Suppose that $c$ is transcendental over $\mathbb{Q}.$ Show that $\sqrt{c}$ and $c + \sqrt{c}$ are also transcendental.

Question

My solution (a bit rough):

(a) A polynomial that $\sqrt{c}$ satisfies is $x^2 - c$. However, since $c$ is not algebraic over $\mathbb{Q},$ $c$ is not in $\mathbb{Q}$ otherwise, $x-c$ would be a polynomial that $c$ satisfies in the rationals. So $x^2 - c$ is not in $\mathbb{Q}[x]$ and so $c$ is not algebraic over $\mathbb{Q}$ (not sure about that statement... why could there not be another polynomial that is satisfies, is it because it is the minimal polynomial?).

(b) A polynomial that $c + \sqrt{c}$ satisfies is $p(x) = x^3 - cx^2 -cx -2c.$ For reasons mentioned above, this is not in $\mathbb{Q}[x]$ either, so c is not algebraic over $\mathbb{Q}.$ (Also because it is the minimal polynomial and the minimal polynomial must divide everything that c satisfies? Really not sure about how to finish up here).

Answer 1

(a) Let $f\in\mathbb Q[x]$. Then $f(\sqrt c)=a(c)+b(c)\sqrt c$ with $a(c),b(c)\in\mathbb Q[c]$. If $f(\sqrt c)=0$ and $f\ne0$ it follows that $a(c)$ and $b(c)$ are not both zero, and $a^2(c)=b^2(c)c$. But the last relation is not possible for degree reasons.

(b) Try a similar approach.

...................

If you are familiar with algebraic extensions, you can use (a) to solve (b). Suppose that $c+\sqrt c$ is algebraic over $\mathbb Q$. Then the field extension $\mathbb Q\subset\mathbb Q(c+\sqrt c)$ is algebraic. On the other side, the field extension $\mathbb Q(c+\sqrt c)\subset\mathbb Q(\sqrt c)$ is clearly algebraic ($\sqrt c$ is a root of the polynomial $X^2+X-(c+\sqrt c)$), so $\mathbb Q\subset\mathbb Q(\sqrt c)$ is also algebraic. In particular, $\sqrt c$ is algebraic over $\mathbb Q$, a contradiction with (a).

Answer 2

Your attempt doesn't work.

You can observe that $\mathbb{Q}(c)\subseteq\mathbb{Q}(\sqrt{c})$, because obviously $c\in\mathbb{Q}(\sqrt{c})$. Can you finish?

What about the second part? Suppose $a=c+\sqrt{c}$ is algebraic over $\mathbb{Q}$. Then $$ c=a^2-2ac+c^2 $$ and therefore $c$ is algebraic over $\mathbb{Q}(a)$. Now the dimension formula ends the argument.