Looking for an Alternative Proof of a Bound on the Number of Maximal Ideals in an Integral Extension Lying over a Maximal Ideal in the Base Ring 
Consider integral domains $R$ and $T$ such that $R \subseteq T.$ Given an element $a \in T$ that satisfies a monic polynomial of degree $d$ with coefficients in $R,$ consider the subring $S = R[a] = \{r_0 + r_1 a + \cdots + r_{d - 1} a^{d - 1} \,|\, r_i \in R\}$ of $T.$ If $Q$ is a maximal ideal of $R,$ prove that there are $\leq d$ maximal ideals $P \subseteq S$ with $P \cap R = Q.$

One can prove this in the following manner. (Credit is due to Souvik Dey for the following argument.)

Consider a maximal ideal $Q$ of $R$ and a prime ideal $P$ of $S$ such that $P \cap R = Q.$ On the level of sets, we have that $Q \subseteq P,$ hence there is a containment of ideals $QS \subseteq PS = P$ of $S.$ We conclude that every prime ideal $P$ of $S$ such that $P \cap R = Q$ contains the ideal $QS,$ hence we may consider the set $\operatorname{Spec}(S / QS) = \{P \,|\, P \text{ is a prime ideal of } S / QS \}.$ We claim that $$\begin{align*} \operatorname{Spec}(S / QS) = \operatorname{MaxSpec}(S / QS) &= \{M \,|\, M \text{ is a maximal ideal of } S / QS \} \text{ and } \\ \\ \lvert \operatorname{Spec}(S / QS) \rvert &\leq d \end{align*}.$$ Considering that $S = R \langle a_0, \dots, a_{d - 1} \rangle$ is an $R$-module and $Q$ annihilates $S / QS,$ it follows that $S / QS$ is an $R/Q$-vector space by hypothesis that $Q$ is a maximal ideal of $R.$ We note that the dimension of $S / QS$ over $R / Q$ is $\leq d$ since it can be generated by the $d$ elements $a_0 + QS, \dots, a_{d - 1} + QS$ over $R,$ hence $S / QS$ is Artinian as a ring. Consequently, $S / QS$ has finitely many prime ideals $P_1, \dots, P_n,$ and they are all maximal, hence we have that $\operatorname{Spec}(S / QS) = \operatorname{MaxSpec}(S / QS).$ Further, we have that $P_i + P_j = S / QS$ so that $P_i^r + P_j^r = S / QS$ for each pair of integers $1 \leq i < j \leq n$ and any positive integer $r.$ By the Chinese Remainder Theorem, it follows that $$\frac{S / QS}{(P_1 \cdots P_n)^r} = \frac{S / QS}{P_1^r \cdots P_n^r} \cong \prod_{i = 1}^n \frac{S / QS}{P_i^r}.$$ Considering that $S / QS$ is Artinian, the nilradical $\operatorname{Rad}(0)$ is nilpotent, i.e., there exists an integer $r \gg 0$ such that $\operatorname{Rad}(0)^r = 0.$ We have therefore that $$0 = \operatorname{Rad}(0)^r = (P_1 \cap \cdots \cap P_n)^r = (P_1 \cdots P_n)^r = P_1^r \cdots P_n^r,$$ from which it follows by our above isomorphism that $$\frac S {QS} \cong \prod_{i = 1}^n \frac{S / QS}{P_i^r}.$$ Each quotient ring $(S / QS) / P_i^r$ is a nonzero $R/Q$-vector space (as each quotient ring is annihilated by $Q$), hence its dimension over $R/Q$ is $\geq 1.$ We have therefore that $$n \leq \sum_{i = 1}^n \dim \frac{S / QS}{P_i^r} = \dim \prod_{i = 1}^n \frac{S / QS}{P_i^r} = \dim \frac S {QS} \leq d,$$ from which it follows that $\lvert \operatorname{Spec}(S / QS) \rvert = \lvert \operatorname{MaxSpec}(S / QS) \rvert = n \leq d,$ as desired.

I am searching for a more elementary proof of the initial fact. If possible, I would like to prove this without relying on the structure of $S$ as an $R$-module or the fact that $S / QS$ is Artinian. Ultimately, I would simply appreciate any thoughts or suggestions. Thanks in advance.
 A: Here is a different proof, based on the fact that if $A$ is a finite dimensional $k$ algebra for $k$ a field, with $n$ distinct maximal ideals $\mathfrak{m}_i$, and $k$ algebra maps $\lambda_i: A\rightarrow A/\mathfrak{m_i}\rightarrow K$ for $K$ a field, then these $\lambda_i$ are $K$ linearly independent. Summarised as "Field homomorphisms are linearly independent".
First lets prove this, consider a linear dependence of homomorphisms, $\sum_{i=1}^n \alpha_i \lambda_i(x)=0$ for all $x\in A$, $\alpha_i \in K$. Since the kernel $\mathfrak{m}_i$ of $\lambda_i$ is a maximal ideal, we will be done if we can find some $x$ in $\bigcap_{j\neq i}\mathfrak{m}_j\setminus \mathfrak{m_i}$, since this will force $\alpha_i=0$. For this, take an element $\beta_j$ in each $\mathfrak{m_j}\setminus \mathfrak{m_i}$ (nonempty by maximality), and multiply them together. This element satisfies our condition by construction.
So with this fact, consider now the maximal ideal $Q$ of $R$ inside $R[\alpha]$. Since $\alpha$ had monic minimal polynomial of degree $d$, we have that $R[\alpha]$ is a free $R$ module of rank $d$. So tensoring over $R$ with $R/Q$ yields the algebra $R/Q[\bar{\alpha}]$, which is a $R/Q$ vector space of dimension $d$. Observe that each maximal ideal $P_i$ in $R[\alpha]$ containing $Q$ yields a maximal ideal in this quotient, which yields a finite extension of $R/Q$, and we can find a large enough common field $K$ to recieve an embedding from each of these.
So we are in the situation of the lemma, so since the $K$ vector space $Hom_k(R/Q[\bar{\alpha}],K)$ is at most $d$ dimensional (over $K$), we can have at most $d$ maximal ideals $P_i$ extending $Q$.
Edit:
I realise that this still relies on the structure of $S=R[\alpha]$ as an $R$ module, but this fact, that adjoining a root of a monic polynomial yields a free module of rank=degree is of a lower level (and more fundamental than) the rest of the proof in my opinion.
