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Can one define the notation Can one define the notation $\mathbb{Q}(\sqrt{5},\sqrt{2})$?

The question is really similar to this one :

field extension-notation problem $\mathbb{Q}(\sqrt{2})=\mathbb{Q}[\sqrt{2}]$

I would say it means "the smallest field that contains all elements of $\mathbb{Q}$ ,contains these two numbers, and is a subfield of $\mathbb{R}$" but i don't understand what in this notation precise that i'm searching a field in $\mathbb{R}$ in particular ,

Arlon Fredolster
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1 Answers1

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There is no reason to think about these being subfields of some larger field. $\Bbb Q(\sqrt 5, \sqrt 2)$ means

The smallest field that contains all elements of $\Bbb Q$, and also $\sqrt 2$ and $\sqrt 5$

That's it. No need to specify that it must also be a subfield of something bigger.

We have that $\Bbb R$ contains all the elements of $\Bbb Q$ and also $\sqrt5$ and $\sqrt2$, so naturally it will contain $\Bbb Q(\sqrt5, \sqrt2)$. So we are looking at a subfield of $\Bbb R$. But there is no need to require this to be the case as part of the definition. It just happens automatically. If you did include it in the definition, you'd get into trouble when you try to use that definition to decode $\Bbb Q(i, \sqrt2)$. Or the field $\Bbb Q(x)$ of rational functions in the variable $x$ with rational coefficients.

Arthur
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  • Side note: We are speaking, as is tradition, "up to isomorphsim". Otherwise this discussion would be really, really messy, and we'd have to be a lot more precise about what we mean when we say "$\Bbb Q$", or "$\sqrt2$", or "$\Bbb Q$ is a subfield of $\Bbb R$". – Arthur Apr 08 '21 at 13:48
  • But don't we need to know what are $\sqrt{2},\sqrt {5}$ if not elements if $\mathbb{R} $? The notion of an algebraic element always assumes fields $K, F$ with $F\subseteq K$ and we say an element $a\in K$ is algebraic over $F$ if... Also please treat me as novice in field theory in your replies. – Paramanand Singh Apr 08 '21 at 16:08
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    @ParamanandSingh We can construct such elements directly. For instance, by starting with the polynomial ring $\Bbb Q[x]$, and dividing out by the ideal $(x^2-2)$. The end result turns out to be $\Bbb Q(\sqrt2)$, where $x+(x^2-2)$ takes on the role of $\sqrt2$. No need to involve the inclusion $\Bbb Q\subseteq \Bbb R$ at all. That's not to say we can't involve the inclusion. It's just not necessary, and therefore not part of the formal definition. – Arthur Apr 08 '21 at 16:35
  • Thanks for clarification. +1 – Paramanand Singh Apr 08 '21 at 16:45