Show that $K = \mathbb{Q}(\sqrt{p} \ | \ \text{p is prime} \}$ is an algebraic and infinite extension of Q Show that $K = \mathbb{Q}(\sqrt{p} \ | \ p\text{ is prime}  )$ is a algebraic and infinite extension on Q.
For every prime $p$, consider the polynomial $f(x)=x^2−p$. Then $f(x) \in \mathbb{Q}(\sqrt{p}\,∣\,p \text{ is prime})[x]$. Therefore, the extension is algebraic.
Moreover, taking $B=\mathbb{Q}(\sqrt{p} \ | \ p\text{ is prime} )$ is a infinite basis for this space.
Is that correct?
 A: Your argument that it's an algebraic extension is fine.  To show that it is an extension of infinite degree, certainly it would suffice to find an infinite basis.  As pointed out in the comments, the set $\{\sqrt{p} \ | \ \text{p is prime} \}$ is not a valid basis, but its elements are linearly independent (which does suffice). To avoid bases, here is another way of going about this:
Suppose that $K = \mathbb{Q}[\sqrt{p} \ | \ \text{p is prime} \}]$ were an extension of finite degree.  Then a simple implication of Galois correspondence  tells us that $K$ has only finitely many subfields of degree $2$ over $\mathbb{Q}$.  Notice that $\mathbb{Q}[\sqrt{p}] \subset K$ for all primes $p$, and $\mathbb{Q}[\sqrt{p}]$ is of degree $2$ over $\mathbb{Q}$.  Since there are an infinite number of primes but only a finite number of subfields of degree $2$, then we must be able to find a pair of distinct primes  $p_1$ and $p_2$ such that $\mathbb{Q}[\sqrt{p_1}] \cong \mathbb{Q}[\sqrt{p_2}]$.  
Let $\phi$ be an isomorphism between these fields.  Note that $\phi$ necessarily fixes $\mathbb{Q}$, and we have $\phi(\sqrt{p_1}) = a + b\sqrt{p_2}$ for some $a, b \in \mathbb{Q}$.  Further, $a \neq 0$ since otherwise $\phi(\sqrt{p_1})^2 = \phi(p_1) = b^2p_2$, which is not possible since $\phi$ fixes $\mathbb{Q}$, and certainly $p_1 \neq b^2p_2$.  Similarly $b \neq 0$ since otherwise $\phi(\sqrt{p_1})^2 = \phi(p_1) = a^2$, which is again not possible for the same reason.
Hence, we must have $\phi(\sqrt{p_1}) = a + b\sqrt{p_2}$ for some $a, b \neq 0$.  From this, we get $\phi(p_1) = a^2 + p_2b^2 + 2ab\sqrt{p_2}$, and since $\phi$ fixes $\mathbb{Q}$, we have $p_1 = a^2 + p_2b^2 + 2ab\sqrt{p_2}$.  With some rearranging, we see that this implies that $\sqrt{p_2}$ is rational--a contradiction!
