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In the Dummit and Foote 3ed chapter on field extensions (ch. 13), it is stated as a theorem (6) that $ F(\alpha) \cong F[x]/(p(x))$ where $\alpha$ is a root of $p(x)$ and goes on to state that any field over $F$ which contains a root contains a subfield isomorphic to the extension of $F$ constructed in an earlier theorem. The earlier theorem (3), however, refers to a quotient $K$ such that once again $ K \cong F(x) / (p(x))$, it does not refer to $K$ as an extension, which is slightly confusing. Is it legitimate then to consider the quotient $F(x) / (p(x))$ in order to construct elements of the extension $F(\alpha)$ so that, for example if $F = \mathbb{Q}$ and $p(x) = x^2 -2$ then the elements of the extension would be of form $a + b \sqrt{2}$ with $ \{a,b \in \mathbb{Q}\} $ ?

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  • $\begingroup$ Yes. In fact, that very procedure is a common exercise which is to show $\mathbb{C} \cong \mathbb{R}[X]/(X^2-1)$. $\endgroup$ – FireGarden Mar 28 '14 at 22:12
  • $\begingroup$ From a notation standpoint, then, is the notation for the extension $K/F$ ($K$ over $F$) which contains $\alpha$ the same as the notation $F(\alpha)$? $\endgroup$ – Rohit Khera Mar 28 '14 at 22:23
  • $\begingroup$ Well, not necessarily; $K/F$ could have a different root of $p(X)$ adjoined, but it would be isomorphic. They would be the same if you knew $K$ contained specifically $\alpha$. $\endgroup$ – FireGarden Mar 28 '14 at 22:33
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    $\begingroup$ @FireGarden whoops: $x^2+1$ right? :) $\endgroup$ – rschwieb Mar 28 '14 at 22:36
  • $\begingroup$ @rschwieb Indeed! Silly mistake. $\endgroup$ – FireGarden Mar 28 '14 at 22:37
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Notice that the polynomials of degree less than one form an isomorphic copy of F. In this way, F is embedded in K so that it's an extension of F.

Furthermore, this did not come up in the discussion so far, but the polynomial must be irreducible over F, or else the quotient won't be a field, hence certainly not an extension field.

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