The quotient of an integral extension is also an integral extension I would like to prove the following fact:
Suppose that $K$ is a field and that $A$ is a ring and an integral extension of $K$. Given a prime ideal $\mathcal{P} \subseteq A$, then the quotient $A / \mathcal{P}$ is also an integral extension of $K$.
This is my attempt: every element in $A / \mathcal{P}$ is of the form $[a]$, with $a \in A$. Then, since $A$ is integral over $K$, given $a \in A$ there exists an integral equation for $a$ with coefficients in $K$:
$$a^m+k_1a^{m-1}+ \ldots +k_{m}=0, ~~ k_i \in K ~~ \forall i=1,\ldots, m.$$
Passing to equivalence classes in the quotient we get
  $$[a]^m+[k_1][a]^{m-1}+ \ldots +[k_{m}]=[0].$$
If we show that for any $i$, the only element contained in $[k_i]$ is $k_i$, then the above expression is an integral relation for $[a]$ with coefficients in $K$ and we are done. From here, I suspect that I should procede in this way: given $k_i' \in [k_i]$, we have $$k_i' - k_i \in K \cap \mathcal{P } ~~ (*)$$ which is a prime ideal since it is the contraction of a prime ideal. But the only prime ideal in $K$ is $(0)$, so $k_i'=k_i$.
My problem is probably quite stupid, but I find some difficulties in justifying this last step: in $(*)$ I assume that the class of an element of $K$ contains only elements of $K$, but is this statement true at all? How should I justify it?

I think the approach may be too elementary. You want to show that the map $K\to A\to A/\mathcal{P}$ is injective. Morphisms from a field to another ring are always injective. Indeed, let $f:K\to B$ be just about any ring morphism and assume that $f(a)=f(b)$ and $a\ne b$, hence $t:=a-b$ is invertible and so you get the following contradiction:
$$
0 = 0 \cdot f(t^{-1}) = (f(a) - f(b))\cdot f(t^{-1}) = f(t)\cdot f(t^{-1})= f(1) = 1.
$$
This should set your mind at ease in a more general sense: Over a field, you can never run into the sort of problem you are trying to avoid.
