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Partitioning the roots of $X^n-1$ by their exact orders givesInductively, the product over proper divisors is a monic polynomial in $\mathbb{Z}[X]$. Division of $X^n-1$ by this monic polynomial stays in $\mathbb{Z}[X]$ and has zero remainder, proving that ${\Phi }_n\in \mathbb{Z}[X]$.

An element $x$ of an $A$-algebra is an integral element precisely when $A[x]$ is a finite $A$-module. More generally, a finite $A$-submodule containing one and stable under multiplication by $x$ proves integrality by the determinant trick. Conversely a monic polynomial equation reduces all powers of $x$ to finitely many generators.

A nonzero quotient of an $r$-variable Laurent polynomial ring is finite over an embedded Laurent polynomial ring in $m\le r$ variables. Given a nonzero Laurent relation, choose $N$ so that weights $(N,N^2,\dots ,N^{r-1},1)$ separate its finite exponent support. The displayed invertible monomial substitution makes its extreme $Y_r$ coefficients units in the remaining Laurent variables. Normalize the relation to a monic polynomial with unit constant term; both $Y_r$ and $Y_r^{-1}$ are integral over the image of the $(r-1)$-variable subring. Induction and transitivity of finite extensions prove the assertion, over finite fields as well as infinite fields. An infinite-dimensional quotient has $m\ge 1$.

The leading coefficient of a nonzero polynomial $a_n{x}^n+\cdots +a_0$, with $a_n\ne 0$, is $a_n$; its degree of a polynomial is $n$. A monic polynomial has leading coefficient one. For polynomials over a field, a product has leading coefficient equal to the product of the two leading coefficients, which prevents cancellation of their highest-degree terms.

For any commutative ring $A$ and a monic polynomial $F$ of positive degree $D$, monic polynomial division over a ring gives unique representatives of degree less than $D$. Thus the classes of $1,T,\dots ,T^{D-1}$ form a basis of a module over $A$. This is a finite free algebra, hence flat, and the map $A\to A[T]/(F)$ is injective. The monic hypothesis makes leading-degree arguments valid even when $A$ has zero divisors.

To normalize $k[a_1,\dots ,a_n]$ over an arbitrary field, choose a nonzero polynomial relation and a base $M$ larger than all its exponents. The substitutions $b_i=a_i-a_n^{M^i}$ give distinct top weights to its monomials, so its highest power of $a_n$ has a nonzero scalar coefficient. It becomes a monic polynomial equation, making the algebra finite over $k[b_1,\dots ,b_{n-1}]$. Induction proves the Noether normalization lemma, including over finite fields.