4
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
1
$\begingroup$

$F[x]$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.