q/c(f)=u/v$ is a unit in $R$, call it $w$; we have $q=wc(f)$ and $f^{\ast }=w{g}^{\ast }$.

2. There are two factorizations of $f(x)g(x)$ in $R[x]$:
   $$\begin{array}{rl}fg& =c(fg)(fg{)}^{\ast },\\ fg& =c(f)f^{\ast }c(g)g^{\ast }=c(f)c(g)f^{\ast }{g}^{\ast }.\end{array}$$
   Since the product of primitive polynomials is primitive, each of these is a factorization as in (1); the uniqueness assertion there says that $c(fg)$ is an associate of $c(f)c(g)$ and $(fg{)}^{\ast }$ is an associate of $f^{\ast }{g}^{\ast }$.

3. If $q\in R$, then it is obvious that $f=q{g}^{\ast }\in R[x]$. Conversely, if $f(x)\in R[x]$, then there is no need to clear denominators, and so $c(f)=d\in R$, where $d$ is the gcd of the coefficients of $f(x)$. Thus $f=d{f}^{\ast }$. By uniqueness, there is a unit $w\in R$ with $q=wd\in R$.

4. Since $bf=h{g}^{\ast }$, we have $bc(f)f^{\ast }=c(h)h^{\ast }{g}^{\ast }=c(h)(hg{)}^{\ast }$. By uniqueness, $f^{\ast }$, $(hg{)}^{\ast }$ and $h^{\ast }{g}^{\ast }$ are associates, and so $g^{\ast }\mid f^{\ast }$, But $f=c(f)f^{\ast }$, and so $g^{\ast }\mid f$.

Definition 1.9.3 Let $R$ be a UFD with $Q=\mathrm{Frac}(R)$. If $f(x)\in Q[x]$, there is a factorization $f=c(f)f^{\ast }$, where $c(f)\in Q$ and $f^{\ast }\in R[x]$ is primitive. We call $c(f)$ the content of $f$ and $f^{\ast }$ the associated primitive polynomial.

Corollary 1.9.1 Let $k$ be a field, and let
$$f(x,y)=y^n+\frac{g_{n-1}(x)}{h_{n-1}(x)}{y}^{n-1}+\cdots +\frac{g_0(x)}{h_0(x)}\in k(x)[y],$$
where each $g_i/h_i$ is in lowest terms. If $f^{\ast }(x,y)\in k[x][y]$ is the associated primitive polynomial of $f$, then
$$\underset{i}{max}\{\mathrm{deg}(g_i),\mathrm{deg}(h_i)\}\le {\mathrm{deg}}_x(f^{\ast })\text{ and }n={\mathrm{deg}}_y(f^{\ast }),$$
where ${\mathrm{deg}}_x(f^{\ast })$ (or ${\mathrm{deg}}_y(f^{\ast })$) is the highest power of $x$ (or $y$) occurring in $f^{\ast }$.

Proof.

Let $l$ be the lcm of $h_0,\dots ,h_{n-1}\in k[x]$. Then, the associated primitive polynomial is
$$f^{\ast }(x,y)=lf(x,y)=l{y}^n+l\frac{g_{n-1}}{h_{n-1}}+\cdots +l\frac{g_0}{h_0}\in k[x,y].$$
Since $l$ is the lcm, there are $u_i\in k[x]$ with $l=u_i{h}_i$ for all $i$. Hence, each coefficient $c(g_i/h_i)=u_i{g}_i\in k[x]$. If $m={\mathrm{deg}}_x(f^{\ast })$, then
$$m=max\{\mathrm{deg}(l),\mathrm{deg}(c(g_i/h_i))\}=max\{\mathrm{deg}(l),\mathrm{deg}(u_i{g}_i)\},$$
for $l$ is a coefficient of $f^{\ast }$. Now $h_i\mid l$ for all $i$, so that $\mathrm{deg}(h_i)\le \mathrm{deg}(l)\le m$.

Also $\mathrm{deg}(g_i)\le \mathrm{deg}(u_i{g}_i)\le m$. We conclude that $\underset{i}{max}\{\mathrm{deg}(g_i),\mathrm{deg}(h_i)\}\le m={\mathrm{deg}}_x(f^{\ast })$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

Theorem 1.9.1 (Gauss) If $R$ is a UFD, then $R[x]$ is also a UFD.

Proof.

• First, let’s prove that every $f(x)\in R[x]$, neither zero nor a unit, is a product of irreducibles by induction.

  The base step $\mathrm{deg}(f)=0$ is true, because $f$ is a constant, hence lies in $R$, and hence is a product of irreducibles (for $R$ is a UFD).

  For the inductive step $\mathrm{deg}(f)>0$, we have $f=c(f)f^{\ast }$, where $c(f)\in R$ and $f^{\ast }(x)$ is primitive.

  Now $c(f)$ is either a unit or a product of irreducibles, by the base step.

  If $f^{\ast }$ is irreducible, we are done. Otherwise, $f^{\ast }=gh$, where neither $g$ nor $h$ is a unit.

  Since $f^{\ast }$ is primitive, however, neither $g$ nor $h$ is a constant; therefore, each of these has degree less than $\mathrm{deg}(f^{\ast })=\mathrm{deg}(f)$.

  By indective hypothesis, each of $g$ and $h$ is a product of irreducibles. Therefore, $f$ is a product of irreducibles.

• Then, let’s prove $R[x]$ is a UFD by Proposition 1.9.1; that is let $p$ be an irreducible, then $(p)$ is a prime ideal in $R[x]$.

  Let $fg\in (p)$.

  • Suppose that $\mathrm{deg}(p)=0$.

  Now $f=c(f)f^{\ast }(x),g=c(g)g^{\ast }(x)$, where $f^{\ast },g^{\ast }$ are primitive and $c(f),c(g)\in R$.

  Since $p\mid fg$, we have $p\mid c(f)c(g)f^{\ast }{g}^{\ast }$. Write $f^{\ast }{g}^{\ast }=\sum _i{a}_i{x}^i$, where $a_i\in R$, so that $p\mid c(f)g(f)a_i$ in $R$ for all $i$.

  Now $f^{\ast }{g}^{\ast }$ is primitive, so there is some $i$ with $p\nmid a_i$ in $R$. For $R$ is a UFD, $p\mid c(f)$ or $p\mid c(g)$.

  Thus, $p\mid f$ or $p\mid g$.

  • Suppose that $\mathrm{deg}(p)>0$ and $p\nmid f$.

  Let $(p,f)=\{s(x)p(x)+t(x)f(x);s(x),t(x)\in R[x\}$; then $(p,f)$ is an ideal in $R[x]$ containing $p$ and $f$.

  Choose $m(x)\in (p,f)$ of minimal degree. If $Q=\mathrm{Frac}(R)$ is the fraction field of $R$, then the division algorithm in $Q[x]$ gives polynomials $q(x),r(x)\in R[x]$ with
  $$f=m{q}^{\prime }+r^{\prime },$$
  where either $r^{\prime }=0$ or $\mathrm{deg}(r^{\prime })<\mathrm{deg}(m)$. Clearing denominators, there is a constant $b\in R$ and polynomials $q(x),r(x)\in R[x]$ with
  $$bf=qm+r,$$
  where $r=0$ or $\mathrm{deg}(r)<\mathrm{deg}(m)$.

  Since $m\in (p,f)$, there are polynomials $s(x),t(x)\in R[x]$ with $n=sp+tf$; hence $r=bf-qm\in (q,f)$. Since $m$ has minimal degree in $(p,f)$, we must have $r=0$; that is $bf=mq$, and so $bf=c(m)m^{\ast }q$. But $m^{\ast }$ is primitive, and $m^{\ast }\mid bf$, so that $m^{\ast }\mid f$ for Lemma 1.9.3(4).

  A similar argument, replacing $f$ by $p$, we have $m^{\ast }\mid p$. Since $p$ is irreducible, we have $m^{\ast }$ is a unit or $m^{\ast }$ and $p$ are associates. If $m^{\ast }$ were an associate of $p$, then $p\mid f$ contrary to out assumption that $p\nmid f$. Thus $m^{\ast }$ must be a unit. Therefore, $m\in R$.

  We have $mg=(sf+pt)g=sfg+ptg$, so that $p\mid mg$. Since $p$ is an irreducible, we have $p$ is primitive. By Lemma 1.9.3, we have $p\mid g$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

Corollary 1.9.2 If $k$ is a field, then $k[x_1,\dots ,x_n]$ is a UFD.

Corollary 1.9.3 If $k$ is a field, then $p=p(x_1,\dots ,x_n)\in k[x_1,\dots ,x_n]$ is irreducible if and only if $(p)$ is a prime ideal in $k[x_1,\dots ,x_n]$.

Corollary 1.9.4 (Gauss’s Lemma) Let $R$ be a UFD, let $Q=\mathrm{Frac}(R)$, and let $f(x)\in R[x]$. If $f=GH$ in $Q[x]$, then there is a factorization
$$f=gh\text{ in }R[x],$$
where $\mathrm{deg}(g)=\mathrm{deg}(G)$ and $\mathrm{deg}(h)=\mathrm{deg}(H)$; in fact, $G$ is a constant multiple of $g$ and $H$ is a constant multiple of $h$. Therefore, if $f$ does not factor into polynomials of smaller degree in $R[x]$, then $f$ is irreducible in $Q[x]$.

Proof.

By Lemma 1.9.3(1), the factorization $f=GH$ in $Q[x]$ gives $q,q^{\prime }\in Q$ with
$$f=q{G}^{\ast }{q}^{\prime }{H}^{\ast }\text{ in }Q[x],$$
where $G^{\ast },H^{\ast }\in R[x]$ are primitive; thus, $G^{\ast }{H}^{\ast }$ is primitive.

Since $f\in R[x]$, by Lemma 1.9.3, the equation $f=(q{q}^{\prime })(G^{\ast }{H}^{\ast })$ shows that $q{q}^{\prime }\in R$.

Therefore, a factorization of $f$ in $R[x]$ is $f=(q{q}^{\prime }{G}^{\ast })H^{\ast }$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

Proposition 1.9.6 Let $k$ be a field, and view $f(x_1,\dots ,x_n)\in k[x_1,\dots ,x_n]$ as a polynomial in $R[x_n]$, where $R=k[x_1,\dots ,x_{n-1}]$:
$$f(x_n)=a_0(x_1,\dots ,x_{n-1})+a_1(x_1,\dots ,x_{n-1})x_n+\cdots +a_m(x_1,\dots ,x_{n-1})x_n^m.$$
If $f(x_n)$ is primitive and cannot be factored into two polynomials of lower degree in $R[x_n]$, then $f(x_1,\dots ,x_n)$ is irreducible in $k[x_1,\dots ,x_n]$.

Proof.

Suppose that $f(x_n)=g(x_n)h(x_n)$ in $R[x_n]$; by hypothesis, the degree of $g$ and $h$ in $x_n$ cannot both be less than $\mathrm{deg}(f)$; say, $\mathrm{deg}(g)=0$.

It follows, because $f$ is primitive, that $g$ is a unit in $k[x_1,\dots ,x_{n-1}]$. Therefore, $f(x_1,\dots ,x_n)$ is irreducible in $R[x_n]=k[x_1,\dots ,x_n]$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

Corollary 1.9.5 If $k$ is a field and $g(x_1,\dots ,x_n),h(x_1,\dots ,x_n)\in k[x_1,\dots ,x_n]$ are relatively prime, then $f(x_1,\dots ,x_n,y)=yg(x_1,\dots ,x_n)+h(x_1,\dots ,x_n)$ is irreducible in $k[x_1,\dots ,x_n,y]$.