Example of Transcendental degree of a polynomial

Question

I'm studying algebraic geometry and I tried to understand th concept of Transcendental degree of a polynomial like the following.

Wiki
The field of rational functions in $n$ variables $K(x_1, \ldots, x_n)$ (i.e., the field of fractions of the polynomial ring $K[x_1, \ldots, x_n]$) is a purely transcendental extension with transcendence degree $n$ over $K$; we can, for example, take $\{ x_1, \ldots, x_n \}$ as a transcendence base.

I tried to come up with some examples and my professor wrote on the blackboard the following examples which I couldn't quite get;

For example, consider the polynomial $ f(x, y) = x^2 + y^2 - 1 $. This polynomial involves two variables, $ x $ and $ y $, which are algebraically independent over the real numbers. The transcendence degree of the field extension $ \mathbb{R}(x, y)/\mathbb{R} $ is 2, because $ x $ and $ y $ are algebraically independent and there are no polynomial relations between them over $ \mathbb{R} $.

In contrast, if you have a polynomial like $ g(x) = x^2 - 2 $, the transcendence degree of the field extension $ \mathbb{R}(x)/\mathbb{R} $ is 1, because although $ x $ is algebraically independent over $ \mathbb{R} $, it is algebraic over $ \mathbb{R} $ (since it satisfies the polynomial equation $ x^2 - 2 = 0 $).

Are the above examples correct? I thought $f(x, y)$ in the first example can have solutions for $x^2 + y^2 = 1$ in $\mathbb{R}$.

Answer 1

Your misunderstanding seems to come from confusing the polynomial $f(x, y) = x^2 + y^2 - 1$ with the polynomial equation $f(x, y) = 0 \iff x^2 + y^2 = 1$.

There are obvious inclusions $\mathbb{R} \subset \mathbb{R}(x) \subset \mathbb{R}(x, y)$. Keeping this in mind, $g(x) = x^2 - 2$ is an element of $\mathbb{R}(x)$ but not of $\mathbb{R}$, and $f(x, y)$ is an element of $\mathbb{R}(x, y)$ but not of $\mathbb{R}(x)$. Instead of $f$ and $g$ one could look at the polynomials $x^2 + 1$ or $x^2 + y^2 + 2$, which have no real solutions, but this distinction has nothing to do with the field extensions or their transcendence degrees.

... it is algebraic over $\mathbb{R}$ (since it satisfies the polynomial equation $x^2 − 2 = 0$).

This looks wrong to me. In the field of rational functions, $x \in \mathbb{R}(x)$ is not algebraic over $\mathbb{R}$. There is a ring $\mathbb{R}[x]/g(x)$ associated to $g$ in which $x$ (or to be precise the image of $x$) is algebraic, and indeed satisfies the equation $x^2 - 2 = 0$. But this ring also has nothing to do with $\mathbb{R}(x)$, and in fact is not itself a field since $g = (x - \sqrt{2})(x + \sqrt{2})$ is not irreducible over $\mathbb{R}$.

The rest of the quoted passages seem correct.