# Field extensions contradicting tower law

$$[\mathbb{R}(x):\mathbb{R}(x+\frac{1}{x})]=2$$ since $$x$$ satisfies a quadratic polynomial over $$\mathbb{R}(x+\frac{1}{x})$$. Similarly, $$[\mathbb{R}(x):\mathbb{R}(x^2+\frac{1}{x^2})]=4$$. But, as $$(x+\frac{1}{x})^2=(x^2+\frac{1}{x^2})+2$$, so $$\mathbb{R}(x):\mathbb{R}(x^2+\frac{1}{x^2}):\mathbb{R}(x+\frac{1}{x})$$ which clearly contradicts Tower law. What is wrong here?

• ? What actually are you asserting after "so"? – Lord Shark the Unknown Mar 31 at 6:34
• I got confused. I got it. Thanks@LordSharktheUnknown – Shanghaikid Mar 31 at 7:47

## 1 Answer

I can't see what you are asserting, but here are some facts: $$|k(x):k(x+1/x)|=2,$$ $$|k(x):k(x^2+1/x^2)|=4,$$ $$|k(x+1/x):k(x^2+1/x^2)|=2.$$ Here $$k$$ is any field, and $$x$$ is transcendental over $$k$$.