Suppose that c is transcendental over Q. Show that c and c+c are also transcendental. My solution (a bit rough): (a) A polynomial that c satisfies is x2−c. However, since c is not algebraic over Q, c is not in Q, x−c would be a polynomial that c satisfies in the rationals. So x2−c is not in Q|x| and so c is not algebraic over 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+c satisfies is p(x)=x3−cx2−cx=2c. For reasons mentioned above, this is not in Q[x] either, so c is not algebraic over 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).

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Suppose that c is transcendental over Q. Show that  c and c+c are also transcendental.  My solution (a bit rough):   (a) A polynomial that c satisfies is x2−c. However, since c is not algebraic over Q, c is not in Q, x−c would be a polynomial that c satisfies in the rationals. So x2−c is not in Q|x| and so c is not algebraic over 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+c satisfies is p(x)=x3−cx2−cx=2c. For reasons mentioned above, this is not in Q[x] either, so c is not algebraic over 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).

Nathen Lamb · Accepted Answer

a) Let f∈Q[x]. Then f(c)=a(c)+b(c)c with a(c),b(c)∈Q[c]. If f(c)=0 and f≠0 it follows that a(c) and b(c) are not both zero, and a2(c)=b2(c)c. But the last relation is not possible for degree reasons.   (b) Try a similar approach.

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

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