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 $R \left[{X}\right]$ and $S \left[{X}\right]$ be the rings of polynomial forms over $R$ and $S$ respectively in the indeterminate $X$.

Then the map $\overline{\phi}: R \left[{X}\right] \to S \left[{X}\right]$ given by:

$\overline{\phi} \left({a_0 + a_1 X + \cdots + a_n X^n}\right) = \phi \left({a_0}\right) + \phi \left({a_1}\right) X + \cdots + \phi \left({a_n}\right) X^n$

is a ring homomorphism.

Proof using evaluation homomorphism

Induced Homomorphism of Polynomial Forms/Proof 1

Proof by direct verification

Let $f = a_0 + \cdots + a_n X^n$, $g = b_0 + \cdots + b_m X^m \in R \left[{X}\right]$.

We have:

\(\displaystyle \overline{\phi} \left({f}\right) \overline{\phi} \left({g}\right)\) \(=\) \(\displaystyle \left[{\phi \left({a_0}\right) + \cdots + \phi \left({a_n}\right) X^n}\right] \left[{\phi \left({b_0}\right) + \cdots + \phi \left({b_m}\right) X^m}\right]\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{r \mathop = 0}^{m+n} \sum_{i \mathop + j \mathop = r} \phi \left({a_i}\right) \phi \left({b_j}\right) X^r\) by the definition of multiplication of polynomial forms
\(\displaystyle \) \(=\) \(\displaystyle \sum_{r \mathop = 0}^{m+n} \phi \left({\sum_{i \mathop + j \mathop = r} a_i b_j}\right) X^r\) because $\phi$ is a homomorphism
\(\displaystyle \) \(=\) \(\displaystyle \overline{\phi} \left({f g}\right)\)


\(\displaystyle \overline{\phi} \left({f}\right) + \overline{\phi} \left({g}\right)\) \(=\) \(\displaystyle \left[{\phi \left({a_0}\right) + \cdots + \phi \left({a_n}\right) X^n}\right] + \left[{\phi \left({b_0}\right) + \cdots + \phi \left({b_m}\right) X^m}\right]\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 0}^{\max \left\{ {m, n}\right\} } \left[ \phi \left({a_i}\right) + \phi \left({b_i}\right) \right] X^i\) by the definition of addition of polynomial forms
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 0}^{\max \left\{ {m, n}\right\} } \phi \left({a_i + b_i}\right) X^i\) because $\phi$ is a homomorphism
\(\displaystyle \) \(=\) \(\displaystyle \overline{\phi} \left({f + g}\right)\)

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


Also see