Two sets of polynomials with distinct roots build the ring of polynomials. Definitions:


$i \in K$
$U_{i}:=\{f\in K[X] |f(i)=0 \}$
$K[X]$ is the ring of polynomials


HINTS:


K[X] is a vector space
Every $U_{i}$ is a vector subspace of $K[X]$


Question:

(i) With $s \neq t$ show, that $U_{s} + U_{t} = K[X]$

(ii) Is this sum direct?

I tried:

I thought it is easy to show, that $U_{s} + U_{t} \subseteq K[X]$ because every $U_{i}$ is allready a subset of $K[X]$ their sum ends up being in $K[X]$ as well.  My Problems are in the other direction... $"\supseteq"$

How can I show that $U_{s} + U_{t}$ build the complete vectorspace $K[X]$ ?

I tried to think of an $N:=\{U_{i}|i \in K\}$ and wanted to show that $dim_{N}(K[X])=2$. This would proof that there always exists a linear combination of $u, v \in N$ with $u \neq v$ with $a_{1}v + a_{2}u = K[X]$. But I am at the stage of confusion.
Hint
For $f\in K[X]$ chose $g\in K[X]$ such that $g(s) = f(s)$ and $g(t) = -f(t)$. Now look at
$$f = \frac12 (f + g) + \frac12(f - g)$$
Here $\frac12 := (1+1)^{-1}$ in the field $K$ (We must thus assume $\mathrm{char}(K) \ne 2$).
If $K = \mathbb F_2$ (up to isomorphism the only field with characteristic $2$), it should be very easy for you since $K[X]$ is particularly easy.
Hint
For $f\in K[X]$ chose $g\in K[X]$ such that $g(s) = f(s)$ and $g(t) = -f(t)$. Now look at
$$f = \frac12 (f + g) + \frac12(f - g)$$
Here $\frac12 := (1+1)^{-1}$ in the field $K$ (We must thus assume $\mathrm{char}(K) \ne 2$).
If $K = \mathbb F_2$ (up to isomorphism the only field with characteristic $2$), it should be very easy for you since $K[X]$ is particularly easy.