Explain to me the difference between the notation $\mathbb{Q}( \sqrt2) $and$ \mathbb{Q}[ \sqrt2]$ Please explain to me the difference between the notation $\mathbb{Q}( \sqrt2) $and$ \mathbb{Q}[ \sqrt2]$.
I know that these two fields are equal. But what difference do the different brackets imply? I know the square brackets give numbers of the form $a+b\sqrt2$. Through what action do the round brackets generate the field?
 A: The notation with the rectangular brackets is at first a ring: $\mathbb{Q}$-linear combinations of $1$ and $\sqrt{2}$. However this ring is actually a field since $\frac{1}{a+b\sqrt{2}}= \frac{a-b\sqrt{2}}{a^2-2b^2}$ is again a $\mathbb{Q}$-linear combination of $1$ and $\sqrt{2}$. So every non-zero element has an inverse and hence we have a field. The notation with the round brackets is use for denoting a field generated by $\mathbb{Q}$ and $\sqrt{2}$. So here the two coincide. Note that if you use a transcendental number, say $\pi$, or a look at polynomials in $X$, then for example $\mathbb{Q}[\pi] \subsetneq \mathbb{Q}(\pi)$ and $\mathbb{Q}[X] \subsetneq \mathbb{Q}(X)$ respectively.
A: $\mathbb{Q}[\sqrt{2}]$ gives you all polynomials with $\sqrt{2}$ as the variable, including $a+b\sqrt{2}+c\sqrt{2}^2+d\sqrt{2}^3$.  This is the same as the set of $a+b\sqrt{2}$ of course because $\sqrt{2}^2$ is rational.
$\mathbb{Q}(\sqrt{2})$ is the set of rational functions, which are ratios of polynomials of $\sqrt{2}$.  That is $(a+b\sqrt{2})/(c+d\sqrt{2})$.  This is the same because
$$\frac{a+b\sqrt{2}}{c+d\sqrt{2}}=\frac{(a+b\sqrt{2})(c-d\sqrt{2})}{c^2-2d^2}$$ which can be written as a polynomial in $\sqrt{2}$.
A: Whenever you have an integral domain $Q$ you can form its field of fractions $F(Q)$ by considering all formal fractions $a/b$, $a,b\in Q$, $b \neq 0$. (You have to take some equivalence classes to ensure things like $(na/nb) = (a/b)$, namely via the relation $a/b = c/d$ if and only if $ad = bc$.) You define multiplication the usual way : $(a/b)\cdot(c/d) = ac/bd$, and similarly with addition.
In this case $\mathbb{Q}[\sqrt{2}]$ is an integral domain since it is a subring of $\mathbb{C}$. So you can form its field of fractions, and it is usually denoted $\mathbb{Q}(\sqrt{2})$. For another example, the field of fractions of $\mathbb{Q}[x]$ is $\mathbb{Q}(x)$, the field of rational functions (with $\mathbb{Q}$-coefficients).
