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When I studied algebra, we talked about fields such as $\mathbb{Q}[\sqrt{2}]$, the rational numbers with the square root of two adjoined to the field. Structures like these are called field extensions and are themselves fields.

My question is when we talk about these structures, are we saying we add the square root of two and its (multiplicative) inverse, or just the square root of two and any numbers necessary to keep the closure property of our field satisfied? If the latter is the case, is this part of what makes transcendental numbers so significant? If we only add the transcendental number to the field, and all numbers necessary to keep the closure, there is no way to obtain an inverse for that number. This is unlike the square root of 2, whose inverse can be written as $\frac{\sqrt{2}}{2}$, which would be in the extension by closure.

We didn't go into a lot of details in the algebra class I took so these are just things I've been thinking about on my own.

KCd
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mphy
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    Careful! The fact that $\Bbb Q[\sqrt 2]$ is a field is a consequence of $\sqrt 2$ being algebraic. $\Bbb Q[\pi]$ is not a field. Of course, $\Bbb Q(\pi)$ is. – Pedro Dec 25 '14 at 04:32
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    Adjoining an element to create a larger field requires including enough new terms to have a field, so a multiplicative inverse must be in the new field. The distinction (I would not call it the significance) between algebraic and transcendental numbers is that when you adjoin (not "add") an algebraic element to $\mathbf Q$ you have a field as soon as you are closed under adding and multiplying, but this is not true when adjoining a transcendental number, e.g., $\mathbf Q[\sqrt{2}] = \mathbf Q(\sqrt{2})$ but $\mathbf Q[e] \not= \mathbf Q(e)$. – KCd Dec 25 '14 at 04:33
  • $\mathbb{Q}(\sqrt{2})$ is the smallest field containing $\sqrt{2}$. So you're essentially taking $\mathbb{Q}$, then adding $\sqrt{2}$, and from there add everything else that you can get of the form $a + b\sqrt{2}$ where $a, b \in \mathbb{Q}$, and then repeat this indefinitely. If our added element wasn't something like it is, we'd need another term for its inverse. But you're essentially augmenting $\mathbb{Q}$ by adding on $\sqrt{2}$, and then making it satisfy your field axioms by adding the necessary elements. – AJY Dec 25 '14 at 04:33
  • Another distinction is that adjoining an algebraic element to $\mathbf Q$ creates a finite-dimensional field extension, while adjoining a transcendental element to $\mathbf Q$ creates an infinite-dimensional field extension. – KCd Dec 25 '14 at 04:36
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    Another point no one has addressed is that adjoining an element (of a larger ring) as well as its inverse is insufficient to create a field. For instance, $\Bbb Q[\pi,\pi^{-1}]$ is still not a field, because it does not contain $(1+\pi)^{-1}$. If $T$ is transcendental over a field $K$, then the field $K(T)$ is more or less the field of rational functions in the variable $T$ with coefficients taken from $K$. – anon Dec 25 '14 at 06:38
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    See this answer for an explanation of ring adjunction notation $,R[x],$ and $,R[\alpha].\ $ – Bill Dubuque Dec 29 '14 at 01:36

2 Answers2

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In the case of $\mathbb{Q}[\sqrt{2}]$, we have not only $\sqrt{2}$ and its multiplicative inverse, but everything necessary to retain closure under the operations. There is a notational point to make here:

  • $F[a]$ is defined to be the set $\{f(a) \ | \ f(x) \in F[x]\}$.
  • $F(a)$ is defined to be the "smallest" extension field of $F$ that contains $a$.

But notice that we are talking about $\mathbb{Q}[\sqrt{2}]$ as a field as in the second point! What gives?

Theorem: When $a$ is algebraic over a field $F$, then $F[a] = F(a)$.

Proof:

Since $F[a]$ is a ring, most field properties already hold. What is left is to demonstrate the existence of multiplicative inverses. To do this, we take advantage of the Euclidean algorithm:

Let $f(x) \in F[x]$ be the minimal polynomial for $a$. Every polynomial without $a$ as a root will correspond to a nonzero element in $F[a]$, and moreover, every such polynomial will be relatively prime to $f(x)$. That is, given such a $g(x)$, there exists polynomials $h(x)$ and $k(x)$ such that:

$$f(x)h(x) + g(x)k(x) = 1$$

Since $a$ is a root of $f(x)$, evaluating the above at $a$ gives:

$$g(a)k(a) = 1$$

So given any nonzero $g(a) \in F[a]$, there exists some $k(a)$ that serves as its multiplicative inverse. This is to say: every nonzero element in $F[a]$ has a multiplicative inverse. We can conclude that, if $a$ is algebraic over $F$, then $F[a]$ is a field and $F[a] = F(a)$.

Final comments:

What makes an algebraic adjunction to a field special? Unlike transcendental adjunctions, algebraic ones are finite. That is, if $a$ is algebraic over $F$, then $F[a]$ can be viewed as a vector space over $F$ spanned by finitely many basis "vectors".

For example, $\mathbb{Q}[\sqrt{2}]$ is a finite extension of degree $2$, meaning any basis contains $2$ basis vectors. One possible basis is $\{1, \sqrt{2}\}$, and so $\mathbb{Q}[\sqrt{2}] = \{a + b\sqrt{2} \ | \ a, b \in \mathbb{Q} \}$.

Kaj Hansen
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The notation "$A[x]$" refers to polynomials in $x$ with coefficients in $A$ and so could in theory not contain inverses. In practice, you can show that an algebra over a field $k$ that is a finite dimensional $k$-vector space and also an integral domain must itself be a field and so we do contain inverses. You are then correct to point out that in cases where we are not dealing with a finite dimensional vector space (such as with a transcendental extension) it may indeed occur that the resulting object is not a field.

Alexander
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  • Unless you specify $x$ is a variable, no. We could say, "let $B/A$ be an extension of rings, and take $x\in B$. Let $B'=A[x]$...". Some authors prefer $A[X]$ for such reason. – Pedro Dec 25 '14 at 04:37
  • But then $B'$ contains precisely polynomial expressions in $x$, though some of those expressions may in fact be equivalent. Right? – Alexander Dec 25 '14 at 04:39
  • Correct. ${}{}{}$ – Pedro Dec 25 '14 at 05:00
  • Yeah, though my wording is somewhat iffy, I agree. I will upvote the other answer. – Alexander Dec 25 '14 at 05:18
  • You could just add "For some authors ," at the beginning of your answer. =) – Pedro Dec 25 '14 at 05:20