ring theory - Show that $E_1$ is algebraic over $K.$

I found a problem in proving a theorem in field theory which states that Every field has an algebraic closure. Here's the sketch of the proof in the lecture notes given by our instructor.

Let $K$ be a field. I want to find out the algebraic closure of $K.$ For that we consider indeterminates $X_f$ for each irreducible polynomial $f \in K[X].$ Now let us consider the polynomial ring $$R = K\left [X_f\ |\ f \in \text {Irr} \left (K[X] \right ) \right].$$ Let $I$ be an ideal of $R$ defined as $$I = \left \langle f \left (X_f \right )\ |\ f \in \text {Irr} \left (K[X] \right ) \right \rangle.$$ Then $I$ is a proper ideal of $R$ and hence by Krull's lemma $I$ is contained in a maximal ideal say $M.$ Let $E_1 = R/M.$ Then $E_1$ is an algebraic extension of $K$ such that every irreducible polynomial in $K[X]$ has a zero in $E_1.$ Continuing the same argument as before we get a chain of fields $$E_0 \subseteq E_1 \subseteq E_2 \subseteq E_3 \subseteq \cdots$$ with the property that every irreducible polynomial in $E_i[X]$ has a zero in $E_{i+1}$ and $E_{i+1}$ is an algebraic field extension of $E_i$ for each $i=0,1,2,. \cdots$ (where $K = E_0).$ Take $E = \cup_{i \geq 0} E_i.$ Then any irreducible polynomial in $E[X]$ has a zero in $E$ and since each $E_i$ is an algebraic field extension of $K$ so is $E.$ Hence $E$ is an algebraic closure of $K.$

In this sketch of the proof I understood everything except the part algebraic extension. I don't understand why $E_{i+1}$ is an algebraic extension of $E_i.$ I observed that it is enough to show that $E_1$ is an algebraic extension of $K.$ But why is it so?

Can anybody please help me in this regard? Any suggestion will be highly appreciated.

Thank you very much for your valuable time.

Answer

For each irreducible $f \in K[X]$, note that $\overline{X_f} := X_f + M \in R/M$ is algebraic over $K$, since $f(X_f) \in M$.

Now $R/M$ is generated (as a $K$-algebra) by all the $\overline{X_f}$, i.e. each element is a sum of products of elements of $K$ and elements of the form $\overline{X_f}$, each of which is algebraic over $K$. So, every element of $R/M$ is algebraic, since sums and products of algebraic elements are algebraic.