Do $\Bbb Q (\sqrt 2)$ and $\Bbb Q [\sqrt 2]$ mean the same?

I'm trying to refer to the field of the real numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are rationals.

E: I'm sorry, my question was unclear, I was using $\sqrt 2$ as an example number, but from what I read in the answers, if I choose a number like $\pi$ then $\Bbb Q (\pi)$ and $\Bbb Q [\pi]$ would be different, correct?

  • $\begingroup$ Personally, I have only seen the use of the latter for what you described. $\endgroup$ – anakhro Jun 9 '15 at 19:46
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    $\begingroup$ Yes. The different meaning of the notations manifests itself only if the "new" element is transcendental over the field, where the former means smallest field containing the element and the latter means smallest ring containing the element (i.e. rational functions vs polynomials). $\endgroup$ – Pavel Čoupek Jun 9 '15 at 19:51
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    $\begingroup$ @PavelČoupek That should be an answer $\endgroup$ – Hagen von Eitzen Jun 9 '15 at 19:58

Yes, they are the same field, although the notation $\Bbb{Q}(\sqrt{2})$ is preferred.

The reason is that if $a$ is a number or a variable and $F$ is a field, then $F[a]$ denotes the ring of polynomial elements in $a$. These are all the elements of the form $$ \gamma_n a^n + \gamma_{n-1} a^{n-1} + \dotsb + \gamma_0 $$ with $n$ any non-zero integer and $\gamma_n,\dotsb,\gamma_0 \in F$. Note that this ring is always a domain.

On the other hand, the notation $F(a)$ denotes the field of fractions of $F[a]$, and its elements are of the form $$ \frac{f}{g} $$ with $f,g \in F[a]$ and $g \neq 0$. This is the smallest field that contains $F[a]$.

The thing is, whenever $a$ is algebraic over $F$ the two are the same. This is because if $a$ is algebraic then it is a root of an irreducible polynomial $\varphi \in F[X]$. You can check that $$ F[X]/(\varphi) \simeq F[a] $$ and this is already a field because $\varphi$ irreducible implies that the ideal $(\varphi)$ is maximal.

It is a nice exercise to check that the elements of this field are exactly the polynomials in $a$ of degree at most $\deg(\varphi)-1$. In particular, this is why $F(a)$ is a vector space of dimension $\deg(\varphi)$ over $F$ whenever $a$ is algebraic over $F$.

Note that when $a$ is transcendental over $F$, then $F[a] \subsetneq F(a)$. Indeed, in this case $F[a]$ is isomorphic to $F[X]$ as a ring, but $X$ is not invertible in $F[X]$.

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If you have a ring $B$, a subring $A\subseteq B$ and $b\in B$, then

$A[b]$ denotes the smallest subring of $B$ which contains both $A$ and $b$.

On the other hand, if $E$ is a field, $F\subseteq E$ a subfield and $e\in E$, then

$F(e)$ denotes the smallest subfield of $E$ which contains both $A$ and $b$.

The inclusion $\mathbb Q\subseteq\mathbb C$ is at the same time an inclusion of rings and of fields, so we can consider $\mathbb Q[\sqrt2]$ and $\mathbb Q(\sqrt2)$. It is a non-trivial fact that these two subrings of $\mathbb C$ are equal. On the other hand, $\mathbb Q[\pi]$ and $\mathbb Q(\pi)$ are not equal.

In short, the two notations do mean different things in general, which happen to coincide in your special case.

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  • $\begingroup$ It should be remarked that if $A$ is a ring and $X$ is a variable (the difference is mostly psycohological, but still), then the notations $A[X]$ and $A(X)$ mean something quite different from what I described in my answer. $\endgroup$ – Mariano Suárez-Álvarez Jun 9 '15 at 20:43

In principle $\Bbb Q [\sqrt 2]$ means the ring of the expressions of the type $a + b \sqrt 2$, and $\Bbb Q (\sqrt 2)$ is the field of fractions of $\Bbb Q [\sqrt 2]$, which for us will simply mean fractions of numbers from $\Bbb Q [\sqrt 2]$, i.e. fractions like $\frac {a + b \sqrt 2} {c + d \sqrt 2}$. Remarkably, though, fractions like this can be rewritten as

$$\frac {(a + b \sqrt 2) (c - d \sqrt 2)} {(c + d \sqrt 2) (c - d \sqrt 2)} = \frac {(ac - 2bd) + (bc - ad) \sqrt 2} {c^2 - 2 d^2} = \frac {ac - 2bd} {c^2 - 2 d^2} + \frac {bc - ad} {c^2 - 2 d^2} \sqrt 2 ,$$

which is again precisely of the form $a' + b' \sqrt 2$.

This shows that in fact $\Bbb Q (\sqrt 2) \subset \Bbb Q [\sqrt 2]$ and since trivially $\Bbb Q [\sqrt 2] \subset \Bbb Q (\sqrt 2)$, the two coincide.

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    $\begingroup$ Actually, $\mathbb Q[\sqrt2]$ means, in principle, the ring of expressions of the type $a+b\sqrt2+c(\sqrt2)^2+\cdots+z(\sqrt2)^k$ for all $k$s, and that happens to coincide with what you describe in this particular case. Your description does not work for $\mathbb Q(\sqrt[3]2)$. $\endgroup$ – Mariano Suárez-Álvarez Jun 9 '15 at 20:40
  • $\begingroup$ @MarianoSuárez-Alvarez: I didn't make a full theory of $\Bbb Q (x)$, I only answered the question asked by the OP using the fact that $(\sqrt 2) ^2 \in \Bbb Q$. Let's not be pedantic in such an elementary context. $\endgroup$ – Alex M. Jun 9 '15 at 20:43
  • $\begingroup$ That is not being pedantic: it is being correct. $\endgroup$ – Mariano Suárez-Álvarez Jun 9 '15 at 20:44
  • $\begingroup$ @MarianoSuárez-Alvarez: What you are saying is like "let $K \le L$ be an extension of degree $n$, with $n=2$". Pedestrians like me would probably cut to the chase and say "an extension of degree 2". $\endgroup$ – Alex M. Jun 9 '15 at 20:47
  • $\begingroup$ No, I am not saying that at all. I am saying that $\mathbb Q[\sqrt2]$ is by definition the ring generated by $\sqrt2$ and $\mathbb Q$, which is what it is. Saying that what you say I am saying would not be pedantic: it would be silly, rather. $\endgroup$ – Mariano Suárez-Álvarez Jun 9 '15 at 20:54

An algebraic structure is specified by two components: a signature (a list of functions and predicates) and a list of axioms. For example, one way to define a group is as follows:

  1. Signature: multiplication (a binary operation), inverse (a unary operation) and identity (a constant). We denote these by concatenation, $^{-1}$ and $e$, respectively.
  2. Axioms: The usual group axioms: $\forall abc (a(bc) = (ab)c)$; $\forall a (aa^{-1} = a^{-1}a = e)$; $\forall a (ea=ae=e)$.

An instance of the algebraic structure is an underlying set and a collection of functions and predicates matching the signature and satisfying the axioms. For example, the group $\mathbb{Z}_2$ (or rather, one realization of it) has the underlying set $\{0,1\}$, multiplication is defined by $0^2=1^2=0,01=10=1$, inverse is defined by $0^{-1}=0,1^{-1}=1$, and the identity by $e = 0$. We usually identify algebraic structures up to isomorphism.

In your case, $\mathbb{Q}[\sqrt{2}]$ is a ring while $\mathbb{Q}(\sqrt{2})$ is a field. This means that there is a type mismatch when asking whether $\mathbb{Q}[\sqrt{2}] = \mathbb{Q}(\sqrt{2})$. How serious this type mismatch is depends on the exact definitions of ring and field and on how formal you want to be. Here are two reasonable options:

  1. You can define a ring as an algebraic structure with addition, multiplication, negation, and additive and multiplicative identities, satisfying the ring axioms. For a field, you add reciprocal to the signature and the field axioms to the list of axioms. In this case, rings and fields have different signatures.

  2. You can define ring as in option 1, and field as having the same signature as ring. In order to state the field axioms, you define the reciprocal as the unique element $x^{-1}$ such that $x^{-1}x=xx^{-1}=1$ if $x \neq 0$, and (say) $0^{-1}=0$.

Under the second option, you can argue that $\mathbb{Q}[\sqrt{2}]$ is a ring that satisfies the additional axioms for fields, and so it can be considered a field.

Under the first option, technically $\mathbb{Q}[\sqrt{2}] \neq \mathbb{Q}(\sqrt{2})$ since they have a different signature; however, there is a unique way to define the reciprocal so that the resulting structure satisfies the field axioms, and once you do that, then again the structures become identical. So $\mathbb{Q}[\sqrt{2}]$ can be "typecast" to a field, and then it becomes identical with (or rather, isomorphic to) $\mathbb{Q}(\sqrt{2})$. This typecast is harmless since for every ring there is at most one way of looking at it as a field, and so no ambiguity arises.

Summarizing, our usual sloppy conventions allow us to regard $\mathbb{Q}[\sqrt{2}]$ and $\mathbb{Q}(\sqrt{2})$ as representing the same structure. However, if we are being very formal (say when using a proof assistant), then we might have to explicitly argue that $\mathbb{Q}[\sqrt{2}]$ satisfies the field axioms, or even to argue that the reciprocal exists.

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  • $\begingroup$ While this is perfectly correct, of course, I fear it will only help confuse the OP or anyone with the same doubt a bit more :-) $\endgroup$ – Mariano Suárez-Álvarez Jun 11 '15 at 1:02
  • $\begingroup$ The idea that setting up a general context in which a question finds its most natural answer and answering it there is helpful is a professional deformation mathematicians often suffer :-D $\endgroup$ – Mariano Suárez-Álvarez Jun 11 '15 at 1:03
  • $\begingroup$ @MarianoSuárez-Alvarez Well, the "expected" answer already appeared here, but this subtlety nobody had commented about. $\endgroup$ – Yuval Filmus Jun 11 '15 at 3:05

Yes, they are same, but it's because $\mathbb{Q}$ is a field, and they are not same if you replace it by an arbitrary ring, which is not a field. In general, bracket gives the meaning "smallest ring containing the element in the bracket and the given ring", while paranthesis means "smallest field containing the element in the paranthesis and the given ring." So, when the given ring is itself a field, they're basically same.

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    $\begingroup$ It is not because $\mathbb Q$ is a field —for example, $\mathbb Q[\pi]$ and $\mathbb Q(\pi)$ are actually different. $\endgroup$ – Mariano Suárez-Álvarez Jun 9 '15 at 20:30
  • $\begingroup$ I've never seen, say, $A(x)$ to mean the field of fractions of $A[x]$ when $A$ is not a field. Could you give me a reference? Also, the first sentence is ambiguous and the last one isn't correct: they are the same because $\Bbb{Q}$ is a field and $\sqrt{2}$ is algebraic. For example, $\Bbb{Q}(\pi)$ is a field, but $\Bbb{Q}[\pi]$ is not. $\endgroup$ – A.P. Jun 9 '15 at 20:34
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    $\begingroup$ Your conclusion is correct but the explanation is wrong: that fact that $\Bbb Q$ is a field is not the crux of the issue here, even though it is important. The core fact is that $\sqrt 2$ is algebraic over $\Bbb Q$. $\endgroup$ – Alex M. Jun 9 '15 at 20:39
  • $\begingroup$ Yes, you guys are definitely right, we use that $\sqrt{2}$ is algebraic, sorry about my false statement. $\endgroup$ – vgmath Jun 10 '15 at 1:00

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