Question About Notation In Field Theory.

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

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

• 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 $\mathbb{Q}[i]$? – Mark Mar 20 '14 at 21:20
• 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 $K$. – Daniel Fischer Mar 20 '14 at 21:22
• @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! – Mark Mar 20 '14 at 21:30
• @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. – Daniel Fischer Mar 20 '14 at 21:48
• @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. :) – 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.