Universal Property of Polynomial Ring/Free Monoid on Set
Theorem
Let $R, S$ be commutative and unitary rings.
Let $\family {s_j}_{j \mathop \in J}$ be an indexed family of elements of $S$.
Let $\psi: R \to S$ be a ring homomorphism.
Let $R \sqbrk {\set {X_j: j \in J} }$ be a polynomial ring.
Then there exists a unique evaluation homomorphism $\phi: R \sqbrk {\set {X_j: j \in J} } \to S$ at $\family {s_j}_{j \mathop \in J}$ extending $\psi$.
Proof
Let $Z$ be the set of all multiindices indexed by $J$.
Let $k_j$ be the $j$th component of a multiindex $k$.
Let $\ds f = \sum_{k \mathop \in Z} a_k \prod_{j \mathop \in J} X_j^{k_j}$ be a polynomial over $R$.
Define:
- $\ds \map \phi f = \sum_{k \mathop \in Z} \map \psi {a_k} \prod_{j \mathop \in J} s_j^{k_j}$
It is clear that $\phi$ extends $\psi$.
If $\ds g = \sum_{k \mathop \in Z} b_k \prod_{j \mathop \in J} X_j^{k_j}$, then:
| \(\ds \map \phi {f + g}\) | \(=\) | \(\ds \map \phi {\sum_{k \mathop \in Z} \paren {a_k + b_k} \prod_{j \mathop \in J} X_j^{k_j} }\) | Definition of Addition of Polynomial Forms | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop \in Z} \paren {\map \psi {a_k + b_k} } \prod_{j \mathop \in J} s_j^{k_j}\) | Definition of $\phi$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop \in Z} \paren {\map \psi {a_k} + \map \psi {b_k} } \prod_{j \mathop \in J} s_j^{k_j}\) | Definition of Ring Homomorphism | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop \in Z} \map \psi {a_k} \prod_{j \mathop \in J} s_j^{k_j} + \sum_{k \mathop \in Z} \map \psi {b_k} \prod_{j \mathop \in J} s_j^{k_j}\) | Ring Axioms of $S$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \phi f + \map \phi g\) | Definition of $\phi$ |
Therefore $\phi$ preserves addition.
Also:
| \(\ds \map \phi {f g}\) | \(=\) | \(\ds \map \phi {\sum_{k \mathop \in Z} \paren {\sum_{p \mathop + q \mathop = k} a_p b_q} \prod_{j \mathop \in J} X_j^{k_j} }\) | Definition of Multiplication of Polynomial Forms | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop \in Z} \map \psi {\sum_{p \mathop + q \mathop = k} a_p b_q} \prod_{j \mathop \in J} s_j^{k_j}\) | Definition of $\phi$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop \in Z} \paren {\sum_{p \mathop + q \mathop = k} \map \psi {a_p} \map \psi {b_q} } \prod_{j \mathop \in J} s_j^{k_j}\) | Definition of Ring Homomorphism | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\sum_{k \mathop \in Z} \paren {\map \psi {a_k} } \prod_{j \mathop \in J} s_j^{k_j} } \paren {\sum_{k \mathop \in Z} \paren {\map \psi {b_k} } \prod_{j \mathop \in J} s_j^{k_j} }\) | Ring Axioms of $S$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \phi f \map \phi g\) | Definition of $\phi$ |
This shows that $\phi$ is a homomorphism.
Now suppose that $\phi'$ is another such homomorphism.
For each $j \in J$, $\phi'$ must satisfy $\map {\phi'} {X_j} = s_j$ and $\map {\phi'} r = \map \psi r$ for all $r \in R$.
In addition $\phi'$ must be a homomorphism, so we compute:
| \(\ds \map {\phi'} {\sum_{k \mathop \in Z} a_k \prod_{j \mathop \in J} X_j^{k_j} }\) | \(=\) | \(\ds \sum_{k \mathop \in Z} \map {\phi'} {a_k \prod_{j \mathop \in J} X_j^{k_j} }\) | $\phi'$ preserves Ring Addition | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop \in Z} \map {\phi'} {a_k} \prod_{j \mathop \in J} \map {\phi'} {X_j}^{k_j}\) | $\phi'$ preserves Ring Product | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop \in Z} \map \psi {a_k} \prod_{j \mathop \in J} s_j^{k_j}\) | as $\map {\phi'} {X_j} = s_j$ and $\map {\phi'} r = \map \psi r$ for all $r \in R$ |
and therefore $\phi' = \phi$.
This concludes the proof.
$\blacksquare$
Remarks
- The requirement that the rings be commutative is vital. A fundamental difference for polynomials over non-commutative rings is additional difficulty identifying polynomial forms and functions using this method.