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In preparing for an upcoming course in field theory I am reading a Wikipedia article on field extensions. It states that the complex numbers are a field extension of the reals. I understand this since $\mathbb R(i) = \{ a + bi : a,b \in \mathbb R\}$.
Then the article states that the reals are a field extension of the rationals. I do not understand how this could be. What would you adjoin to $\mathbb Q$ to get all the reals? The article doesn't seem to say anything more about this. Is there a way to explain this to someone who has yet to take a course in field theory?

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    $\begingroup$ See here. $\endgroup$
    – rogerl
    Commented Dec 28, 2015 at 23:03
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    $\begingroup$ it is a non algebraic extension it just a question of definitions $\endgroup$ Commented Dec 28, 2015 at 23:04
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    $\begingroup$ If one was truly inclined to do this, one could first take the field of rational functions (with integer coefficients) in uncountably many variables. If you take the algebraic completion of that, I think you'd end up with a field isomorphic to $\mathbb R$. This makes the idea of "extension" clear, but it's a nasty construction (and I'm not 100% sure it works) (The variables represent transcendental numbers, and using the axiom of choice, you can find a set of transcendentals from which no real is algebraically independent) $\endgroup$ Commented Dec 28, 2015 at 23:13
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    $\begingroup$ An extension is not necessarily a simple extension, or a finite extension for that sake (Wikipedia links). $\endgroup$ Commented Dec 29, 2015 at 16:21
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    $\begingroup$ You could adjoin all of $\Bbb{R}$... $\endgroup$ Commented Dec 29, 2015 at 22:49

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Saying "the reals are an extension of the rationals" just means that the reals form a field, which contains the rationals as a subfield. This does not mean that the reals have the form $\mathbb{Q}(\alpha)$ for some $\alpha$; indeed, they do not. You have to adjoin uncountably many elements to the rationals to get the reals.

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  • $\begingroup$ I realize Im late to the party but I recently started learning this subject. I understand your argument but am thrown off by an observation. Wouldnt you agree that all real numbers are an infinite sum of rationals? $\pi = 3.1415\ldots = 3 + \frac{1}{10}+\frac{4}{100} +\frac{1}{1000}+\frac{5}{10000}+\cdots $. Each of those addends are a product of 1 and an element out of $\mathbb{Q}$. So being very loose with notation, $\mathbb{R} = \mathbb{Q}(1)$. I suppose the argument fails if not every real number is expressible in an infinite string of decimals?! $\endgroup$ Commented Apr 30, 2020 at 0:31
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    $\begingroup$ @CogitoErgoCogitoSum Infinite sums aren't part of the field structure. $\mathbb{R}$ is not the smallest field containing $\mathbb{Q}$ (since $1\in\mathbb{Q}$ it's redundant to write "$\mathbb{Q}(1)$"); that's just $\mathbb{Q}$ itself. Every real is of course expressible as an infinite string decimals, but that doesn't mean $\mathbb{Q}$ isn't a field. $\endgroup$ Commented Apr 30, 2020 at 1:24
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    $\begingroup$ I know its redundant to say $Q(1)$. Thats why I said "loose with notation". Is that really the only criticism you have? I didnt say Q wasnt a field. Where are you getting any of this? I was just trying to point out that every real is an (infinite) sum of rationals. I didnt know it was forbidden to take infinite sums. Arent the algebraics taken on whole an infinite extension upon the rationals? The set of all algebraics necessarily includes real numbers that are the sums of infinitely many different roots, do they not? They have to be for closure. $\endgroup$ Commented Apr 30, 2020 at 2:22
  • $\begingroup$ @CogitoErgoCogitoSum I don't understand what you're getting at here. For $\alpha$ a real, "$\mathbb{Q}(\alpha)$" is the smallest field containing $\alpha$. Since there's no requirement that fields be closed under infinite sums in any sense the fact that all reals are infinite sums of rationals isn't relevant. (If you mean something else by "$\mathbb{Q}(\alpha)$" you should explain what you mean.) "The set of all algebraics necessarily includes real numbers that are the sums of infinitely many different roots, do they not?" No they don't. You're mixing up two very different concepts here. $\endgroup$ Commented Apr 30, 2020 at 2:28
  • $\begingroup$ At this point I'm not really sure what you're claiming, or disputing in my answer. Can you state precisely what your point is? (Maybe you're asking about the relation between algebraic and topological closure? $\overline{\mathbb{Q}}$ is algebraically closed, but is not a topologically closed subfield of the topological field $\mathbb{C}$ (at least, with the usual topology).) $\endgroup$ Commented Apr 30, 2020 at 2:31
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For $\mathbb{R}$ to be field extension of $\mathbb{Q}$, all we need is that $\mathbb{R}$ is a field containing $\mathbb{Q}$ as a subfield. That's definitely true.

The construction is a bit more delicate and analytic in nature: $\mathbb{R}$ is the completion of $\mathbb{Q}$ and is substantially larger. Since $\mathbb{R}$ is uncountable, it's not an extension of finite degree, meaning that you cannot write $\mathbb{R} = \mathbb{Q}(a_1, a_2, ..., a_n)$ for some finite sequence of symbols $a_i$ (nor even a countable sequence). You have to adjoin uncountably many symbols.


If you're interested in a way to construct reals from rationals, take a look at Dedekind cuts.

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  • $\begingroup$ As an addendum that I'm not sure about, I seem to recall reading an almost completely algebraic construction of $\mathbb{R}$ from $\mathbb{Q}$ in terms of some sequence of field extensions - probably in Lang or Hungerford. If someone can confirm / correct this, I'd appreciate it. $\endgroup$
    – user296602
    Commented Dec 28, 2015 at 23:10
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    $\begingroup$ The reals are a fundamentally analytic or topological thing, there relevant stems from being complete, and this is decidedly non-algebraic. But there are two things you might recall: a construction of the reals that uses a bit of algebraic jargon and even results. Something like the set of all Cauchy sequence of rationals is a commutative ting and the subset of sequences that have limit zero are a maximal ideal. The quotionent is a field and these are the real numbers. Or, it is was about the algebraic closure of $Q$. $\endgroup$
    – quid
    Commented Dec 28, 2015 at 23:13
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    $\begingroup$ @quid Ah, I think you're right - it was about $\mathbb{C}$ containing the algebraic closure while avoiding any complex analytic methods. $\endgroup$
    – user296602
    Commented Dec 28, 2015 at 23:15
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You are confusing field extension with an algebraic field extension. In the former case, you only need a non trivial field homomorphism from the rationals to the reals, which will be injective.

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The field extension is of a quite high degree. You need to adjoin infinitely many elements (more precisely continuum many). Nobody can give you an explicit list, at least not a non-redundant one.

A field $L$ being an extention of the field $K$ just means that $K \subset L$ and the operations on elements of $K$ are the same when considered in $K$ and in $L$.

So it really just says the rational are a subset of the reals, and it does not matter whether I add and multiply two rationals thinking of them as rationals or as reals.

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Several people here have already noted that, while $\mathbb{Q}$ is a subfield of $\mathbb{R}$, this fact isn't "caused" by the same explanation as $\mathbb{R}$ being a subfield of $\mathbb{C}$. Pithily, $\mathbb{C}$ is an algebraically complete algebraic extension of $\mathbb{R}$ while $\mathbb{R}$ is a metric-complete metric extension of $\mathbb{Q}$.

If $N$ is a norm on a ring $R$, we can define Cauchy sequences in $R$ with respect to $N$ as those $(x_n)$ with $\forall \delta \in \mathbb{R}^+ \exists N\in\mathbb{N} \forall m,\,n \in \mathbb{N} (m,\,n > N \to N(x_m-x_n)<\delta)$. We can also define null sequences in $R$ with respect to $N$, viz. $\forall \delta \in \mathbb{R}^+ \exists N\in\mathbb{N} \forall n \in \mathbb{N} (n > N \to N(x_n)<\delta)$. We call Cauchy sequences $(x_n),\,(y_n)$ equivalent if $(x_n-y_n)$ is a null sequence. We can think of equivalent Cauchy sequences as "having the same limit", even if that limit does not exist in $R$. We say $R$ is metric-complete if it contains its Cauchy sequences' limits; for the choice $N(x)=|x|$, $\mathbb{R}$ is metric-complete but $\mathbb{Q}$ does not. Indeed, just as we may identify $\mathbb{C}=\mathbb{R}[i]$ with $i^2=-1$, we can identify $\mathbb{R}$ with the set of equivalence classes on $\mathbb{Q}$ with $N(x)=|x|$. Each real number is one such equivalence class. For example, $\sqrt{2}$ is the set of Cauchy sequences $(x_n)$ in $\mathbb{Q}$ for which $x_n^2 \to 2$.

If you want something to which to compare the $\mathbb{Q}$-to-$\mathbb{R}$ extension, you can consider the $p$-adic numbers. These are obtained the same way, but with a different choice of $N$. The trivial norm $$N\left(x\right)=\left\{ \begin{array}{cc} 0 & x=0\\ 1 & x\neq0 \end{array}\right.$$ obtains only eventually constant Cauchy sequences, which are equivalent iff their eventually constant values match, so $\mathbb{Q}$ is metric-complete with respect to this choice. It can be shown the only other norms on $\mathbb{Q}$ are the $p$-adic norms; for fixed $p\in\mathbb{P}$ define $$\text{ord}_p x=\inf {k\in\mathbb{Z}|p^k x\in\mathbb{Z}},\,N(x)=p^{-\text{ord}_p x}.$$Now we get different Cauchy sequences and different equivalence classes of them (as we have different null sequences), and the $p$-adic numbers $\mathbb{Q}_p$ differ from the reals (as well as the $q$-adic numbers for $q\in\mathbb{P}$ with $p\neq q$).

All metric completions of $\mathbb{Q}$ are also field extensions. They can all be algebraically extended by introducing an imaginary unit; there are also complex $p$*-adic numbers* $\mathbb{C}_p$.

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    $\begingroup$ +1 for the pith in the first paragraph. One question: I'm assuming $\mathbb{P}$ refers to some "well known" set, but I haven't see that before. What does $\mathbb{P}$ denote? $\endgroup$ Commented Dec 29, 2015 at 14:54
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    $\begingroup$ The set of primes. $\endgroup$
    – J.G.
    Commented Dec 29, 2015 at 16:08

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