# Is there algebraic functions with infinitely many roots?

For example, a rational function is zero if and only if its numerator (which is a polynomial) is zero. Thus, a rational function which is not identically zero have only a finite number of roots.

Is the same conclusion valid for smooth algebraic functions? If so, what would a proof or a source?

Edit (in response to the comments). I'm particularly interested in a real-valued function of a real variable given explicitly by a formula obtained from the elementary algebraic operations (addition, subtraction, multiplication, division, roots).

• Do you mean univariate functions ? – Peter Jun 21 at 11:51
• What is your definition of smooth algebraic functions? Note that multivariable polynomials can have infinitely many roots. – N. S. Jun 21 at 11:52
• With non-standard meanings of "algebraic" we can do this. I say $x - \sqrt{|x|^2}$ is not "algebraic", but if you say it is, then that is an answer for you. – GEdgar Jun 21 at 11:58
• @N.S. See my edit. – Pedro Jun 21 at 12:06
• " a real-valued function of a real variable given explicitly by a formula obtained from the elementary algebraic operations (addition, subtraction, multiplication, division, roots)." ... so you do allow $x - \sqrt{x^2}$, which is zero for $x \ge 0$ and nonzero for $x<0$. – GEdgar Jun 21 at 12:13

Say $$F$$ is algebraic in the sense that it is real-analytic and satisfies $$P_n(x)F(x)^n + P_{n-1}(x)F(x)^{n-1}+\dots+P_1(x)F(x)+P_0(x) = 0 \tag{1}$$ for all $$x$$, where $$P_0,\dots,P_n$$ are polynomials, and the left side of $$(1)$$ is irreducible, that is: it cannot be factored onto two nonconstant expressions of the same form. Can $$F$$ have infinitely may zeros? If $$\{x_k\}_{n=1}^\infty$$ are zeros of $$F$$, plug into $$(1)$$ to conclude they are also zeros of $$P_0$$. Since $$P_0$$ is a polynomial, $$P_0 = 0$$. Then $$(1)$$ can be factored, where one factor is $$F(x)$$.
There are even non-zero polynomials $$f(x)$$ having infinitely many roots. This can happen when we do not consider polynmials over fields, but, say, over the real algebra of quaternions $$\mathbb{H}$$. The polynomial $$f(x) = x^2+1$$ has infinitely many roots in $$\mathbb{H}[x]$$.