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Is there any difference between $\mathbb{Q}(\sqrt2)$ and $\mathbb{Q}[\sqrt2]$

I used to be very comfortable with the definition of $\mathbb{Q}[\sqrt2]$ but once I got to Simple Field extensions, a new notation was introduced, that of $\mathbb{Q}(\sqrt2)$ which got me a little confused.

Thank you

Quester
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  • Possible duplicate of http://math.stackexchange.com/questions/472497/field-extension-notation. – lhf Jan 31 '14 at 00:41

3 Answers3

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By definition, $\mathbb{Q}(\sqrt 2)$ is the field generated by $\mathbb{Q}$ and $\sqrt 2$, while $\mathbb{Q}[\sqrt 2]$ is the ring generated by these (compare with the polynomial ring $\mathbb{Q}[x]$). However, since $\sqrt 2$ is algebraic over $\mathbb{Q}$, these coincide: In fact, we can compute that

$$\sqrt 2 \left(\frac{\sqrt 2}{2}\right) = 1$$

so that $\sqrt 2$ has a multiplicative inverse in $\mathbb{Q}[\sqrt 2]$.

For a different scenario, note that $\mathbb{Q}[\pi]$ is properly contained in $\mathbb{Q}(\pi)$, since $\pi$ has no multiplicative inverse within the ring $\mathbb{Q}[\pi]$.

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    Just a remark: a priori, it isn't sufficient to check that $\sqrt 2$ is invertible in $\mathbf Q[\sqrt 2]$ to conclude that $\mathbf Q[\sqrt 2]$ is a field and thus equal to $\mathbf Q(\sqrt{2})$. One should do so for every nonzero $a+b\sqrt 2$. – Bruno Joyal Jan 31 '14 at 00:44
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$\mathbb{Q}(\sqrt{2})$ is the smallest field containing $\mathbb{Q}$ and $\sqrt{2}$. $\mathbb{Q}[\sqrt{2}]$ if, by definition, the ring consisting of all the elements $p(\sqrt{2})$ for $p$ any polynomial in $\mathbb{Q}[x]$. It is not too hard to show that in fact $\mathbb{Q}[\sqrt{2}] = \{a + b\sqrt{2} : a,b\in \mathbb{Q}\}$. So clearly you have $\mathbb{Q}[\sqrt{2}] \subseteq \mathbb{Q}(\sqrt{2})$.

So if you can find an inverse to any general (non-zero) element $a + b\sqrt{2}$, then you have showed that every (non-zero) element has an inverse and so $\mathbb{Q}[\sqrt{2}]$ is in fact a field and $\mathbb{Q}(\sqrt{2}) = \mathbb{Q}[\sqrt{2}]$.

Thomas
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Much more generally, if $r$ is algebraic over $\mathbb{Q}$, then $$ \mathbb{Q}[r]=\mathbb{Q}(r) $$ Of course, it's sufficient to show that any non zero element in $\mathbb{Q}[r]$ has its inverse in $\mathbb{Q}[r]$. But it's known that an element $s\in\mathbb{Q}[r]$ is again algebraic over $\mathbb{Q}$, so, if $s\ne0$, it has a minimum polynomial $$ p(X)=a_0+a_1X+\dots+a_{n-1}X^{n-1}+X^n\in\mathbb{Q}[X] $$ with $a_0\ne0$ (otherwise the polynomial wouldn't be minimal for $s$). Therefore $$ 0=s^{-1}(a_0+a_1s+\dots+a_{n-1}s^{n-1}+s^n) $$ and so $$ s^{-1}=-a_0^{-1}(a_1+a_2s+\dots+a_{n-1}s^{n-2}+s^{n-1})\in\mathbb{Q}[s] \subseteq\mathbb{Q}[r]. $$

egreg
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