I read the following from the Wikipedia article about algebraic varieties:

Proven around the year 1800, the fundamental theorem of algebra establishes a link between algebra and geometry by showing that a monic polynomial in one variable with complex coefficients (an algebraic object) is determined by the set of its roots (a geometric object).

Can someone explain why "the set of its roots" are called a geometric object?

  • $\begingroup$ It's a set of points in the complex plane. $\endgroup$
    – user98602
    Feb 17, 2014 at 22:51
  • $\begingroup$ This is not really visible in the one-dimensional case. For polynomials in more than one variable, you get more interesting geometric objects. $\endgroup$ Feb 18, 2014 at 17:10

4 Answers 4


The set of roots of a complex polynomial in one variable in just a finite subset of the plane, which might not seem to be interesting geometrically-but certainly geometry has always studied points in the plane-"point" being one of Euclid's undefined terms. Perhaps it's also helpful to notice that examples of sets of roots of polynomials in two variables over the real numbers include ellipses, parabolas and hyperbolas, some more strikingly geometric objects.

  • 2
    $\begingroup$ If the downvoter would care to explain his or her criticism, I'd appreciate it very much. $\endgroup$ Feb 17, 2014 at 23:53
  • $\begingroup$ No wonder I upvoted your answer and the number still shows "0". I thought there's something wrong with the system. $\endgroup$
    – user9464
    Feb 18, 2014 at 0:00
  • 2
    $\begingroup$ Nah. You can click on the number itself and see the breakdown into up- and downvotes. $\endgroup$ Feb 18, 2014 at 0:03

As an example, eventually, constructible points in the plane (unruled straightedge and compass) turned out to be algebraic and characterized through tower of fields.


I think this is a more modern view of why the roots of a polynomial are geometric objects.

Let's suppose, for simplicity, that we are talking about some irreducible polynomial $f(T)\in\mathbb{Q}[T]$. Consider then the field $K=\mathbb{Q}(\alpha)$ where $\alpha$ is a root of $f$. Then, modern algebraic geometry has a way of associating a geometric object to this situation. Namely, we associate to the extension $K/\mathbb{Q}$ the mapping of points $\text{Spec}(K)\to\text{Spec}(\mathbb{Q})$.

That said, since geometry happens only in the algebraically closed setting, we are still not seeing the geometric picture. To remedy this we base change to this setting by considering a geometric point $\text{Spec}(\overline{\mathbb{Q}})\to\text{Spec}(\mathbb{Q})$. I leave it to you to check that if the following diagram is fibered:

$$\begin{matrix}X & \to & \text{Spec}(K)\\ \downarrow & & \downarrow\\ \text{Spec}(\overline{\mathbb{Q}}) & \to & \text{Spec}(\mathbb{Q})\end{matrix}$$

then $X$ can be identified with a discrete set of points, one corresponding to each root of $f$.

Thus, the roots of $f$ can be seen as the geometric fiber of $\text{Spec}(K)\to\text{Spec}(\mathbb{Q})$. More generally, you can check that if $L/K$ is a finite Galois extension, then the geometric fiber of the associated mapping of spectra is a disjoint union of points, where the points correspond to the Galois group of $L/K$. In this setting, we can even view the Galois group as being some sort of deck transformations--acting (transitively!) on the geometric fiber.


As long as you have to ask this question, I assume you've never seen a course on algebraic geometry or advanced abstract algebra, so I'll try to explain by not using "fancy" words, here I go...

First of all, roots of polynomial equations are not "algebraic objects", in any book on abstract algebra we define polynomials and then we define polynomial functions (try not to think of graphs of functions here) and then we define zeros of polynomial functions, that is, roots of a polynomial equation. No mention of any geometry here!

BUT... there is a specific context where this extract from Wikipedia is right, but this is "just" a matter of interpretation.

Consider for example the following polynomials:

$$(i)\ x^2+y^2-1 = 0$$ $$(ii)\ y-x^2 = 0$$ $$(iii)\ y^2-x^3+x = 0$$

When we try to find solutions for these equations in the set $\mathbb{R}^2$, that is, the real plane, the solutions have a "nice" format, they are very "geometric", for example, the set of solutions of $(i)$ is a circle, $(ii)$ is a parabola and $(iii)$ is one example of what we call elliptic curves.

Now consider the following polynomials:

$$(I)\ x^2+y^2=0$$ $$(II)\ x^2+y^2+1=0$$

What about now ? The only "point" in $\mathbb{R}^2$ that is solution of $(I)$ is the origin, $(0,0)$, even worse is the case for $(II)$, there is no solution in $\mathbb{R}^2$.

But if we search for solutions in the complex plane $\mathbb{C}^2$ we find that:

$$(I)\ x^2+y^2=0 \Rightarrow (x+iy)(x-iy)=0$$ $$(II)\ x^2+y^2+1=0 \Rightarrow u = ix, \ v = iy \Rightarrow u^2+v^2-1=0$$

now have infinitely many solutions and we "restore" the geometry of the solutions set, for example $(I)$ is the union of two lines and $(II)$ is a circle!

So by using complex numbers we always (by the fundamental theorem of algebra) get "nice" sets of solutions for polynomial equations, and in this sense zero of polynomials are geometric objects.

Final remark: Above I said that this is "just" a matter of interpretation, but this slightly change in the view point give us one of the most glorious and exciting branch of mathematics, this give us algebraic geometry.

  • $\begingroup$ This is a very good answer, +1. $\endgroup$ Feb 18, 2014 at 17:13

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