Every finite-variable polynomial ring over a field is a UFD, with prime irreducibles and principal height-one primes

For $r=0$ the ring is the field $k$ by [L6]. Every nonzero element of a field is a unit by [L7], so there are no nonzero nonunits at all; the existence and uniqueness clauses of [L1] hold vacuously, and by [L2] there is no irreducible element, so the statement about irreducibles is vacuous too. This is the base case of the induction on $r$.

Let $R$ be any UFD and $p\in R$ irreducible. Then $p$ is prime: if $p\mid ab$ with $a,b\mathrm{\ne }0$, factor $a={\prod }_i{a}_i$, $b={\prod }_j{b}_j$ and write $ab=pc$ with $c\mathrm{\ne }0$ factored as ${\prod }_k{c}_k$, as [L1] allows whenever the element in question is a nonzero nonunit, the unit cases being immediate from $p\mid ab$; comparing the two products of irreducibles for $ab$ and $pc$ by the uniqueness clause of [L1] shows that $p$ is associate to one of the $a_i$ or one of the $b_j$, hence divides $a$ or $b$.

Now let $R$ be a UFD, $K=\mathrm{Frac}(R)$ by [L4], and $0\mathrm{\ne }f\in R[x]$. For each associate class $C$ of irreducibles in $R$, let $v_C(a)$ be the exponent of that class in the factorization of a nonzero $a\in R$; it is well-defined by the uniqueness clause of [L1]. Let $\mathcal{C}(f)$ be the finite set of such classes appearing in the factorizations of the nonzero coefficients of $f$, and for each $C\in \mathcal{C}(f)$ choose one representative $p_C$. Set $e_C=\min \{v_C(a):a\mathrm{\ne }0\text{ is a coefficient of }f\}$ and $c(f)={\prod }_{C\in \mathcal{C}(f)}{p}_C^{e_C}$. This finite product is a common divisor of the coefficients, every common divisor divides it up to associates, and $f^{\ast }:=f\mathrm{/}c(f)$ is primitive in the sense of [L3]; thus $f=c(f)f^{\ast }$. For $\lambda \in K^{\times }$, define $v_C(\lambda )=v_C(a)-v_C(b)$ when $\lambda =a\mathrm{/}b$ with nonzero $a,b\in R$; this is well-defined, and $v_C(\lambda )=0$ for every associate class exactly when $\lambda$ is a unit of $R$. Note also that an irreducible $f$ of positive degree is primitive: a nonunit constant $r\in R$ dividing all coefficients would write $f=r\cdot (f\mathrm{/}r)$ with both factors nonunits, since $f\mathrm{/}r$ has positive degree by [L7]. The content $c(f)$ factors into irreducibles of $R$ by [L1], and each such constant is irreducible in $R[x]$, because a factorization of a constant in $R[x]$ has both factors constant by [L7] and so is a factorization in $R$. For the primitive part, [L5] gives a factorization $f^{\ast }=u{g}_1\cdots g_n$ in $K[x]$, with $u\in K^{\times }$ and each $g_j$ irreducible in $K[x]$; if $n=0$, then $f^{\ast }$ is a nonzero constant in $R$ and, being primitive, is a unit of $R$. For each $j$ choose $a_j\in R\setminus \{0\}$ with $a_j{g}_j\in R[x]$ and write $a_j{g}_j=d_j{h}_j$, where $d_j\in R$ is a content and $h_j\in R[x]$ is primitive; then $g_j=(d_j\mathrm{/}{a}_j)h_j$, so $h_j$ is a nonzero scalar multiple of $g_j$ in $K[x]$, hence irreducible in $K[x]$ and in $R[x]$ by [L3]. By [L3] the product $h_1\cdots h_n$ is primitive, and $f^{\ast }=\lambda h_1\cdots h_n$ with $\lambda =u{\prod }_j(d_j\mathrm{/}{a}_j)\in K^{\times }$. Since both $f^{\ast }$ and $h_1\cdots h_n$ are primitive, comparison of the least exponents in each associate class gives $v_C(\lambda )=0$ for every $C$, so $\lambda$ is a unit of $R$. Thus $f$ is a product of irreducibles of $R[x]$.

Assume now that $r\ge 1$ and that $k[x_1,\dots ,x_{r-1}]$ is a UFD in which every irreducible element is prime; this is the induction hypothesis.

For uniqueness, let $p_1\cdots p_m=q_1\cdots q_n$ be two factorizations of the same element $f\mathrm{\ne }0$ of $R[x]$ into irreducibles. Multiply the positive-degree factors of each side together, using the primitivity noted in 1.3; the result is primitive by [L3] on each side, so comparing contents as in 1.3 shows that the two sides' constant factors are associates and that the products of positive-degree factors are associates of one another. Each positive-degree factor is irreducible in $K[x]$ by [L3], and $K[x]$ is a UFD by [L5], so the two products of positive-degree factors agree up to order, associates in $K[x]$, and a unit scalar; that scalar is a unit of $R$ by the exponent comparison of 1.3, so they agree up to order and associates in $R[x]$. The constant factors are products of irreducibles of $R$ and agree up to order and associates by the uniqueness clause of [L1]. Hence $R[x]$ satisfies the uniqueness clause of [L1], and with 1.3 it is a UFD.

Every irreducible element of $R[x]$ is prime. Let $p\in R[x]$ be irreducible. If $\mathrm{deg}p=0$ then $p\in R$ is irreducible in $R$, hence prime in $R$ by 1.2; if $p\mid ab$ in $R[x]$, then the induced map $R[x]\to (R\mathrm{/}(p))[x]$ of [L7] kills $ab$, and $(R\mathrm{/}(p))[x]$ is a domain by [L7] because $R\mathrm{/}(p)$ is a domain for the prime element $p$; so all coefficients of $a$ or all coefficients of $b$ lie in $(p)$, that is, $p\mid a$ or $p\mid b$. If $\mathrm{deg}p\ge 1$, then $p$ is primitive by 1.3, hence irreducible in $K[x]$ by [L3], hence prime in the UFD $K[x]$ by 1.2; if $p\mid ab$ in $R[x]$, then $p\mid ab$ in $K[x]$, so after possibly swapping $a,b$ we have $a=pq$ for some $q\in K[x]$. Write $q=\lambda q^{\ast }$ with $\lambda \in K^{\times }$ and $q^{\ast }\in R[x]$ primitive, by the content construction of 1.3 applied to a polynomial clearing the denominators of $q$; then $a=\lambda (p{q}^{\ast })$, and the product $p{q}^{\ast }$ is primitive by [L3]. Comparing contents in the equality $a=\lambda (p{q}^{\ast })$ shows that $\lambda$ is associate to the content $c(a)\in R$ of step 1.3: clearing the denominators of $\lambda$ by some $u\mathrm{\ne }0$ gives $ua=(\lambda u)(p{q}^{\ast })$, whose left side has content $uc(a)$ and whose right side has content $\lambda u$ up to units because $p{q}^{\ast }$ is primitive, the exponents $v_p$ of step 1.3 being additive in a constant factor. Hence $\lambda \in R$ and $q=\lambda q^{\ast }\in R[x]$, so $a=pq$ with $q\in R[x]$ and $p\mid a$ in $R[x]$.

In any UFD $A$, every prime ideal of height one is generated by an irreducible element. By [L8], $\mathrm{ht}(\mathfrak{p})=1=\dim A_{\mathfrak{p}}$ means that inside $\mathfrak{p}$ there is a strict chain of primes of length one and none of length two; in particular $\mathfrak{p}$ contains a nonzero element, so choose $0\mathrm{\ne }f\in \mathfrak{p}$. Factoring $f$ into irreducibles and using that $\mathfrak{p}$ is prime, some irreducible factor $p$ lies in $\mathfrak{p}$ by [L1] and [L2]; then $p$ is prime by 1.2, so $(p)$ is a nonzero prime ideal contained in $\mathfrak{p}$. Were $(p)⊊\mathfrak{p}$, the strict chain $0⊊(p)⊊\mathfrak{p}$ of primes of $A$ would, by the inclusion-preserving bijection of [L8], give a chain of length two in $\mathrm{Spec}{A}_{\mathfrak{p}}$, contradicting $\dim A_{\mathfrak{p}}=1$. Hence $\mathfrak{p}=(p)$.

By [L6] the ring $k[x_1,\dots ,x_r]$ is the polynomial ring $A[x_r]$ over $A=k[x_1,\dots ,x_{r-1}]$, which is a UFD by step 1.4. Steps 1.3 and 2.1 therefore make $k[x_1,\dots ,x_r]$ a UFD, and step 2.2 shows that every irreducible element of it is prime, using the primitivity results quoted in those steps. So the UFD clause and the irreducible-is-prime clause hold for this $r\ge 1$ whenever they hold for $r-1$.

The base case 1.1 and the induction step 3.1 prove the UFD clause and the irreducible-is-prime clause for every $r\ge 0$. Finally, if $\mathfrak{p}\subseteq k[x_1,\dots ,x_r]$ is a prime ideal of height one, then $k[x_1,\dots ,x_r]$ is a UFD by 3.1, so 2.3 exhibits an irreducible element generating $\mathfrak{p}$; for $r=0$ the case is vacuous, a field having no nonzero prime ideal. This proves all three clauses of the statement.