Polynomial is Linear Combination of Monomials

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

Let $R \sqbrk X$ be a polynomial ring over $R$ in the variable $X$.

Let $P \in R \sqbrk X$.

Then $P$ is a linear combination of the mononomials of $R \sqbrk X$, with coefficients in $R$.

Outline of Proof
We show that the subset $S$ of linear combinations of mononomials forms a subring.

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

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

Proof
Let $S \subset R \sqbrk X$ be the subset of all elements that are linear combinations of mononomials.

Let $\iota: S \to R \sqbrk X$ denote 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 $g: R \sqbrk X \to S$ with $\map g X = X$.

By Inclusion Mapping on Subring is Homomorphism, $\iota: S \to R \sqbrk X$ is a ring homomorphism.

By Composition of Ring Homomorphisms is Ring Homomorphism, $\iota \circ g: R \sqbrk X \to R \sqbrk X$ is a ring homomorphism.

By construction, $\map {\paren {\iota \circ g} } X = X$.

By Universal Property of Polynomial Ring, there exists a unique ring homomorphism $h : R \sqbrk X \to R \sqbrk X$ with $\map h X = X$.

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

By Identity Mapping is Ring Automorphism, the identity mapping $I$ on $R \sqbrk X$ is one too.

By uniqueness, $\iota \circ g = I$.

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

By Inclusion Mapping is Surjection iff Identity, $S = R \sqbrk X$.

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

Also see

 * Mononomials Form Basis of Polynomial Ring