Let $n > 0$ be some integer and let $G$ be a splitting field of the set of all polynomials of degree at most $n$ over a field $F$. I need to prove that $G$ is not an algebraic closure of $F$ in the following two cases:

  1. $F = \mathbb{Q}$
  2. $F = \mathbb{Z}_{p}$ for prime $p$ (NOT the $p$-adic numbers! I'm talking here about $\mathbb{Z}/p\mathbb{Z}$, the integers modulo $p$.)

However, I have no idea how to get started with either of them. In fact, the idea that $G$ wouldn't be the algebraic closure of $F$ doesn't make much sense to me, since by definition an algebraic closure is an algebraically closed algebraic extension, and in both cases, $G$ seems to be exactly that.

Could someone please explain to me why it makes sense that $G$ would not be the algebraic closure of these fields, as well as some hints on how to show it? Thank you for your time and patience.

Please see the comment thread immediately following this question. Some hints have been given to me, but I am having difficulty seeing how to implement them. For example, say I exhibit a polynomial $f$ of degree $p > n$ such that $f$ is irreducible over, say, $F = \mathbb{Q}$. How do I know that $G$ is not a splitting field for $f$ even though $f$ is of greater degree that what $G$ is designed to take care of? Being a splitting field is necessary (although certainly not sufficient) for something to be algebraically closed, right?

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    $\begingroup$ I would try to find a prime $P>n$ and a polynomial $f(x)\in F[x]$ of degree $P$ which is irreducible over $F$. $\endgroup$ – Batominovski Aug 22 '17 at 23:11
  • $\begingroup$ @Batominovski is the nuance that I'm missing here that in order for $G$ to be an algebraic closure of a field $F$, it is necessary for EVERY polynomial. regardless of degree to split there? $\endgroup$ – ALannister Aug 22 '17 at 23:14
  • $\begingroup$ Yes, all degrees must be accounted for. However, there are some special cases like the extension $\mathbb{R}<\mathbb{C}$, where $\mathbb{C}$ is merely the splitting field of all polynomial over $\mathbb{R}$ of degree at most $2$, and it is already algebraically closed. $\endgroup$ – Batominovski Aug 22 '17 at 23:17
  • $\begingroup$ Hint: Eisenstein's criterion. Also you probably mean $\mathbb Q_p$, not $\mathbb Z_p$. $\endgroup$ – Ravi Aug 22 '17 at 23:21
  • $\begingroup$ @ALannister Well, that's not a field. $\endgroup$ – Ravi Aug 22 '17 at 23:41

Observe $G=F(X)$, where $X$ is the set of all elements of degree $\le n$ in $\overline{F}$ and $F(X)$ is field of all elements expressible as rational expressions in the elements of $X$. This is because these polynomials all split in $F(X)$, and conversely if all these polynomials split in some extension then it contains $F(X)$, thus making it minimal,

Now, suppose $q>n$ is prime. Can $f(x_1,\cdots,x_m)\in F(X)$ have degree $q$ over $F$? Consider the tower of extensions $F(x_1,\cdots,x_m)/\cdots/F(x_1,x_2)/F(x_1)/F$ and think about degrees.

  • $\begingroup$ what does $\overline{F}$ denote? $\endgroup$ – ALannister Aug 23 '17 at 1:26
  • $\begingroup$ Algebraic closure. $\endgroup$ – anon Aug 23 '17 at 1:26
  • $\begingroup$ Every element $\alpha$ of $F(x_1,\cdots,x_m)$ has degree $[F(\alpha):F]$ which divides $[F(x_1,\cdots,x_m):F]$. $\endgroup$ – anon Aug 25 '17 at 23:52
  • $\begingroup$ and since $[F(x_{1},\cdots , x_{m}):F]$ is not divisible by $q$? We know this how? $\endgroup$ – ALannister Aug 25 '17 at 23:56
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    $\begingroup$ @ALannister I said "There are only finitely many polynomials of degree $\le n$ over a finite field $F$, so $G/F$ will wind up having finite degree" and if $G/F$ has finite degree when $F=\Bbb F_p$ is finite, then $G$ is finite. $\endgroup$ – anon Aug 26 '17 at 23:06

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