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Actually, I've not studied deeply field theory and I'm reading an article about the field theory, especially field extension.

Suppose that $\mathbb{F}$ be a field. And the book treats $\mathbb{F}(\theta)$ as the field extension of $\mathbb{F}$.

Furthermore, even $\mathbb{Q}(x,\theta_1,\theta_2)$ appears on here.

I guessed that $\mathbb{F}(x)$ means the extension field of $\mathbb{F}$ which has $x$ as its element. To guarantee the uniqueness, it would be better to put one more condition, $\mathbb{F}\cup\{x\}$ cannot be a subset of every proper subfield of $\mathbb{F}(x)$.

Did I guess correctly?

If you want to know full context, visit here.

Thank you.

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  • $\begingroup$ If $x\not\in F$ and $K/F$ is an arbitrary nontrivial field extension then $F\cup\{x\}$ is not a subset of the proper subfield $F$ of $K$, so your condition is applicable to any extension $K$ (hence doesn't guarantee anything, let alone uniqueness). $\endgroup$ – anon Jul 4 '13 at 0:21
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If $K$ is an extension of $F$ and $\theta\in K$, then by definition, $F(\theta)$ is the smallest subfield of $K$ containing both $F$ and $\theta$, that is, the intersection of all subfields of $K$ containing $F$ and $\theta$). More generally, the single element $\theta$ can be replaced with an arbitrary set of elements in $K$, say $S$, and then $F(S)$ is the smallest subfield of $K$ containing $F$ and $S$.

I should add that, you are correct when you say that no proper subfield of the extension $F(\theta)/F$ contains both $F$ and $\theta$. This follows immediately from the definition of $F(\theta)$ as the smallest subfield of $K$ containing $F$ and $\theta$.

If $\theta$ is algebraic over $F$, with minimal polynomial $f(X)\in F[X]$, then the unique $F$-algebra map $X\mapsto\theta:F[X]\rightarrow F(\theta)$ gives an isomorphism of $F$-algebras $F[X]/(f(X))\cong F(\theta)$. If $\theta$ is transcendental, the aforementioned map is injective and extends to an isomorphism $F(X)\cong F(\theta)$ (note that $F(X)$ is by definition the field of fractions of $F[X]$, the polynomial ring, but this is consistent with the notational convention I described in the previous paragraph with $K=F(X)$ and $\theta=X$, because the smallest subfield of $F(X)$ containing $F$ and $X$ is $K$ itself).

One should be careful with the notation $F(\theta)$ when there is no extension $K$ containing the element $\theta$, because it isn't always well-defined. Usually when you write something like $F(\theta)$, $\theta$ is meant to be some root of a polynomial $f(X)\in F[X]$, perhaps your favorite polynomial, but if $f(X)$ is not irreducible, the notation $F(\theta)$ is problematic because $f(X)$ might have irreducible factors of different degree. When the polynomial $f(X)$ is irreducible, the notation is unambiguous if taken to mean $K:=F[X]/(f(X))$. Indeed, this will be a field because of the irreducibility of $F$, the element $\theta=X+(f(X))$ will be root of $f(X)$ in $K$, and, as can be checked, $K=F(\theta)$.

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"The" field extension of $\mathbb{F}$ doesn't make any sense. In general, fields have many extensions. $\mathbb{F}(\theta)$ is the smallest extension of $\mathbb{F}$ containing the element $\theta$; similarly, $\mathbb{F}(\theta,\alpha,\delta)$ is the smallest extension of $\mathbb{F}$ containing the elements $\theta$, $\alpha$, and $\delta$.

It's important to note that $\mathbb{F}(\theta)$ isn't just $\mathbb{F}\cup \{\theta\}$, as that wouldn't be a field. When $\theta$ is algebraic over $\mathbb{F}$, $\mathbb{F}(\theta)$ is the set of all elements of the form $$a_0+a_1\theta+a_2\theta^2+\cdots +a_{n-1}\theta^{n-1}$$ where $a_i\in\mathbb{F}$ and $n$ is the degree of $\theta$ over $\mathbb{F}$. All of these elements must be included in the extension so that everything still has a multiplicative inverse.

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