# The Unapologetic Mathematician

## Polynomials with Too Few Roots

Okay, we saw that roots of polynomials exactly correspond to linear factors, and that a polynomial can have at most as many roots as its degree. In fact, there’s an expectation that a polynomial of degree $n$ will have exactly $n$ roots. Today I want to talk about two ways this can fail.

First off, let’s work over the real numbers and consider the polynomial $p=X^2-2X+1$. A root of $p$ will be a real number ${x}$ so that $x^2-2x+1=0$, but a little playing around will show us that $x=1$ is the only possible solution. The degree of the polynomial is two, so we expect two roots, but we only find one. What went wrong?

Well, we know from the fact that $x=1$ is a root that $X-1$ is a factor of $p$. Our division algorithm shows us that we can write $p=(X-1)(X-1)$. The factor that gives us the root $x=1$ shows up twice! But since we’ve already counted the root once the second occurrence doesn’t do anything.

To remedy this, let’s define the “multiplicity” of a root ${x}$ to be the number of times we can evenly divide out a factor of $(X-x)$. So in our example the root ${1}$ has multiplicity $2$. When we count the roots along with their multiplicities, we get back exactly the degree.

So do multiplicities handle all our problems? Unfortunately, no. An example here over the real numbers is the polynomial $p=X^2+1$. A root of this polynomial would be a real number ${x}$ with $x^2=-1$. But since the square of any real number is nonnegative, this can’t be satisfied. So there exist polynomials with fewer roots than their degrees would indicate; even with no roots at all!

Now, some fields are well-behaved. We say that a field is “algebraically closed” if every polynomial over that field has a root ${x}$. In that case we can divide out by $(X-x)$ to get a polynomial of one degree less, which must again have a root by algebraic closure. We can keep going until we write the polynomial as the product of a bunch of linear factors — the number is the same as the degree of the polynomial — and leave one field element left once we get down to degree zero. Thus over an algebraically closed field every polynomial has exactly as many roots as its degree indicates.. if you count them with their multiplicities!

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Posted by John Armstrong | Algebra, Polynomials, Ring theory

## 14 Comments	»

1. Excuse the nitpick, but more usual is “algebraically closed”.

Comment by | August 6, 2008 | Reply

2. Sorry, I thought I cleared that.. typed the wrong thing, noticed it later, and failed to correct…

Comment by | August 6, 2008 | Reply

3. When you write “a polynomial can have at most as many roots as its degree”, maybe you should recall that you consider polynomials over a field, say in the proof of this statement where you use integrality, and mention the degree-two polynomial (X-2)(X-3) with four roots in Z/6Z.

Comment by | August 7, 2008 | Reply

4. Benoit, this is true, but I’ve been pretty consistent in this whole section that I’m working over a field, and not over a general ring.

Comment by | August 7, 2008 | Reply

5. [...] Complex Numbers Yesterday we defined a field to be algebraically closed if it always has exactly as many roots (counting multiplicities) as we expect from its degree. But [...]

Pingback by | August 7, 2008 | Reply

6. [...] into linear factors because the complex numbers are algebraically closed. But we also know that real polynomials can have too few roots. Now, there are a lot of fields out there that aren’t algebraically closed, and I’m not [...]

Pingback by | August 14, 2008 | Reply

7. [...] Until further notice, I’ll be assuming that the base field is algebraically closed, like the complex numbers [...]

Pingback by | February 2, 2009 | Reply

8. [...] of the eigenvalues are not distinct. Worse, we could be working over a field that isn’t algebraically closed, so there may not be roots at all, even counting duplicates. But still, in the generic case [...]

Pingback by | February 10, 2009 | Reply

9. [...] we can use our definition of multiplicity for roots of polynomials to see that a given value of has multiplicity equal to the number of [...]

Pingback by | February 19, 2009 | Reply

10. [...] indeed, some real polynomials have no roots. But all is not lost! We do know something about factoring real polynomials. We can break any one [...]

Pingback by | March 13, 2009 | Reply

11. [...] together now. Start with a linear endomorphism on a vector space of finite dimension over an algebraically closed field . If you want to be specific, use the complex numbers [...]

Pingback by | April 9, 2009 | Reply

12. [...] so we may find things a little more complicated now. We will, however, have to assume that is algebraically closed and that no multiple of the unit in is [...]

Pingback by | August 25, 2012 | Reply

13. [...] recall that any linear endomorphism of a finite-dimensional vector space over an algebraically closed field can be put into Jordan normal form: we can find a basis such that its matrix is the sum of [...]

Pingback by | August 28, 2012 | Reply

14. [...] recall that any linear endomorphism of a finite-dimensional vector space over an algebraically closed field can be put into Jordan normal form: we can find a basis such that its matrix is the sum of [...]

Pingback by | August 28, 2012 | Reply

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