what does $\mathbb{Q}(\sqrt{3})$ mean?
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This is the extension field of $\mathbb{Q}$ obtained by adjoining the square root of $3$. It can be realized as the quotient $\mathbb{Q}[x]/(x^2-3)$, but more directly it can be realized as the set of all numbers $a+b\sqrt{3}$ with $a$ and $b$ rational subject to the usual addition and multiplication rules.
Matt Samuel
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In general, if you have two fields $K$ and $F$ with $F\supseteq K$, and $\alpha\in F\smallsetminus K$, $K(\alpha)$ is the smallest subfield of $F$ containing $\alpha$ and $K$. In the case of $\mathbb{Q}(\sqrt3)$, $K=\mathbb{Q}$, $\alpha=\sqrt3$, and $F$ is typically taken to be the reals, $\mathbb{R}$, but could be taken to be the field of constructible numbers, or of algebraic complex numbers. The result would probably be the same, i.e. the set of all $a+b\sqrt3$ with $a,b\in\mathbb{Q}$.
As a side note, $K[\alpha]$ (with square brackets rather than round ones) denotes the smallest subring of $F$ containing $K$ and $\alpha$. These two need not be equal, but providing a counterexample goes beyond the question's scope, it seems.
MickG
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