Induced Homomorphism of Polynomial Forms

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Let $R$ and $S$ be commutative rings with unity.

Let $\phi: R \to S$ be a ring homomorphism.

Let $R \sqbrk X$ and $S \sqbrk X$ be the rings of polynomial forms over $R$ and $S$ respectively in the indeterminate $X$.

Then the map $\overline \phi: R \sqbrk X \to S \sqbrk X$ given by:

$\map {\overline \phi} {a_0 + a_1 X + \cdots + a_n X^n} = \map \phi {a_0} + \map \phi {a_1} X + \cdots + \map \phi {a_n} X^n$

is a ring homomorphism.

Proof using evaluation homomorphism

Induced Homomorphism of Polynomial Forms/Proof 1

Proof by direct verification


$f = a_0 + \cdots + a_n X^n$
$g = b_0 + \cdots + b_m X^m \in R \sqbrk X$

We have:

\(\ds \map {\overline \phi} f \map {\overline \phi} g\) \(=\) \(\ds \sqbrk {\map \phi {a_0} + \cdots + \map \phi {a_n} X^n} \sqbrk {\map \phi {b_0} + \cdots + \map \phi {b_m} X^m}\)
\(\ds \) \(=\) \(\ds \sum_{r \mathop = 0}^{m + n} \sum_{i \mathop + j \mathop = r} \map \phi {a_i} \map \phi {b_j} X^r\) Definition of Multiplication of Polynomial Forms
\(\ds \) \(=\) \(\ds \sum_{r \mathop = 0}^{m + n} \map \phi {\sum_{i \mathop + j \mathop = r} a_i b_j} X^r\) because $\phi$ is a homomorphism
\(\ds \) \(=\) \(\ds \map {\overline \phi} {f g}\)


\(\ds \map {\overline \phi} f + \map {\overline \phi} g\) \(=\) \(\ds \sqbrk {\map \phi {a_0} + \cdots + \map \phi {a_n} X^n} + \sqbrk {\map \phi {b_0} + \cdots + \map \phi {b_m} X^m}\)
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 0}^{\max \set {m, n} } \sqbrk {\map \phi {a_i} + \map \phi {b_i} } X^i\) Definition of Addition of Polynomial Forms
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 0}^{\max \set {m, n} } \map \phi {a_i + b_i} X^i\) because $\phi$ is a homomorphism
\(\ds \) \(=\) \(\ds \map {\overline \phi} {f + g}\)

Thus $\overline \phi$ is a ring homomorphism.


Also see