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Is $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ the same as $\mathbb{Q}(\sqrt{2},\sqrt{3})$?

I mean, $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ can be viewed as: $\mathbb{Q}[\sqrt{2}][\sqrt{3}]$, as polynomials of $\sqrt{3}$ having coefficients of $\mathbb{Q}[\sqrt{2}]$?, and $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is the smallest field containing $\mathbb{Q}, \sqrt{2}\,\sqrt{3}$. Since $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ is a field we must have that $\mathbb{Q}(\sqrt{2},\sqrt{3}) \subseteq \mathbb{Q}[\sqrt{2},\sqrt{3}]$?, but since all the terms in $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ are made with sums and multiplications of $\mathbb{Q},\sqrt{2},\sqrt{3}$, we must have that $\mathbb{Q}[\sqrt{2},\sqrt{3}] \subseteq \mathbb{Q}(\sqrt{2},\sqrt{3})$?

Can someone confirm that this is correct?

But do we then also have that:

$F[\alpha]=F(\alpha)$, always? and

$F[\alpha_1,\alpha_2]=F(\alpha_1,\alpha_2)$?

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  • $\begingroup$ Your question asks: can any rational function of $\sqrt{2}$ and $\sqrt{3}$ be written as a polynomial of $\sqrt{2}$ and $\sqrt{3}$? The elementary technique is called "rationalize the denominator". Does that work in this case? $\endgroup$ – GEdgar May 11 '15 at 15:49
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Yes, and your reasoning is more or less correct.

Given a field $F$, $F[\alpha] = F(\alpha)$ if and only if $F[\alpha]$ is a field if and only if $\alpha$ is algebraic over $F$.

Note that if $\alpha$ is transcendental, then $F[\alpha]$ is isomorphic to the polynomial ring $F[x]$, which is not a field.

Now if $F$ is a field and $\alpha_1$ and $\alpha_2$ are both algebraic over $F$, $$F[\alpha_1,\alpha_2]\cong F[\alpha_1][\alpha_2] \cong F(\alpha_1)[\alpha_2]\cong F(\alpha_1)(\alpha_2)\cong F(\alpha_1,\alpha_2),$$ since $\alpha_2$ is also algebraic over $F(\alpha_1)$.

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  • $\begingroup$ Thank you very much, it was really cool the sentence: $F[\alpha] = F(\alpha)$ if and only if $F[\alpha]$ is a field if and only if $\alpha$ is algebraic over $F$. The last equivalence does $\alpha$ algebraic $\rightarrow$ $F[\alpha]$ field, follow from the fact the if $\alpha$ algebraic we know that $F(\alpha)$ is a field, and it is spanned by $1,\alpha,..,\alpha^{n-1}$ over F, so we then know that all higher terms in $F[\alpha]$ like $f\alpha^{10n}$ will be reduced to lower terms, and we get only elements in $F(\alpha)$ so they must be the same, and hence $F[\alpha]$ is a field? But I do[cont $\endgroup$ – user119615 May 11 '15 at 16:18
  • $\begingroup$ struggle with the other implication $F[\alpha]$ field $\rightarrow$ $\alpha$ algebraic over F. Can you please explain this one? $\endgroup$ – user119615 May 11 '15 at 16:19
  • $\begingroup$ Yes, the fact that $\alpha$ algebraic $\rightarrow$ $F[\alpha]$ is a field is the argument involving powers of $\alpha$ spanning given in aes's answer, and more abstractly in lhf's answer. I briefly explained the other direction ($F[\alpha]$ a field $\rightarrow$ $\alpha$ algebraic) in my answer, but as the contrapositive ($\alpha$ transcendental $\rightarrow$ $F[\alpha]$ is not a field). $\endgroup$ – Alex Kruckman May 11 '15 at 16:22
  • $\begingroup$ Thanks I see. My last question is then if this is a correct proof of that statement: Let $\phi: F[x] \rightarrow F[\alpha]$ be defined in an obvious way, by renaming elements, obviously it is onto, and it is 1-1 since if it wasn't then we could easily make a nonzero polynomial in $\alpha$ with coefficients in F, which is zero, and hence $\alpha$ is not transcendental? $\endgroup$ – user119615 May 11 '15 at 16:32
  • $\begingroup$ @user119615 correct! $\endgroup$ – Alex Kruckman May 11 '15 at 17:17
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If $F$ is a field and $\alpha$ is algebraic over $F$, then $F[\alpha] = F(\alpha)$ always.

Explanation: $c_n \alpha^n + \cdots + c_0 = 0$ so $-\frac{1}{c_0}\left(c_n \alpha^{n-1} + \cdots + c_1\right) \cdot \alpha = 1$, so $\alpha$ is invertible in $F[\alpha]$. (If $c_0 = 0$ then factor some power of $\alpha$ out of the equation.) Similar logic finds an inverse for any expression involving a linear combination of $\alpha$ and its powers (i.e. the general element of $F[\alpha]$).

But if $\alpha$ is transcendental, then $F[\alpha] \cong F[t]$ is not a field, and is a strict subring of $F(\alpha) \cong F(t)$.

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  • $\begingroup$ Thank you very much, I see that I forgot to think about why $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ is really a field, but I can show it with your technique. But I just have one question. You say that if $c_0$ is zero, then factor it out, but is it not impossible for $c_0$ to be zero?, because the polynomial you are working with is supposed to be irreducible, so if $c_0$ was zero, then it would be reducible?, or am I wrong maybe? $\endgroup$ – user119615 May 11 '15 at 16:01
  • $\begingroup$ @user119615 - the assumption in this answer is simply that $\alpha$ satisfies a polynomial equation. Of course it satisfies an irreducible polynomial, but the proof goes through without this assumption. $\endgroup$ – Mark Bennet May 11 '15 at 16:06
  • $\begingroup$ @user119615 Yes, if the polynomial is irreducible, then $c_0$ is never zero. I just added it so I didn't have to mention irreducibility. $\endgroup$ – aes May 11 '15 at 16:06
  • $\begingroup$ @MarkBennet Thanks for the sign correction! $\endgroup$ – aes May 11 '15 at 16:07
  • $\begingroup$ Thank you very much for the help aes. $\endgroup$ – user119615 May 11 '15 at 16:23
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More generally, if $A$ is a domain which is a finite-dimensional algebra over a field $F$, then $A$ is a field.

Indeed, take $a \in A$, $a\ne 0$, and consider $\phi: A \to A$ given by $\phi(x)=ax$. Then $\phi$ an $F$-linear transformation. Moreover, $\phi$ is injective because $A$ is a domain. Since $A$ is finite-dimensional algebra over $F$, $\phi$ injective implies $\phi$ surjective, and so $1$ is in the image of $\phi$ and $a$ has an inverse in $A$. This inverse can be found by expressing $\phi$ with respect to a basis of $A$ and solving a linear system.

$\mathbb{Q}[\sqrt{2},\sqrt{3}]$ is clearly a domain because it is contained in $\mathbb{R}$.

$\mathbb{Q}[\sqrt{2},\sqrt{3}]$ is also a finite-dimensional algebra over $\mathbb{Q}$. Indeed, $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ is the set of all polynomial expressions in $\sqrt{2}$ and $\sqrt{3}$ with coefficients in $\mathbb{Q}$, and so a basis is $\{ 1, \sqrt{2}, \sqrt{3}, \sqrt{6} \}$ because all powers greater than $1$ can be reduced.

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Another way is to notice that $\mathbb Q(\sqrt2, \sqrt3)=\mathbb Q(\sqrt2+\sqrt3),$ and $(\sqrt2+\sqrt3)^{-1}=\sqrt3-\sqrt2\in\mathbb Q(\sqrt2, \sqrt3)=\mathbb Q(\sqrt2+\sqrt3),$ so $$\mathbb Q(\sqrt2, \sqrt3)=\mathbb Q(\sqrt2+\sqrt3)=\mathbb Q[\sqrt2+\sqrt3].$$
Now it follows easily that $\mathbb Q(\sqrt2,\sqrt3)=\mathbb Q[\sqrt2,\sqrt3].$ In fact, this works for every finite separable extension, mutatis mutandis.
Hope this helps.

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  • $\begingroup$ Thanks for the help awllower. $\endgroup$ – user119615 May 11 '15 at 16:22
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I would prefer to think of it this way: suppose that $\alpha_1,\ldots, \alpha_n$ are algebraic over $R$, so that they all lie in the algebraic closure $\overline R$. Then

  • $R[\alpha_1, \ldots, \alpha_n]$ is the smallest subring of $\overline R$ containing $R$ and the elements $\alpha_1,\ldots, \alpha_n$.
  • $R(\alpha_1,\ldots, \alpha_n)$ is the field of fractions of $R[\alpha_1, \ldots, \alpha_n]$.

This is exactly what the definition of "polynomials in the $\alpha_i$" is saying. Starting with a ring $R$, we want to form a ring that is as small as possible and contains $R$ and $\alpha_1, \ldots, \alpha_n$.

If we add the $\alpha_i$ elements to $R$, then what is the smallest number of elements that we need to add so as to be left with a ring - i.e. so that $R[\alpha_1, \ldots, \alpha_n]$ is closed under addition and multiplication? We must add all numbers formed by adding and multiplying by the $\alpha_i$ - i.e. all polynomials in the $\alpha_i$.

The condition that $\alpha_1,\ldots, \alpha_n\in \overline R$ simply captures the fact that we are assuming that we already know how to add and multiply the $\alpha_i$.


In particular, if $R$ is a field, and $\alpha_1, \ldots, \alpha_n$ are algebraic over $R$, then $R[\alpha_1, \ldots, \alpha_n]$ is also a field, and is therefore equal to $R(\alpha_1, \ldots, \alpha_n)$.

However, if $R$ is not a ring, or if $\alpha_1,\ldots, \alpha_n$ are not algebraic, then this need not be true. For example,

  • $\mathbb Z[\sqrt 2] \ne \mathbb Z(\sqrt 2)$
  • $\mathbb Q[X] \ne \mathbb Q(X)$
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  • $\begingroup$ Thanks for the help Mathmo123. $\endgroup$ – user119615 May 11 '15 at 16:23

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