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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.

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2 Answers 2

$\mathbb{Q}$ 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 $$\begin{align*} F[S] &= \{ p(s_1,\ldots,s_n)\mid n\in\mathbb{N}, p(x_1,\ldots,x_n)\in F[x_1,\ldots,x_n], s_1,\ldots,s_n\in S\}\\ F(S) &= \left.\left\{ \frac{p(s_1,\ldots,s_n)}{q(s_1,\ldots,s_n)}\right| n\in\mathbb{N}, p,q\in F[x_1,\ldots,x_n], q(s_1,\ldots,s_n)\neq 0\right\}. \end{align*}$$

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$\mathbb{Q}[x]$ 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.

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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
3  
@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

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