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 monomials of $R \sqbrk X$, with coefficients in $R$.
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Proof using specific model
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Proof using universal property
Outline of Proof
We show that the subset $S$ of linear combinations of monomials 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 monomials.
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.
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