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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).

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  • $\begingroup$ Do you mean univariate functions ? $\endgroup$ – Peter Jun 21 at 11:51
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    $\begingroup$ What is your definition of smooth algebraic functions? Note that multivariable polynomials can have infinitely many roots. $\endgroup$ – N. S. Jun 21 at 11:52
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    $\begingroup$ 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. $\endgroup$ – GEdgar Jun 21 at 11:58
  • $\begingroup$ @N.S. See my edit. $\endgroup$ – Pedro Jun 21 at 12:06
  • $\begingroup$ " 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 $x<0$. $\endgroup$ – GEdgar Jun 21 at 12:13
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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 $F(x)$.

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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 $\mathbb{H}[x]$.

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