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:
 * $\ker\bar\phi = \left\{{ a_0 + a_1X + \cdots + a_nX^n \in R\left[{X}\right] : \phi\left(a_i\right) = 0 \text{ for } i = 0,\ldots,n }\right\}$

Or more concisely,
 * $\ker\bar\phi = \left(\ker\phi\right) \left[{X}\right]$

Proof
Let $P\left(X\right) = a_0 + a_1X + \cdots + a_nX^n \in R\left[{X}\right]$.

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

We have by definition that
 * $\bar{\phi}(a_0 + a_1 X + \cdots + a_n X^n) = \phi(a_0) + \phi(a_1)X + \cdots + \phi(a_n)X^n = 0$

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

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

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

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

This concludes the proof.