About a couple of weeks ago, Noah Snyder liveblogged at the Secret Blogging Seminar on two topology talks given by Jacob Lurie. You may want to learn more about the contents of the talks by clicking the appropriate link. The thing is when I came to know about Noah’s liveblog, the thought that immediately sprang to my mind was the one involving his elementary proof of the well-known Mason-Stothers theorem.
In this post, I wish to present Noah Snyder’s proof of the Mason-Stothers Theorem, which is the polynomial version of the yet unproven (and well-known) ABC Conjecture in number theory. I will follow it up with a problem and its solution using the aforesaid theorem. For a detailed and wonderful exposition on the ABC conjecture, you may want to read an article, titled The ABC’s of Number Theory, written by Noam Elkies for The Harvard College Mathematics Review.
First, a brief history. Though this theorem on polynomials was proved by Stothers in 1981, it didn’t attract much attention until 1983 when it was rediscovered by Mason. Noah, in 1998 (while still in high school), gave perhaps the most elegant elementary proof.
The proof below is the version given in Serge Lang’s Undergraduate Algebra, which it seems has quite a number of typos.
Okay, now some terminology. If and are polynomials, then
- denotes the degree of ,
- denotes the number of distinct zeros of , and
- denotes the .
To illustrate, suppose and . Then, and . And, and . Also, .
Let us first prove a couple of useful lemmas before stating the theorem and its proof.
Lemma 1: If is a polynomial, then has repeated roots iff and have a common root.
Proof: If has a root of multiplicity , then , where . Therefore,
.
Now, if , then , which is the same thing as saying, if does not have a repeated root, then and don’t have a common root. And, if , then has root of multiplicity at least . This is same as saying, if has a repeated root , then and have a common root . And, this completes our proof.
Lemma 2: If is a polynomial, then .
Proof: Suppose and has distinct roots , with multiplicities , respectively. Then, . Now, from the proof of lemma above, we note that . Therefore,
. And, we are done.
Okay, we are now ready to state the theorem.
Mason-Stothers Theorem: If are relatively prime polynomials, not all constant, such that , then .
Proof: We first note that
.
Indeed, we have . Therefore,
. And, we are done.
Also, note that at least two of the polynomials and are non-constants, for if any two polynomials are constants, then this forces the third to be a constant as well. So, without any loss of generality, assume and are non-constant polynomials. Now, we note that , for otherwise, , and since and are relatively prime, this would imply , which leads to a contradiction!
Now, we observe that and divide the left hand side of , and divides the right hand side of , which is equal to the left hand side. And, since and are relatively prime, we conclude that
divides .
The above implies that
,
which implies
Now, applying lemma to , we obtain,
Using the above in , we get
since are relatively prime polynomials.
Due to symmetricity, the above arguments can similarly be repeated for and to get similar inequalities for and . And, this concludes our proof.
( Earlier, I had mentioned I would pose a problem and also give its solution that would use the above theorem. I will do that in my next post.)
8 comments
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March 4, 2008 at 4:51 am
Ben Webster
Not that it really matters, but you might want to wait until Noah actually gets around to graduating to start calling him “Dr.” Otherwise, it won’t seem fun by the time it’s official.
March 4, 2008 at 5:01 am
Vishal
Thanks for that piece of info, Dr Webster! 🙂 For some reason, I completely missed it. I have now made the appropriate changes.
March 4, 2008 at 5:47 am
A problem on relatively prime polynomials « Vishal Lama’s blog
[…] Math Problem Solving Tags: mason-stothers theorem, noah snyder Yesterday, I wrote a post on the Mason-Stothers theorem and presented an elementary proof of the theorem given by Noah Snyder. As mentioned in that post, I […]
February 28, 2009 at 10:23 pm
Zygmund
Nice post. Your exposition of the proof is much better than the one in Serge Lang’s Algebra, which is where I learned first about the theorem, but this has clarified it.
One application is to prove the analog of Fermat’s last theorem for polynomials. Let n >= 3, and let x(t), y(t), z(t) be relatively prime polynomials in F[t] where F is an algebraically closed field of characteristic zero. Then if x(t)^n + y(t)^n = z(t)^n, the three polynomials are constants. To see this, assume wlog that z(t) has the biggest degree, or move around the terms and multiply by roots of -1. Since F[t] is a principal ideal domain, assume moreover that x(t), y(t), and z(t) are relatively prime. We have then
n deg z(t) <= N_0( x(t)y(t)z(t)) – 1 <= deg x + deg y + deg z – 1 2, this is a contradiction.
March 6, 2009 at 10:22 am
Vishal Lama
Thanks, Zygmund! I will admit that I was inspired by Emil Artin’s style when I wrote the post.
March 7, 2009 at 12:58 pm
Todd Trimble
It’s interesting that Lang was Artin’s student. Although one gets the definite impression that Artin took more time polishing his textbook writing than Lang ever did. (There are various legends that Lang wrote some of his more famous books, like the first edition of Algebra or Algebraic Number Theory, in a single weekend. Impressive as that would be, to some extent it would almost seem that way.)
February 6, 2011 at 9:51 pm
Game Changing Conjectures In Mathematics « Gödel’s Lost Letter and P=NP
[…] Stothers in 1981 and independently Richard Mason in 1983 discovered a basic theorem, not a conjecture, about polynomials, that looks like the ABC Conjecture for integers. Since their […]
September 13, 2012 at 2:25 am
The ABC Conjecture And Cryptography « Gödel’s Lost Letter and P=NP
[…] this for a simple proof due to Noah Synder, who found it while still in high school. One cool […]