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The way we define $\mathbb{Q}(\sqrt{2})$ is via the following set,

$F(u) := \{ f(u)/g(u):f(u),g(u)\in F[x],g(u)\neq 0 \}$

So, I am then confused on how one defines the set $\mathbb{Q}(\sqrt{2})(\sqrt{3})$. And how is this set functionally different from $\mathbb{Q}(\sqrt{2},\sqrt{3})$? How would one even define $\mathbb{Q}(\sqrt{2},\sqrt{3})$? (It was not defined explicitly with commas anywhere in my lecture).

Anmol Bhullar
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    What is $u$ ? I assume it is $\sqrt{2}$... Anyway, $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is the smallest field containing $\mathbb{Q},\sqrt{2},\sqrt{3}$, you can check that is is equal to $$ \left{ \frac{f(\sqrt{2},\sqrt{3})}{g(\sqrt{2},\sqrt{3})},f,g,\in\mathbb{Q}[X,Y],g(\sqrt{2},\sqrt{3})\neq 0 \right} $$ or, since $\sqrt{2}$ and $\sqrt{3}$ are algebraic over $\mathbb{Q}$, it is also equal to ${ f(\sqrt{2},\sqrt{3}),f\in\mathbb{Q}[X,Y] }$. Note that this is the same field as $\mathbb{Q}(\sqrt{2})(\sqrt{3})$. – Tuvasbien Apr 17 '23 at 00:00

1 Answers1

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The notation I'm familiar with is that, given a field $F$ and a specific element $\alpha$ of some field extension of $F$ (say, $K$), then $F(\alpha)$ denotes the smallest field containing both $\alpha$ and $F$. That is,

$$F(\alpha) := \bigcap_{\substack{\text{fields $L$} \\ \alpha \in L \\ F \subseteq L}} L$$

One may extend this in the obvious way to $F(\alpha_1,\cdots,\alpha_n)$, the smallest field containing all of $F$ as well as all of the $\alpha_i$.

One may show that $F(\alpha,\beta) = (F(\alpha))(\beta)$.

Given an indeterminant $x$ and field $F$, however, $F(x)$ denotes all rational functions over $F$, i.e.

$$F(x) := \left\{ \frac{a(x)}{b(x)} \, \middle| \, a,b \in F[x], b \ne 0 \right\}$$

and likewise

$$F(x_1,\cdots,x_n) := \left\{ \frac{a(x_1,\cdots,x_n)}{b(x_1,\cdots,x_n)} \, \middle| \, a,b \in F[x_1,\cdots,x_n], b \ne 0 \right\}$$

Given $\alpha$, there is a natural equivalence between $F(\alpha)$ as defined above, and $F(x)$ when each rational function is evaluated at $x=\alpha$.

Frankly, this is something I've always found a bit confusing, but this answer somewhat cleared things up for me.

It may be simplest to consider:

  • $F[x]$ as the smallest ring containing $x$ (indeterminant or named element, since there is a natural way to make them equivalent)
  • $F(x)$ as the smallest field containing $x$
PrincessEev
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