# Extension Fields and Quotients

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 $\{a,b \in \mathbb{Q}\}$ ?

• Yes. In fact, that very procedure is a common exercise which is to show $\mathbb{C} \cong \mathbb{R}[X]/(X^2-1)$. – FireGarden Mar 28 '14 at 22:12
• From a notation standpoint, then, is the notation for the extension $K/F$ ($K$ over $F$) which contains $\alpha$ the same as the notation $F(\alpha)$? – Rohit Khera Mar 28 '14 at 22:23
• 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 $\alpha$. – FireGarden Mar 28 '14 at 22:33
• @FireGarden whoops: $x^2+1$ right? :) – rschwieb Mar 28 '14 at 22:36
• @rschwieb Indeed! Silly mistake. – FireGarden Mar 28 '14 at 22:37