# Kernel of Induced Homomorphism of Polynomial Forms

## Theorem

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

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

Let $K = \ker \phi$.

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

Let $\bar\phi: R \left[{X}\right] \to S \left[{X}\right]$ be the induced morphism of polynomial rings.

Then the kernel of $\bar\phi$ is:

$\ker \bar\phi = \left\{{ a_0 + a_1 X + \cdots + a_n X^n \in R \left[{X}\right] : \phi \left({a_i}\right) = 0 \text{ for } i = 0, \ldots, n }\right\}$

Or, more concisely:

$\ker \bar\phi = \left({\ker \phi}\right) \left[{X}\right]$

## Proof

Let $P \left(X\right) = a_0 + a_1 X + \cdots + a_n X^n \in R \left[{X}\right]$.

Suppose first that $\phi \left({a_i}\right) = 0$ for $i = 0, \ldots, n$.

We have by definition that:

$\bar\phi \left({a_0 + a_1 X + \cdots + a_n X^n}\right) = \phi \left({a_0}\right) + \phi \left({a_1}\right) X + \cdots + \phi \left({a_n}\right) X^n = 0$

That is to say, $P \left({X}\right) \in \ker \bar\phi$.

Conversely, suppose that $P \left({X}\right) \in \ker \bar\phi$.

That is, $\bar\phi \left({P \left({X}\right)}\right) = \phi \left({a_0}\right) + \phi \left({a_1}\right) X + \cdots + \phi \left({a_n}\right) X^n$ is the null polynomial.

This by definition means that for $i = 0, \ldots, n$ we have $\phi \left({a_i}\right) = 0$.

Hence, $P \left({X}\right) \in \left({\ker \phi}\right) \left[{X}\right]$.

This concludes the proof.

$\blacksquare$