# Example of Transcendental degree of a polynomial

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 $$\mathbb{R}$$.

• When you write the extensions $\mathbb{R}(x, y)/\mathbb{R}$ or $\mathbb{R}(x)/\mathbb{R}$, nothing seems to depend on the polynomials $f$ or $g$. Maybe you want to clarify exactly what field extensions your prof is trying to define, or maybe your notation is not quite right. Commented Mar 15 at 5:09
• To be precise, you can't get the field $\mathbb{R}(x)$ out of the polynomial $x^2 - 2$, but $g(x) = x^2 - 2$ is an element of that field. But it is also an element of $\mathbb{R}(x, y)$, and more generally there is an obvious inclusion $\mathbb{R}(x) \subset \mathbb{R}(x, y)$. Commented Mar 15 at 9:23

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.

• hmm, so what's the answer to my question? Commented Mar 15 at 11:10
• @Rowing0914 added details, had missed the last sentence of the quoted examples initially. Commented Mar 15 at 12:03
• Thank you for adding details! Commented Mar 15 at 12:06