Question about construction of an algebraic closure of a field In Constructing algebraic closures by Keith Conrad, the author writes:

Let $K$ be a field. We want to construct an algebraic closure of $K$, i.e., an algebraic extension of $K$ which is algebraically closed. It will be built as the quotient of a polynomial ring in a very large number of variables.
For each nonconstant monic polynomial $f(X)$ in $K[X]$, let its degree be $n_f$ and let $t_{f,1},\dots,t_{f,n_f}$ be independent variables. Let $A=K[\{t_{f,i}\}]$ be the polynomial ring generated over $K$ by independent variables doubly indexed by every nonconstant monic $f(X)\in K[X]$ and $1\le i \le n_f$. This is a very large polynomial ring containing $K$.

Why will it be built as the quotient of a polynomial ring?
How to build it?
Why is $A$ a very large polynomial ring containing $K$?
$t_1, t_2, \dots, t_n$ are variables of $K[X]$
$t_{f_1,i},t_{f_2,i},\dots$ are variables of $A$. $f_j$ and $f_k$ are different polynomial if $j\neq k$.
Why is it containing $K$?
 A: First let me give short answers to the questions you asked.

Why will it be built as the quotient of a polynomial ring? 

Why wouldn't it be built as a quotient of a polynomial ring? There are different constructions to get an algebraic closure. This particular construction is very short and straightforward though. 

How to build it?

You cited Conrad's notes, read on and Conrad will explain how it is built!

Why is it containing $K$?

Well, $A=K[\{t_{f,i}\}]$ is a polynomial ring over $K$, it contains the elements of $K$ as constant polynomials.

Now let me explain the idea behind this construction:
Over an algebraic closure, you want all polynomials of degree $d$ to split into a product of $d$ linear factors. A weaker but equivalent condition is that all monic polynomials split in this way. So what we want from our algebraic closure is for a nonconstant monic polynomial $f$ of degree $n_f$ to split in linear factors
$$
f(X) = \prod_{i=1}^{n_f} (X-r_i),
$$
where $r_i$ will be the roots of $f$.
To achieve this, we just add independent variables for each of the roots we want, so for each nonconstant monic polynomial $f$ we add the independent variables $t_{f,1},\dots,t_{f,n_f}$ and then mod out the maximal ideal containing the elements
$$
f(X) - \prod_{i=1}^{n_f} (X-t_{f,i}).
$$
This construction now ensures we can split the way we want and since we mod out by a maximal ideal, what we obtain is a field again.
The main thing to verify here is that this maximal ideal exists. It could have happened that the elements above generate the whole ring, in which case our quotient would be trivial. This is checked in Lemma 1 in Conrad's notes.
