I consulted a several sources on the internet, and they all begin with the construction of an extension, $F_1$ of $F$ such that each polynomial in $F[x]$ has a root in $F_1$. Specifically, $F_1 = \frac{F[S]}{M}$, the quotient of a multi-variable polynomial ring, denoted by $F[S]$, over the maximal ideal, $M$, containing: $$U = \left\{ \, \sum_{i=0}^n g_if_i(x_i) : g_i \in F[S] \text{ and } f_i(x_i) \in F[x] \, \right\}$$ My question is, in more accurate terms, $\frac{F[S]}{M}$ is an extension of $\{ \, \alpha + M : \alpha \in F \, \}$, the isomorphic copy of $F$ in $\frac{F[S]}{M}$. However, this will not change the fact that every polynomial in $F[x]$ has a root in $\frac{F[S]}{M}$; for each $f \in F[x]$, we have $f(x_f + M) = f(x_f) + M = 0 + M$ since $f(x_f) \in U$. Is my understanding correct?

At this point, we can proceed in one of two ways. One is to pick out the elements in $F_1$ that are algebraic over $F$. Let's called that set $F_1^*$; we can verify that it is a field containing the isomorphic copy of $F$ via a separate proof. Construct a chain $F \subset F_1^* \subset F_2^* \ldots$ via iteration, and the limit $K = \lim_{ \, i \rightarrow \infty} F_i^*$ is our algebraic closure.

My second question is, can we specify what $F_1^*$ looks like?

  • $\begingroup$ Your understanding for the first part looks ok. For specifying what $F_1^\ast$ looks like, it can be very complicated (think: what are all the solutions to polynomials over $\mathbb{Q}$?) In some cases (such as for finite fields), $F_1^\ast$ can be described more easily (there are only finitely many irreducible polynomials over a finite field). $\endgroup$ – Michael Burr Jul 20 '15 at 14:40
  • $\begingroup$ It would improve your presentation in the first part if you clarified that $S$ is the set of $x_i$'s, $i=0,1,2,\ldots$. $\endgroup$ – hardmath Jul 20 '15 at 14:49

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