Incomparability of prime ideals in algebraic ring extensions It is known that if $R\subset S$ is an integral extension of rings. Two distinct primes of $S$ having the same intersection with $R$ are incomparable. It is proved due to the following lemma.

Let $R\subset S$ be domains. If $K(S)$ (the quotient field of $S$) is algebraic over $K(R)$, then any nonzero ideal of $S$ intersects $R$ nontrivially.

For the proof of the incomparability, suppose $Q\subset Q_1\subset S$ are primes with $R\cap Q=R\cap Q_1=P\subset R$, by factoring out $Q$ in $S$ and $P$ in $R$, the result can be deduced by the above lemma, since integral equations persist modulo $P$ and $K(S)$ is thus algebraic over $K(R)$.
However, I wonder if the incomparability property still holds when the ring extension is algebraic?
 A: No, INC generally does not hold for algebraic extensions.
Note that for an integral domain $D$ with field of fractions $K$, every intermediate ring $D \subseteq T \subseteq K$ (aka overring of $D$) is an algebraic extension of $D$.  Indeed, every element $u \in T$ can be written as $a/b$ for some $a, b \in D$ and hence satisfies the (linear) algebraic relation $bu - a = 0$.
A thorough answer to your question lies in an old paper of D.E. Dobbs, On Inc-Extensions and Polynomials with Unit Content.
Following are a couple of the key results:

Theorem For any extension of commutative rings $R \subset T$ and any element $u \in T$, $R \subset R[u]$ has INC if and only if $u$ satisfies a polynomial $f \in R[x]$ such that the coefficients of $f$ generate the unit ideal of $R$.

Such elements satisfying polynomials which coefficients generate the unit ideal are called primitive.

Theorem Let $D$ be an integral domain with field of fractions $K$.  Every element $k \in K$ is primitive iff the integral closure of $D$ in $K$ is a Prüfer domain.

The second theorem is originally due to Gilmer and Hoffman.
As a corollary we get:

Corollary [failure of INC in overrings] Let $D$ be an integral domain such that the integral closure of $D$ in its field of fractions $K$ is not a Prüfer domain.  Let $k \in K$ be a non-primitive element that witnesses this.  Then $D \subset D[k]$ is an algebraic extension of domains that does not satisfy INC.

So, for example, this tells us that if we take any integrally closed domain $D$ that is not a field, then there will be rings $D[x] \subset T \subset K(x)$ such that $D[x] \subset T$ is algebraic and does not have INC.  However, this all a little bit abstract.
To be concrete, let's also produce a class of explicit examples.  Let $A$ be any commutative ring and let $R = A[x,y]$.  Let $T(R)$ denote the total ring of fractions of $R$.  Consider the extension $R \subset R[x/y] \subset T(R)$.  I'll leave it as an exercise to check that $x/y$ is not primitive. Therefore, the extension $R \subset R[x/y]$ must not have INC by the aforementioned theorem.  (HINT: first deduce that $x/y$ primitive would imply $x/y$ integral.  Then note that $x/y$ integral implies - in general! - that $y$ divides some power of $x$, which for indeterminates $x,y$ is absurd.)
