The integral multivariable Laurent polynomial ring is Noetherian and a UFD

For $r=1$ the statement is [F1]. Suppose $r>1$, put $P_r={\mathrm{\Lambda }}_1[t_2,\dots ,t_r]$, and let $S_r$ be the multiplicative set generated by $t_2,\dots ,t_r$. Clearing negative powers identifies ${\mathrm{\Lambda }}_r$ with $S_r^{-1}{P}_r$.

We prove by induction on the number of added variables that ${\mathrm{\Lambda }}_1[t_2,\dots ,t_r]$ is a UFD. The base is [F1]. For the induction step let $R$ be the preceding UFD and $K$ its fraction field. Every nonzero $f\in R[x]$ is its content $c\in R$ times a primitive polynomial $f_0$: since $f$ has finitely many coefficients, take the product over irreducibles of their minimum exponents in those coefficients. Factor $c$ in $R$ and factor $f_0$ in $K[x]$ using [F5]. Clear denominators in each positive-degree irreducible factor and divide by its content to obtain primitive polynomials $q_1,\dots ,q_m\in R[x]$. By [F4], their product $q$ is primitive and is associated to $f_0$ in $K[x]$. Write this scalar as $a\mathrm{/}b$ with coprime $a,b\in R$, so $b{f}_0=aq$. Any irreducible divisor of $a$ then divides every coefficient of $f_0$, and any irreducible divisor of $b$ divides every coefficient of $q$; primitiveness forces $a$ and $b$ to be units. Thus $f_0$ is associated in $R[x]$ to the product of the $q_i$. Each $q_i$ is irreducible in $R[x]$ by [F4], and the content factors in $R$, so every nonzero nonunit of $R[x]$ factors into irreducibles.

By [F2], $P_r$ is Noetherian because ${\mathrm{\Lambda }}_1$ is Noetherian by [F1]. Its localization $S_r^{-1}{P}_r={\mathrm{\Lambda }}_r$ is Noetherian by [F3].

For uniqueness in the same induction step, contents have unique factorization in $R$. Two factorizations of the primitive part become factorizations in the UFD $K[x]$, so their primitive positive-degree factors pair up up to scalar associates. If primitive $f,g\in R[x]$ satisfy $f=(a\mathrm{/}b)g$ for coprime $a,b\in R$, then $bf=ag$; any irreducible factor of $a$ must divide every coefficient of $f$, and any irreducible factor of $b$ must divide every coefficient of $g$. Primitiveness forces $a,b$ to be units, so the scalar is a unit of $R$. This proves uniqueness in $R[x]$ and completes the induction.

Localizing the UFD $P_r$ at $S_r$ preserves unique factorization: factor a numerator in $P_r$ and discard the irreducible factors that divide an element of $S_r$, since precisely those factors become units. Every remaining irreducible stays prime: if it divides a product of localized fractions, clearing denominators gives divisibility of the product of their numerators in $P_r$, so primality in $P_r$ makes it divide one numerator; it cannot divide a cleared denominator because it was not inverted. Uniqueness follows by clearing denominators and using uniqueness in $P_r$. Thus ${\mathrm{\Lambda }}_r$ is a UFD.

If $u=f\mathrm{/}s\in S_r^{-1}{P}_r$ is a unit, then $fg=sq$ for some $q\in S_r$. Every irreducible factor of $f$ therefore divides a product of powers of $t_2,\dots ,t_r$, so it is associated in $P_r$ to one of these variables. The only units already in $P_r$ are the units of ${\mathrm{\Lambda }}_1$, namely $\pm t_1^k$, by degree in $t_2,\dots ,t_r$ and [F1]. Consequently the units of ${\mathrm{\Lambda }}_r$ are exactly $\pm t_1^{k_1}\cdots t_r^{k_r}$.

Steps 2.1, 3.1 and 4.1 prove Noetherianity, unique factorization, and the stated unit description for every $r\ge 1$.