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What's the difference between $\mathbb{Q}[\sqrt{-d}]$ and $\mathbb{Q}(\sqrt{-d})$?

Let $$\mathbb Q[i]=\{a+ib|a, b\in \mathbb Q\}$$ Any nonzero element $a+ib\in\mathbb{Q}[i]$ has an inverse element because $$\frac{1}{a+ib}=\frac{a-ib}{a^2+b^2}\in\mathbb{Q}.$$

Is it true that $\mathbb{Q}(i)=\mathbb{Q}[i]$?


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marked as duplicate by Willie Wong May 24 '12 at 14:02

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up vote 4 down vote accepted

Yes, that is true. Of course, this follows from your explicit description of the inverse.

Also, $i$ has the minimal polynomial $X^2+1$ over $\mathbb{Q}$ and therefore the extension $\mathbb{Q}[i]|\mathbb{Q}$ is finite. Then it follows $\mathbb{Q}(i)=\mathbb{Q}[i]$. This is a general result for finite extensions.

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Why from $[\mathbb{Q}[i]:\mathbb{Q}]<\infty$ follows $\mathbb{Q}[i]=\mathbb{Q}(i)$? – Aspirin May 24 '12 at 12:15

Yes. This is a general fact: if $\,\mathbb{F}\,$ is any field and $\,\alpha\,$ is an element in some field extension, then $\,\alpha\,$ is algebraic over $\,\mathbb{F}\,$ iff $\,\mathbb{F}[\alpha]=\mathbb{F}(\alpha)\,$ .

As $\,i\,$ is a root of the rational pol. $\,x^2+1\,$ , it is alg. over the rationals...

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added The proof of the above is short and easy: – DonAntonio May 24 '12 at 12:07

Yes, for exactly the reason you cited.

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