# Degree of extension $\Bbb Q(\sqrt{5},\sqrt{7})$.

I tried to consider the tower of extension $\Bbb Q\subset \Bbb Q(\sqrt{7})\subset\Bbb Q(\sqrt{5},\sqrt{7})$.

The minimal polynomial of $\Bbb Q(\sqrt{7})$ over $\Bbb Q$ is $x^6-7$ by Eisenstein. But although it is easy to see that it has no root in $\Bbb Q(\sqrt{7})$, how can I formally conclude that $x^4-5$ is irreducible over $\Bbb Q(\sqrt{7})$ and thus we can see the basis of $\Bbb Q(\sqrt{7}, \sqrt{5})$？

I know that for if we have the degree of $\Bbb Q(\sqrt{7})$ and $\Bbb Q(\sqrt{5})$ are coprime, then it can be much simpler. But how to deal with that in this case where they are not coprime. Any help would be appreciate. Thanks so much！

• You might check through linear algebra that the minimal polynomial of $\sqrt{7}+\sqrt{5}$ over $\mathbb{Q}$ has degree $24$. Since $\mathbb{Q}(\sqrt{7}+\sqrt{5})\subseteq\mathbb{Q}(\sqrt{7},\sqrt{5})$ the extensions through $\sqrt{7}$ and $\sqrt{5}$ are independent. Apr 13, 2017 at 12:20

The polynomial $x^4-5$ can be factored over $\mathbb{R}$, which contains $\mathbb{Q}(\sqrt{7})$, as $$x^4-5=(x-\sqrt{5})(x+\sqrt{5})(x^2+\sqrt{5})$$ Thus a factorization over $\mathbb{Q}(\sqrt{7})$ can only be the one above or $(x^2-\sqrt{5})(x^2+\sqrt{5})$. So you just need to show that $\sqrt{5}\notin\mathbb{Q}(\sqrt{7})$.

Suppose $\mathbb{Q}(\sqrt{5})\subseteq\mathbb{Q}(\sqrt{7})$. Then the degree of $\sqrt{7}$ over $\mathbb{Q}(\sqrt{5})$ is $3$ by the dimension formula. The factorization of $x^6-7$ over $\mathbb{R}$ is $$(x^3-\sqrt{7})(x^3+\sqrt{7})= (x-\sqrt{7})(x^2+\sqrt{7}\,x+\sqrt{7}) (x+\sqrt{7})(x^2-\sqrt{7}\,x+\sqrt{7})$$ Since $\sqrt{7}\notin\mathbb{Q}(\sqrt{5})$, you can only get degree three factors as $$(x-\sqrt{7})(x^2-\sqrt{7}\,x+\sqrt{7})$$ or $$(x+\sqrt{7})(x^2+\sqrt{7}\,x+\sqrt{7})$$ and in both cases you'd conclude that $\sqrt{7}\in\mathbb{Q}(\sqrt{5})$, which is impossible.

Suppose $x^4-5$ can be factored as $f(x)g(x)$ over some extension $K$ of $\mathbb{Q}$, $K\subseteq\mathbb{R}$; suppose also that $f(x)$ and $g(x)$ are non constant. Since $x^4-5$ is monic, also $f$ and $g$ can be assumed monic. Continuing like this, we can assume that $x^4-5$ is factored into monic factors, irreducible over $K[x]$.

Let $h(x)\in K[x]$ be one of these factors; its factorization in $\mathbb{R}[x]$ must consist of polynomials in the set $\{x-\sqrt{5},x+\sqrt{5},x+\sqrt{5}\}$, which are the irreducible factors of $x^4-5$ in $\mathbb{R}[x]$, because of uniqueness of factorization in $F[x]$ (for $F$ any field).

Now it's just a matter of checking the various possibilities. A factorization of $x^4-5$ can only be with degrees

• $1$, $1$ and $2$
• $2$ and $2$
• $1$ and $3$

If a degree $1$ factor appears, then $\sqrt{5}\in K$; if a degree $2$ factor appears, then $\sqrt{5}\in K$. In both cases, $\sqrt{5}\in K$.

The same argument applies for the second part of the proof.

• Thanks! But may I please ask how can we see that the factorization over $\Bbb Q(\sqrt{7})$ is either $(x−\sqrt{5})(x+\sqrt{5})(x^2+\sqrt{5})$ or $(x^2-\sqrt{5})(x^2+\sqrt{5})$, which can allow us to conclude that it it factorize then $\sqrt{5}\in \Bbb Q(\sqrt{7})$? Apr 14, 2017 at 3:33
• @ymxu0809 If a factorization exist over $\mathbb{Q}(\sqrt{7}$, this is also a factorization over the reals. We know what the irreducible factors are over the reals, so a factorization over the smaller field must be obtained from one over the larger field. This is used also in the final argument. Apr 14, 2017 at 8:09
• Thanks! I do know that the factorization comes from $\Bbb R$, but how can I see that the only possible ways we can gain the factorization over $\Bbb R$ must involves the existence of $\sqrt{5}$? That is, how to see that there is not possible that we can have $5=abcd$ or $5=abc$ or $5=ab$ where $a,b,c,d\in\Bbb R$ but none of them involves $\sqrt{5}$? Apr 14, 2017 at 23:32
• @ymxu0809 If $x^4-5$ factors in some number field $K\subseteq\mathbb{R}$, then its factorization contains one among $x-\sqrt{5}$, $x+\sqrt{5}$ or $x^2-\sqrt{5}$; thus $\sqrt{5}$ must belong to $K$. Remember that factorization in $\mathbb{R}[x]$ is unique. Apr 14, 2017 at 23:42
• Yes it intuitively makes sence to me... But how can I see that 5 can never be a product of transendental elements or some product of other algebraic number where $\sqrt{5}$ does not appear? Is that the fact that as the factorization is unique, once we have a factorization already, we cannot have another factorization? Apr 14, 2017 at 23:48

In general, we have the following: Let $K/\mathbf Q$ and $L/\mathbf Q$ be two number fields such that there exists a rational prime $p$ which is totally ramified in $K/\mathbf Q$ and unramified in $L/\mathbf Q$. Then, the extensions $K/\mathbf Q$ and $L/\mathbf Q$ are linearly disjoint, that is, $[KL : \mathbf Q] = [K : \mathbf Q][L : \mathbf Q]$.

Proof. Let $\mathfrak q$ be a prime of $LK$ lying over the rational prime $p$. Then, $e_{\mathfrak q | p} \geq [K : \mathbf Q]$, since ramification indices are multiplicative across towers and $p$ is totally ramified in $K/\mathbf Q$. On the other hand, if we let $\mathfrak p$ be the prime of $L$ lying below $\mathfrak q$; then we have that

$$[K : \mathbf Q] \leq e_{\mathfrak q | p} = e_{\mathfrak q | \mathfrak p} e_{\mathfrak p | p} = e_{\mathfrak q | \mathfrak p} \leq [LK : L]$$

However, we obviously have that $[K : \mathbf Q] \geq [LK : L]$; thus it follows that $[K : \mathbf Q] = [LK : L]$, and multiplication by $[L : \mathbf Q]$ on both sides of the equality gives the result.

Now, notice that we have exactly the situation of this claim with the rational prime $p = 5$, which is totally ramified in $\mathbf Q(\sqrt{5})$ but unramified in $\mathbf Q(\sqrt{7})$.

• It seems to be a quite general argument. But as I have not learnt about the terminology ramified, could you please give some explaination on this word? And for the set $LK$, do you mean the set $\lbrace lk: l\in L, k\in K\rbrace$? Apr 14, 2017 at 3:37

To show that the polynomial $X^4-5$ is irreducible over $\mathbb{Q}(\sqrt{7})$ you must assume that it is not irreducible,thus it can be factorized to polynomials of degree:$1* 1* 1* 1$ or $2*2$ or $3*1$ or $2*1*1$.

You have to work by cases but its difficult because you have complicated elements in $\mathbb{Q}(\sqrt{7})$ thus consider the opposite tower.

If you prove it then a basis for the final extension will be:

$A=\{\sqrt{7}^j*\sqrt{5}^i|j=0,1...5 ,i=0,1,2,3,\}$

This is one way but it has many calculations.