Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to get caught up wit my abstract algebra class, but I'm getting lost with notation . Can someone help explain these things to me regarding fields?

$\mathbb{Q}$ is the field of all rational numbers? $\mathbb{Q}[x]$ is the field of rational numbers with polynomials? $\mathbb{Q}(x)$ is ??? An example like $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is ??

Any insight would be great. Thank you in advance.

share|cite|improve this question

$\mathbb{Q}$ is the field of rational numbers.

$\mathbb{Q}[x]$ is the ring of polynomials in one indeterminate $x$ with rational coefficients.

$\mathbb{Q}(x)$ is the field of fractions of $\mathbb{Q}[x]$; it is called the field of rational functions with coefficients in $\mathbb{Q}$. Its elements are of the form $\frac{p(x)}{q(x)}$, where $p(x)$ and $q(x)$ are polynomials with rational coefficients, and where $$\frac{p(x)}{q(x)} = \frac{r(x)}{s(x)}\iff p(x)s(x)=q(x)r(x);$$ etc.

$\mathbb{Q}(\sqrt{2},\sqrt{3})$ is the smallest subfield of $\mathbb{C}$ that contains $\mathbb{Q}$, $\sqrt{2}$, and $\sqrt{3}$. In principle, it will be equal to $$\left.\left\{\frac{p(\sqrt{2},\sqrt{3})}{q(\sqrt{2},\sqrt{3})}\right| p(x,y),q(x,y)\in\mathbb{Q}[x,y], q(\sqrt{2},\sqrt{3})\neq 0\right\},$$ though in fact one can show that every element can be written uniquely as $$a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6},\qquad a,b,c,d\in\mathbb{Q}.$$

In general, if $F\subseteq K$ are fields, and $\alpha\in K$, then $F[\alpha]$ denotes the smallest subring of $K$ that contains $F$ and $\alpha$, and $F(\alpha)$ denotes the smallest subfield of $K$ that contains $F$ and $\alpha$. It is not hard to prove that, as sets, $$\begin{align*} F[\alpha] &= \{ p(\alpha)\mid p(x)\in F[x]\},\\ F(\alpha) &= \{p(\alpha)/q(\alpha)\mid p(x),q(x)\in F[x], q(\alpha)\neq 0\}. \end{align*}$$ Though if $\alpha$ is algebraic over $F$, there are simpler expressions.

Similarly, if $S\subseteq K$ is a subset of $K$ (finite or infinite), then $$\begin{align*} F[S] &= \{ p(s_1,\ldots,s_n)\mid n\in\mathbb{N}, p(x_1,\ldots,x_n)\in F[x_1,\ldots,x_n], s_1,\ldots,s_n\in S\}\\ F(S) &= \left.\left\{ \frac{p(s_1,\ldots,s_n)}{q(s_1,\ldots,s_n)}\right| n\in\mathbb{N}, p,q\in F[x_1,\ldots,x_n], q(s_1,\ldots,s_n)\neq 0\right\}. \end{align*}$$

share|cite|improve this answer

$\mathbb{Q}[x]$ is a ring of polynomials with rational coefficients, $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ is the smallest extension of $\mathbb{Q}$ in which equations $x^2 = 2$ and $x^2 = 3$ have solutions.

share|cite|improve this answer
And $\mathbb Q(x)$ is probably the field of rational functions with coefficients in $\mathbb Q$. – user23211 Mar 29 '12 at 14:03
So $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is a field with all rational elements in addition to $\sqrt{2},\sqrt{3}$ ? – Peter K Mar 29 '12 at 14:08
@PeterK: And $\sqrt{2}+\sqrt{3}$, and $3+\sqrt{2}-\frac{1}{\sqrt{2}-4\sqrt{3}}$. And $\frac{\sqrt{2}}{\sqrt{3}}$. And... – Arturo Magidin Mar 29 '12 at 14:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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