I have a question about notation specifically square brackets $[$ and round brackets $($.

My textbook doesn't explain any of this and I cannot find a reliable source online to confirm the difference.

So my question is: What is the difference between round brackets and square brackets in terms of notation in Field Theory? For example, I see $F(x)$ and $F[x]$ in my textbook and I've always assumed they were the same thing. But apparently they're not. Is there ever a time they're the same?

I wanted to know the difference, it may be a silly question but it's something I want to make sure I understand.


Consider $x$ as an element contained in some extension of $F$; then $F(x)$ is the smallest field that contains both $F$ and $x$. On the contrary $F[x]$ is the smallest ring that contains both $F$ and $x$. Clearly $F[x]\subseteq F(x)$, and in some cases they are equal. You can prove easily (or see it on Morandi's book) that

$$F[x]=\{f(x):\textrm{$f$ is a polynomial with coefficients in $F$}\}$$ $$F(x)=\left\{\frac{f(x)}{g(x)}:\textrm{$f,g$ are polynomials with coefficients in $F$ and $g(x)\neq0$}\right\}$$

  • $\begingroup$ Ah, so to denote the difference. Fields uses square brackets and rings uses round brackets? So in that case, would that mean $\mathbb{Q}(i)$ is the same as $\mathbb{Q}[i]$? $\endgroup$ – Mark Mar 20 '14 at 21:20
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    $\begingroup$ Yes, $\mathbb{Q}[i] = \mathbb{Q}(i)$, @Mark. There's a difference between $R(\alpha)$ and $R[\alpha]$ generally when $R$ is a ring but not a field. For a field $K$, you have $K[\alpha] \neq K(\alpha)$ if and only if $\alpha$ is transcendental over $K$. $\endgroup$ – Daniel Fischer Mar 20 '14 at 21:22
  • $\begingroup$ @DanielFischer So let's say you have a ring R. What's the difference between $R[\alpha]$ and $R(\alpha)$? Is there even a point to consider $R[\alpha]$ since it's not a field? One more question as well, so let's say $K$ is a field and $\alpha$ algebraic. Would that mean $K[\alpha] = K(\alpha)$?. Thanks again! $\endgroup$ – Mark Mar 20 '14 at 21:30
  • $\begingroup$ @Mark We're often interested in rings that aren't fields, $\mathbb{Z}[i]$ or $\mathbb{Z}[e^{2\pi i/3}]$ for example are quite useful rings, as are polynomial rings. For a field $K$ and algebraic $\alpha$, we always have $K[\alpha] = K(\alpha)$. That's because $K[\alpha]$ is then a finite-dimensional $K$-vector space, and has no zero divisors. Therefore multiplication by any nonzero element $\beta$ is an injective $K$-linear map, hence surjective, and that means $\beta$ has a multiplicative inverse. $\endgroup$ – Daniel Fischer Mar 20 '14 at 21:48
  • $\begingroup$ @DanielFischer Woops. I meant to say $R(\alpha)$! Sorry about that. Anyways, your explanation pretty much sums up what I wanted. Thank you once again. :) $\endgroup$ – Mark Mar 20 '14 at 21:54

$F[x]$ denotes the ring of (formal!) polynomials over $F$ in the indeterminate $x.$
$F(x)$ denotes its fraction field - the so-called rational "functions" in $x.$

The same notation is used for adjunctions of elements to rings and fields, i.e. if $\,\alpha\,$ is an element of some extension ring $\,E\,$ then $\,F[\alpha]\,$ denotes the smallest subring $\,R\,$ of $\,E\,$ that contains $\,F\,$ and $\,\alpha.\,$ Equivalently, it is the image of $\,F[x]\,$ under the evaluation map $\,x\mapsto \alpha,\,$ i.e. the set of all elements expressible as polynomials in $\,\alpha\,$ with coefficients in $\,F.\,$ If $\,\alpha\,$ is transcendental (= not algebraic) over $\,F,\,$ i.e. $\,\alpha\,$is not a root of any nonzero polynomial $\,f\in F[x],\,$ then there is a ring isomorphism $\,F[\alpha]\cong F[x].\,$ Thus the polynomial ring $\,F[x]\,$ can be viewed as the ring obtained by adjoining to $\,F\,$ any element that is transcendental over $\,F,\,$ e.g. $\,\Bbb Q[x]\cong \Bbb Q[\pi].\,$ Similarly for the case of fields.


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