Show that $E_1$ is algebraic over $K.$

Question

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 1

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.

Answer 2

It is not necessary to build the chain of fields: $E_1$ is already an algebraic closure of $K$ (so all the later fields $E_i$ actually equal $E_1$). This is a theorem of Robert Gilmer and should be more widely known. See here or here. For a similar construction where it is less technical to see you get an algebraic closure in one step, see here.