$K$ is a field, and $K[t]$ is the set of polynomials over $K$. If we pick a polynomial $m$ in $K[t]$ we could perform arithmetic modulo $m$, in the sense that a polynomial $a$ is congruent to another polynomial $b$ if $a-b$ is divisible by $m$ in $K[t]$.
The author then claims that a polynomial $a$ is congruent modulo $m$ to a unique polynomial in $K[t]$. He gives a proof by contradiction, assuming that $a=r$(mod $m$) and $a=s$(mod $m$), but he then assumes that the degree of $s$ is smaller than the degree of $m$, but gives no explanation of such assumption.
How could one prove that the latter statement that the degree of $s$ is smaller than the degree of $m$?
Alternitavely, if its easier to the repliers, how could one complete the proof in another way?
Besides, wouldn't it be possible to say that
$$a=m+g_1$$ $$a=2m+g_2$$
(...)
$$a=nm+g_n$$
So that $a$ would be congruent to all the (different) $g_i$, contradicting the author's statement?
Where is my mistake, then?
I would really appreciate any help/thoughts.