A good way to describe a kernel of a group homomorphism

Question

Let $\psi : \Bbb R[x] \to \Bbb R$ be the function defined by $\psi (p(x)) = p(3)$, for all $p(x) \in \Bbb R [x]$.

I'm trying to describe the kernel of $\psi$ as simply as possible. Is there a better way to answer this than just saying that it's the set of all polynomials such that $p(3) = 0$? or $\{ p(x) : p(3) = 0 \}$?

Answer 1

Not really. You can say in words that the kernel is the set of all polynomials that have $3$ as a root, or equivalently all polynomials that contain the linear factor $x-3$ in their factorisation.

In abstract algebra, this is also called the kernel of the evaluation map at the point $3$. It does not get a special name.

Answer 2

We first observe that

$\psi: \Bbb R[x] \to \Bbb R, \; \psi(p(x)) = p(3) \tag 0$

is in fact a ring homomorphism, the well-known evaluation homomorphism of $\Bbb R[x]$ at $x = 3$. Thus, we use some ring-theoretic concepts in the following argument.

We have

$\ker \psi = (x - 3) = \{(x - 3)q(x), \; q(x) \in \Bbb R[x] \}, \tag 1$

where $(x - 3)$ is the principal ideal generated by $x - 3$, that is, the set of multiples of $x - 3$ in $\Bbb R[x]$. That this is in fact the case is easily seen using Euclidean division of polynomials, which states that any $p(x) \in \Bbb R[x]$ may be written in the form

$p(x) = (x - 3)q(x) + r \tag 2$

with

$\deg r = 0, \tag 3$

i.e.,

$r \in \Bbb R \tag 4$

is constant. If we evaluate (2) at

$x = 3 \tag 5$

we find

$0 = p(3) = (3 - 3)q(3) + r = 0 + r = r, \tag 6$

whence (2) becomes

$p(x) = (x - 3)q(x), \tag 7$

and clearly, any $p(x) \in \Bbb R[x]$ of this form satisfies

$p(3) = (3 - 3)q(3) = 0q(3) = 0, \tag 8$

thus proving our assertion (1). Indeed, (1) describes $\ker \psi$ in (relatively) simple mathematical terms.

We note in closing that we may replace $3$ with any $a \in \Bbb R$ in the above and obtain the analogous result

$\ker \psi = (x - a) = \{(x - a)q(x), \; q(x) \in \Bbb [x] \}. \tag 9$