# 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 $K$?

• The gist is that it will be the quotient of a polynomials ring by a maximal ideal (and thus a field we get). $\;A\;$ is a polynomial ring in lots of variables: one for each non-constant polynomial in $\;K[X]\;$. The mapping $\;k\to k\in A\;$ is an injection. – DonAntonio Mar 1 '14 at 10:33

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?

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.

• Your proposed construction seems to be way off line (or I misunderstood big time, of course): the variables $\;t_{f,i}\;$ are connected to each non-constant polynomial and its degree : we cannot talk of its roots as we still don't have them (at least not for all polynomials!), otherwise $\;K\;$ would already be alg. closed. This is very close to Artin's trick as mentioned, for example, in Lang's "algebra". – DonAntonio Mar 1 '14 at 10:41
• I said what we want is to have these roots, that's just to motivate where the $t_{f,i}$ come from. Then we adjoin them as abstract independent variables and mod out the ideal that makes those the roots as proposed. I changed the phrasing from "are the roots" to "will be the roots", maybe I wasn't too precise there. Is it fine this way? – Christoph Mar 1 '14 at 10:48
• Well, I think so. I haven't read Conrad's paper and I still cannot understand off the top of my head why so many variables for each polnomial if it is enough just one (as in Artin's). – DonAntonio Mar 1 '14 at 10:51
• Artin constructs a sequence $E_1\subseteq E_2\subseteq \cdots$ of field extensions of $K$ and then takes the union of all. In this construction we need more variables, but we get the algebraic closure in a single step. – Christoph Mar 1 '14 at 11:01
• Well it is a nice though non-trivial exercise to show that in Artin's proof it is also necessary only one step: using his notation, the field $\;\Bbb F[x_f]/M\;$ is already an algebraically closed field. – DonAntonio Mar 1 '14 at 11:06