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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.

Use integral cohomology and let $k\ge 1$. On the projective bundle $Y=\mathbb{P}(E)$ define the complex tautological line bundlePut $t=e(\mathcal{L})\in H^2(Y;\mathbb{Z})$, using the canonical complex orientation. On each fibre, $t$ is the Euler class of the tautological line over ${\mathbb{CP}}^{k-1}$, so $1,t,\dots ,t^{k-1}$ restrict to an integral basis of its cohomology.

Here is the finite-cover Leray-Hirsch theorem proof in this case. For each open set $U\subseteq X$, defineIf $E$ is trivial on $U$, its projective bundle is $U\times {\mathbb{CP}}^{k-1}$ and $\mathcal{L}$ is pulled back from the tautological line on the second factor. The Künneth theorem makes ${\Phi }_U$ an isomorphism, since the fibre has finite free integral cohomology. The same holds for every open subset of $U$.

Compactness of $X$ provides a finite trivializing cover $U_1,\dots ,U_N$. Induct on its size. If the result holds on $V=U_1\cup \cdots \cup U_{N-1}$, it holds on $U_N$ and on $V\cap U_N$, both lying in a trivializing chart. Form the diagram of Mayer–Vietoris sequences for the base, with the finitely many degree shifts on the left, and the total space on the right. Naturality of pullback and multiplication by the global even-degree classes $t^i$ makes the diagram commute. The Five lemma gives the isomorphism on $V\cup U_N$. ThusThis is the claimed free module statement, with its graded degree shifts made explicit.

The module basis expresses $t^k$ uniquely using the lower powers, with homogeneous coefficients. Define the Chern classes $c_j(E)\in H^{2j}(X;\mathbb{Z})$ by the unique relationThe pullbacks are suppressed when $f$ is regarded as a polynomial over $H^{\ast }(X)$. Evaluation at $t$ gives a surjective map $H^{\ast }(X)[T]\to H^{\ast }(Y)$. Since $f(T)$ is monic, monic polynomial division over a ring writes any polynomial as $q(T)f(T)+r(T)$ with degree of $r$ below $k$. If its evaluation is zero, module independence forces every coefficient of $r$ to vanish. Hence the kernel is exactly the ideal generated by $f$, provingThe even-degree coefficients are central in the graded commutative algebra, so this division and ideal statement also apply when the base has odd-degree cohomology. Uniqueness of the coefficients proves their naturality under pullback, by pulling back the relation and using the same module basis. With the hyperplane convention $u=-t$, the relation has the usual all-positive Chern coefficients; the alternating signs here correspond to the tautological line itself.

Now suppose $E={⨁}_i{L}_i$. The sections of a projective bundle $s_i:X\to Y$ choose the line $(L_i{)}_x\subset E_x$. They satisfy $s_i^{\ast }\mathcal{L}\cong L_i$, so $s_i^{\ast }t=e_i$ and pulling back the relation gives $f(e_i)=0$.

To obtain the full factorization over the possibly torsion-containing base ring, also use the associated open chartsThey contain the images of the sections $s_i$ and cover $Y$. Projection identifies $\mathcal{L}{|}_{V_i}$ with ${\pi }^{\ast }{L}_i{|}_{V_i}$, so $t-{\pi }^{\ast }{e}_i$ restricts to zero on $V_i$. The long exact sequence of the pair lifts this class to $H^2(Y,V_i)$. The relative cup product of the $k$ lifted classes lies inIts absolute image is ${\prod }_i(t-{\pi }^{\ast }{e}_i)$, so that product vanishes. The polynomial ${\prod }_i(T-e_i)$ is monic of degree $k$ and lies in the kernel of evaluation. Subtracting the monic generator $f(T)$ leaves degree below $k$, and module independence again makes the difference zero. ThereforeThe open-cover argument proves the factorization without a non-zero-divisor assumption on the differences $e_i-e_j$.

A suitable field with four elements isWrite $\alpha$ for the class of $t$. Monic polynomial division over a ring gives a unique representative $a+b\alpha$, with $a,b\in {\mathbb{F}}_2$: existence follows by division, and uniqueness follows because a nonzero multiple of the degree-two modulus cannot have degree less than two. The four elements are therefore $0,1,\alpha ,\alpha +1$.

The defining relation gives ${\alpha }^2=\alpha +1$, and hence $\alpha (\alpha +1)=1$. Thus the nonzero elements have inversesThe quotient ring is commutative and every nonzero element is a unit, so it is a field. This verifies the field property directly, without quoting a theorem about irreducible-polynomial quotients.

Polynomial division separates a polynomial into a multiple of a chosen divisor and a smaller polynomial remainder. Given $f,g\in F[X]$ over a field $F$, with $g\ne 0$, there are unique polynomials $q,r$ satisfyingHere $q$ is the polynomial quotient. The bound concerns polynomial degree, rather than the sizes of the coefficients.^[ref]

For an original example over the rational numbers, takePolynomial long division first cancels $2X^3$ by subtracting $2Xg$:Subtracting one more copy of $g$ cancels $X^2$:ThusThe final polynomial remainder has polynomial degree one, below the divisor's polynomial degree two. Multiplying out the last identity checks every coefficient, including the constant term $-2+8=6$.

The polynomial degree bound makes this decomposition useful: it replaces a potentially long polynomial with a short polynomial remainder whenever multiples of $g$ can be discarded. At any root of a polynomial of $g$, those multiples vanish under polynomial evaluation. Repeated polynomial division also drives the Euclidean algorithm for finding a greatest common divisor of polynomials.

The coefficient setting matters. Over a commutative ring, the same existence and uniqueness hold when the divisor's leading coefficient is a unit in a ring. A monic polynomial therefore works over the integers, even though they do not form a field; monic polynomial division over a ring explains why zero divisors cause no difficulty in that case.^[ref]

A nonzero divisor alone is insufficient over the integers. Dividing $X$ by $2X+1$ with an integer polynomial quotient and a constant polynomial remainder would require $1=2c$ for an integer $c$, which is impossible. Over the rational numbers the division instead givesThe obstruction is the nonunit leading coefficient $2$. This example separates existence from uniqueness: over an integral domain, the polynomial degree argument still makes any admissible pair unique, if it exists.

For a commutative ring $R$ and $n\ge 1$, this quotient ring imposes $Y^n=0$. Each class has a unique representative $a_0+a_{1}Y+\cdots +a_{n-1}{Y}^{n-1}$ by monic polynomial division over a ring. Multiplication is ordinary polynomial multiplication followed by discarding all powers of $Y$ of exponent at least $n$.