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let $K\subseteq L$ be a field extension and let $K=\mathbb{Q}$ and $L=\mathbb{C}$ also let $\alpha =\sqrt{2}$.

Then $\mathbb{Q}(\sqrt{2})=\mathbb{Q}[\sqrt{2}]$

What is the difference in the above notations? and why do we get them?

$\mathbb{Q}(\sqrt{2})=\{a+b\sqrt{2}|a,b\in\mathbb{Q}\}$

what is $\mathbb{Q}[\sqrt{2}]$?

H.E
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    $\mathbb{Q}[\sqrt{2}]$ is the smallest subring of $\mathbb{C}$ that contains $\mathbb{Q}$ and $\sqrt{2}$. $\mathbb{Q}(\sqrt{2})$ is the smallest subfield containing $\mathbb{Q}$ and $\sqrt{2}$. It turns out that for algebraic numbers, the smallest subring is a field, so the two coincide then. They are different for transcendental numbers, $\mathbb{Q}[\pi] \neq \mathbb{Q}(\pi)$. – Daniel Fischer Nov 01 '13 at 15:57
  • Thank you Daniel. Great explanation – H.E Nov 01 '13 at 16:05

1 Answers1

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For fields $K \subset L$, and an $\alpha\in L$,

  • $K[\alpha]$ denotes the smallest subring of $L$ that contains both $K$ and $\alpha$, and
  • $K(\alpha)$ denotes the smallest subfield of $L$ that contains both $K$ and $\alpha$.

$K[\alpha]$ is the set of all polynomial expressions in $\alpha$,

$$K[\alpha] = \left\lbrace f(\alpha) : f \in K[X]\right\rbrace,$$

and $K(\alpha)$ is the set of all rational functions evaluated in $\alpha$ subject to the restriction that the denominator doesn't vanish in $\alpha$,

$$K(\alpha) = \left\lbrace \frac{f(\alpha)}{g(\alpha)} : f,g\in K[X],\, g(\alpha)\neq 0\right\rbrace.$$

It turns out that for algebraic $\alpha$, the two coincide, $K[\alpha] = K(\alpha)$. In fact, that is a necessary and sufficient condition, $\alpha$ is algebraic over $K$ if and only if $K[\alpha]$ is a field.

Daniel Fischer
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