# Can we always find a primitive element that is a square?

Let $L/\mathbb Q$ be a finite field extension. The Primitive Element Theorem says that there is an element $\alpha \in L$ so that $L=\mathbb Q(\alpha)$. Can I always find an element $\beta \in L$ so that $L=\mathbb Q(\beta^2)$ ?

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This is certainly true if $L$ has odd degree over $\mathbf Q$. Do you see why? – Dylan Moreland Jun 13 '12 at 14:08

$\newcommand{\Q}{\mathbb Q}$

I think this works. Let $K/\Q$ be a finite field extension. Note that $K$ has finitely many subfields, call them $F_1,\dots,F_t$ and let $W$ be their union. Then any element in $K \setminus W$ is necessarily a primitive element. So it suffices to show their exists some $\alpha \in K \setminus W$ that has a square root. Pick $\alpha \in K \setminus W$, note that for $k \in \Q$ we still have $\alpha_k=\alpha + k \in K \setminus W$ otherwise $\alpha \in W$. Consider $\alpha_k^2=(\alpha+k)^2=\alpha^2+2k\alpha+k^2$ suppose that for every $k \in \Q \setminus {0}$ we have $\alpha_k^2 \in W$. Then by the pigeonhole principle we must have some $j$ and $k_1 \neq k_2$ such that $\alpha_{k_1}^2,\alpha_{k_2}^2 \in F_j$. So we have

$$\frac{\alpha_{k_1}^2-\alpha_{k_2}^2-(k_1^2+k_2^2)}{2(k_1-k_2)} \in F_j$$

but

$$\frac{\alpha_{k_1}^2-\alpha_{k_2}^2-(k_1^2+k_2^2)}{2(k_1-k_2)} =\frac{2\alpha(k_1-k_2)}{2(k_1-k_2)}=\alpha.$$

Which would imply that $\alpha \in F_j \subset W$ so we have some $k \in \Q\setminus \{0\}$ such that $(\alpha+k)^2 \in K \setminus W$ and thereby $\Q((\alpha+k)^2)=K$.

I think this could be turned into a constructive argument fairly easy. Take one of the primitive elements given to you by the primitive element theorem then add $t+1$ non-zero rational numbers to it and the square of one of those will be the desired element.

Ammendum: This could also extend to a proof that there always exists an element $\alpha \in K$ such that $\Q(\alpha^n)=K$. Essentially use the same argument except find $t$ such distinct $c_i$ so that $\alpha_{c_i}^n \in F_t$ then notice that the matrix given by $A_{ij}=\binom{m}{j}c_i^j$ has determinant a constant multiple of the Vandermonde matrix $A_{ij}=c_i^j$. Thereby the determinant is nonzero and so we can find a linear combination of the $\alpha_{c_i}^n$ that gives us $\alpha$.

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Thank you very much. It is a nice proof. – Lisa Mainhard Jun 14 '12 at 13:18