Polynomial is Linear Combination of Monomials

Theorem
Let $R$ be a commutative ring with unity.

Let $R[X]$ be a polynomial ring over $R$ in the variable $X$.

Let $P\in R[X]$.

Then $P$ is a linear combination of the monomials of $R[X]$, with coefficients in $R$.

Outline of Proof
We show that the subset of linear combinations of monomials forms a subring.

We use the universal property to exhibit a ring homomorphism $R[X]\to S$.

We use uniqueness to show that it is necessarily a bijection.

Proof
Let $S\subset R[X]$ be the subset of all elements that are linear combinations of monomials.

Let $\iota : S\to R[X]$ be the inclusion mapping.

Suppose for the moment that $S$ is a commutative ring with unity.

Then by Universal Property of Polynomial Ring, there exists a ring homomorphism $g : R[X] \to S$ with $g(X) = X$.

By Inclusion of Subring is Ring Homomorphism and Composition of Ring Homomorphisms is Ring Homomorphism, $\iota \circ g : R[X] \to R[X]$ is a ring homomorphism.

By construction, $(\iota \circ g )(X) = X$.

By Universal Property of Polynomial Ring, there exists a unique ring homomorphism $h : R[X] \to R[X]$ with $h(X) = X$.

We have that $\iota\circ g$ is such a ring homomorphism.

By Identity Mapping is Ring Automorphism, the identity mapping $\operatorname{id}$ on $R[X]$ is one too.

By uniqueness, $\iota \circ g = \operatorname{id}$.

By Identity Mapping is Surjection and Surjection if Composite is Surjection, $\iota$ is a surjection.

By Inclusion Mapping is Surjection iff Identity, $S = R[X]$.

It remains to show that $S$ is a commutative ring with unity.

Also see

 * Monomials Form Basis of Polynomial Ring