# Is $‎‎‎\sqrt[3]{y^3}‎‎‎$ or $\frac{x^2}{x}$ a polynomial?

A polynomial is a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients. Now are

$$‎‎‎\sqrt[3]{y^3}‎‎‎,\quad \frac{x^2}{x}\quad\text{or}\quad\left|x\right|\sqrt[2]{x^2}$$

polynomials?

I think $$‎‎‎\sqrt[3]{y^3}‎‎‎\quad\text{and}\quad\left|x\right|\sqrt[2]{x^2}$$ are polynomials (since $x$ may be any number), but $\frac{x^2}{x}$ is not a polynomial (since $x\neq 0$)

• They are polynomial functions (except that the second one has a removeable singularity at $x = 0$), but the expressions are not polynomial. Commented Jan 22, 2016 at 19:39
• @Dr.MV $y\mapsto \sqrt[3]{y^3}$ is a polynomial function, but $\sqrt[3]{y^3}$ is not a polynomial expression. Commented Jan 22, 2016 at 20:01
• If you use the principal branch of the cube root, for example, $((-1)^3)^{1/3} = e^{i \pi/3} \ne -1$. Commented Jan 22, 2016 at 20:06
• @A.G function $\sqrt[3]{z}$ is multivalued -- at every point $z\ne0$ it takes three distinct values. One of them at $z=-1$ is $-1$, but this is not the principal value. Commented Jan 22, 2016 at 20:36
• This is used in working with complex numbers. If $z = r e^{i\theta}$ is the polar representation of the complex number $z$ with $-\pi < \theta \le \pi$ and $r >0$, the principal branch of $z^\alpha$ is $r^\alpha e^{i\alpha \theta}$. Commented Jan 22, 2016 at 20:37

No, these are not polynomials.

• All polynomials are continuous, $\frac{x^2}{x}$ has a hole at $x=0$.
• All polynomials are single-valued, which fails with $\sqrt[3]{y^3}$ on the complex plane.
• All polynomials are holomorphic. The function $|x|\sqrt{x^2}$ is not.
• @Dr.MV $(1^3)^{1/3}=1^{1/3}\in\{1,\tfrac{-1-i\sqrt3}{2},\tfrac{-1+i\sqrt3}{2}\}$ Commented Jan 22, 2016 at 20:04
• "All polynomials are continuous, $\frac{x^2}{x}$ has a hole at $x=0$". A hole in the classically assumed domain does not make a function non-continuous. $1/x=f(x): \mathbb R\ \{0\}\to\mathbb R$ is very well contionous. Commented Jan 22, 2016 at 20:32
• I'd say all of these are polynomials in $\Bbb R_{>0}$. Don't you agree? Commented Jan 22, 2016 at 20:35
• @YoTengoUnLCD I would say that they are equal to some polynomial in $\mathbb{R}_{>0}$, or as others said I would call them "polynomial functions" rather than polynomials. In this case everything depends on definition of a polynomial. If a polynomial has strict definition of how it is expressed, they are obviously not polynomials. Commented Jan 22, 2016 at 20:40
• Ah, yes. I understand, there's a subtle distinction on the definition of equality as functions (maybe we don't even define a polynomial as a function, but as a finite $k$-tuple of coefficients!). We could even say they something funny like that they belong to the same equivalence class of expression under the "defining the same function" relation. Commented Jan 22, 2016 at 20:48

Well... yes, and no.

The point is: when we say a polynomial, we want so much to entail the fact that it is a finite linear combination of natural powers that it is usual, for example, to define it as a function $f: \mathbb{N} \rightarrow A$ (where $A$ is whatever ring we are on) which is zero for all, except finitely many $a \in A$. (c.f. Lang's Algebra). This is to entail the fact that there exists an "indeterminate", and coefficients. The coefficients are given by $f(n)$ ($f(0)$ is the first coefficient, of what would be $X^0$, $f(1)$ of $X^1$ etc).

If we were going to be nitpicky, even $f: \mathbb{R} \rightarrow \mathbb{R}$ given by $f(x)=x^2$ is not a polynomial. This is what I think @Arthur is trying to entail in his comment. This is what he calls a polynomial function.

However, we of course have no shame on calling this function a polynomial, since there is an obvious equivalence. However, the expressions you mention are not capturing the essence of a polynomial in their form. Personally, I would not refer to them as polynomials, but instead simply as functions. But this is the same thing as refraining from calling a force (on the context of physics) a point on $\mathbb{R}^3$, and instead referring to it as a vector.

In the end of the day, it is all a matter of definitions and communicating oneself.

• The distinction between a polynomial and a polynomial function is important when working with a finite field. For any integer $x$, $x^2 \equiv x \mod 2$, which says that the functions $x \mapsto x^2$ and $x \mapsto x$ on the field $\mathbb Z_2$ (the integers mod 2) are the same, However, the polynomials $X^2$ and $X$ over $\mathbb Z_2$ are different. On an infinite field, two polynomials that agree as functions must be the same polynomial, so the distinction is not so important there. Commented Jan 22, 2016 at 20:47