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In 1847, Lame gave a false proof of Fermat's Last Theorem by assuming that $\mathbb{Z}[r]$ is a UFD where $r$ is a primitive $p$th root of unity.

The best description I've found is in the book Fermat's Last Theorem A Genetic Introduction to Algebraic Number Theory. For the equation $x^n + y^n = z^n$, it says

That is, he planned to show that if $x$ and $y$ are such that the factors $x+y, x+ry, \dots, x+r^{n-1}y$ are relatively prime then $x^n + y^n = z^n$ implies that each of the factors $x+y, x+ry, \dots$ must itself be an $n$th power and to derive from this an impossible infinite descent. If $x+y, x+ry, \dots$ are not relatively prime he planned to show that there is a factor $m$ common to all of them so that $(x+y)/m, (x+ry)/m, \dots, (x+r^{n-1}y)/m$ are relatively prime and to apply a similar argument in this case as well.

Nowhere else can I find any more detail as to how the rest of the argument goes. I don't see how to use infinite descent here, can anyone fill in the details?

I would like to also mention Conrad's great notes on Kummer's proof for regular primes. While this does indeed give a proof, it seems quite sophisticated and I am hoping there is an easier method when one assumes (falsely, in general) that $\mathbb{Z}[r]$ is a UFD.

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The paper (in French) to which Edwards refers in his book may be found here:

http://gallica.bnf.fr/ark:/12148/bpt6k29812/f310.image

My French isn't good enough, however, to elaborate on the argument given above.

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  • $\begingroup$ Thank you for the link! I should have checked references. I can read french, once I have time to give it a quick look over I am all but certain I will accept this answer. $\endgroup$ – RghtHndSd Oct 1 '14 at 19:00
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    $\begingroup$ It does appear to be a complete proof, and also much more complicated than what is usually indicated when it is discussed (at least I got the impression that the descent argument was rather short). I will be editing your answer to incude summary in a few days. $\endgroup$ – RghtHndSd Oct 3 '14 at 0:23
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    $\begingroup$ @RghtHndSd Will you include the mentioned summary? It would be kindly appreciated! $\endgroup$ – Jose Brox Oct 7 '15 at 9:42
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    $\begingroup$ @JoseBrox: After looking through the proof for about an hour or so, I found that it wasn't the French that was difficult but rather the language that the mathematics was expressed in. Since I wanted to use this for a class I was teaching, and it was clear the descent argument would not be short enough to present, I decided to skip it. I have no current intentions of trying to understand it, but maybe sometime in the future. $\endgroup$ – RghtHndSd Oct 7 '15 at 15:30
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Lamé's proof is quite short and easy to follow for a modern reader. I won't give a full translation, just a summary. Since my French is far better than my algebraic number theory, I'll refrain from taking too many shortcuts, and stick to the original structure of the proof, so as not to introduce any mistakes.

The idea of the proof is to prove that $A^n+B^n=C^n$ has no solution in $\mathbb Z[r]$, where $r$ is a primitive root of $r^n=1$. In particular, this shows that it has no solution in $\mathbb Z$.

Lamé immediately points out that it suffices to consider the case $n$ prime. He also mentions at the end of the proof that, due to a technical reason which we'll get to later, the proof doesn't work for $n=3$, so we're talking about an odd prime $n>3$.

§ I - Properties of $\mathbb Z[r]$

Lamé begins by defining more or less explicitly $\mathbb Z[r]$, where $r$ is a primitive root of $r^n=1$. He quickly goes through some basic properties of $\mathbb Z[r]$ and defines some basic concepts:

  1. The representation of an element of $\mathbb Z[r]$ as $\alpha_0 + \alpha_1 r + ... + \alpha_{n-1}r^{n-1}$ (all the $\alpha_i$ integers) is unique up to adding the same integer to each coefficient.

  2. If for some $E$, $A=DE$, where all values are in $\mathbb Z[r]$, we say that $A$ is divisible by $D$.

  3. In order for an element $a\in\mathbb Z[r]$ to be divisible by an element $k\in\mathbb Z$, it is necessary and sufficient that the coefficients of $a$ to be congruent modulo $k$ (which does not depend on the choice of coefficients).

  4. Given a value $A$, we can consider the $n$ values $r^iA$ for different values of $i$. Lamé uses the notation $A^{(i)}=r^iA$, but I kept misreading that as an exponent while reading the paper, so I'll just write $r^i A$. Lamé notes that the $n$-th powers of these numbers are all the same.

  5. If we think of $A$ as a polynomial expression in $r$, we can imagine replacing $r$ with $r^i$. We'll call this value $A_i$. In modern language we're talking about the image of $A$ under the automorphism $r\to r^i$.

  6. We'll call the product of the $A_i$ the modulus of $A$. Since it's a symmetric function of the roots of $r^n=1$, it's an integer. The modulus is invariant under the automorphisms $r\to r^i$ and the modulus of $r^i A$ is the same as the modulus of $A$. The modulus is multiplicative.

  7. Finally, let $z_i=r^i+r^{-i}$.

§II - Factorization of $A^n+B^n$

We're now going to factor $A^n+B^n$ in a rather specific way. We have

$$\prod_{i=0}^{n-1} (B+r^i A) = \sum_{i=0}^n S_i A^iB^{n-i}=A^n+B^n \tag 1$$

Here $S_i$ is the i-th elementary symmetric polnomial in the $r^i$, so that $S_0=1$, $S_1=1+r+r^2+...r^{n-1}$, and $S_n=1rr^2...r^{n-1}$. The first equality is just straight algebra, but the second involves specific properties of the roots $r^i$: since they solve the equation $x^n-1=0$, all of the $S_i$ are zero except $S_0=S_n=1$.

Lamé now rewrites this factorization in a slightly different form:

$$A^n+B^n=\prod_{i=0}^{n-1} (Ar^i + Br^{n-i}) \tag 2$$

To obtain this form, start with the factorization in $(1)$. Take the factor $r^i A + B$, and factor out $r^k$ with $k=-i/2\mod n$. We obtain $r^k(r^i A + B)=Ar^{-k}+Br^k$, which can be written $Ar^j+Br^{n-j}$ for some $j$, as required. As we run through all the values of $i$, $k$ and $j$ both run through all values $0, 1, 2, ..., n-1$, so we have

$$A^n+B^n=r^{-(0+1+2+...+(n-1))}\prod_{j=0}^{n-1} (Ar^j + Br^{n-j})$$

Since the exponent on that factor of $r$ is divisible by $n$, that factor is just $1$, so we have the desired factorization.

Now let's write $M_i$ for the $i$-th factor, that is, $M_i=Ar^i+Br^{n-i}$. These factors satisfy a system of linear equations, namely, for any two indices $i,j$:

$$M_i+M_j=z_{\frac{i-j}2} M_{\frac{i+j}2} \tag 3$$

where the indices are interpreted modulo $n$, and the $z_i$ are the constants which I defined at point 7 of the list above (in §1). This isn't very obvious, but is easy to check.

Lamé now states without proof that thanks to these equation, we can conclude two things about the $M_i$:

  1. If any number $D$ divides any two of the $M_i$, it divides all of them.

  2. If one of the $z_i$ divides any one of the $M_j$, it divides all of them.

Neither of these facts seems obvious to me. If we write (3) as $M_i+M_j=z_kM_l$, then we can say that if $D$ divides both $M_l$ and $M_j$, it divides $M_i$, $i=2l-j\mod n$. Presumably the idea is to then repeat this argument with other equations until we cover all of the indices. It seems to me we need the following lemma:

Lemma (Not mentioned in Lamé's paper): If a subset of $\mathbb Z/n\mathbb Z$ contains at least two elements and is closed under the operation $(k, j)\to 2k-j$ for $k\neq j$, then it is equal to all of $\mathbb Z/n\mathbb Z$.

Note that if $k=j$ we would have $2k-j=k$, so strictly speaking the condition $k\neq j$ is not necessary (it's just that in that case there's no corresponding equation). Similar logic will allow us to justify point (2) if we have the lemma:

Lemma (Also not mentioned): If a subset of $\mathbb Z/n\mathbb Z$ contains at least two elements and is closed under the operation $(k, j)\to 2k+j$, then it is equal to all of $\mathbb Z/n\mathbb Z$.

I'm not sure if either of these lemmas is true. If not, then I don't know how to justify Lamé's claims (1) and (2). In any case, if they are true, then after pulling common factors out of the $M_i$, we obtain a new factorization of $A^n+B^n$:

$$A^n+B^n=k^nm_1m_2...m_{n-1}$$

where the $m_i$ are pairwise coprime in $\mathbb Z[r]$ and are each indivisible by any of the $z_j$. This implies, and I believe this is where the assumption of unique factorization creeps in for the first time, that the power of $z_j$ for any $j$ in the unique prime factorization of $A^n+B^n$ (that is, generally, in a sum of two $n$-th powers) must be a multiple of $n$.

§III - Conclusion

Now that we have a factorization of $A^n+B^n$, suppose we had $A^n+B^n=C^n$. Again implicitly using uniqueness of prime factorization, this implies that in $A^n+B^n=k^nm_1...m_{n-1}$, each of the $m_i$ must itself be an $n$-th power, since they're coprime from one another and from $k$. Thus we would have, say, $m_i=\mu_i^n$, but then, going back to the equations for the $M_i$, we would have:

$$\mu_i^n+\mu_j^n=z_k\mu_l^n \tag 4$$

This is impossible, since by what we obtained at the end of the last section, the power of $z_k$ in the factorization of the left hand side must be a multiple of $n$, but one the right it is one more than multiple of $n$. Here Lamé mentions that it can be shown that the $z_i$ are not themselves $n$-th powers, but I'm not sure if this is important. He also mentions that since for $n=3$, there is only one value of $z_i$ which is $z_i=-1$, a unit, this argument fails in that case, which is why we assumed $n>3$ above.

Lamé concludes by explaining why Fermat's last theorem is a special case of the above. For me this seems obvious since it seems to me we've proven the theorem for $\mathbb Z[r]$, and $\mathbb Z\subset\mathbb Z[r]$, but for some reason Lamé continues with a fairly involved final paragraph. I don't understand the argument or why it's necessary, so I'll just translate it for those who do:

Fermat's theorem, for $n>3$, is only a special case of what has just been proven; since if $A$ and $B$ are integers, that is if they reduce to $\alpha_0$, $\beta_0$, $M_1$ will be integer, as well as $C$, $k$, and $\mu_1$; but $\mu_2, ... \mu_{n-1}$ will remain complex numbers: however, their product must be an integer modulus, that is $\mu_1, ... \mu_{n-1}$ must be the factors of a whole number of the form $Y^2\pm nZ^2$ [Lamé mentions in §1 that moduli are always of this form]; finally, the relations [displayed equation (4)] will still be necessary, and the conclusion of impossibility will be the same.

The article ends with some observations from Liouville in which he points out the flaw in the assumption of unique factorization. Edwards describes these comments in his book, although he makes them out to seem alot ruder than they look to me in the original French. Maybe I'm just not picking up on subtle nineteenth century academic sarcasm.

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    $\begingroup$ Nice work! Both lemmas can be proved by using the following, easy fact for a prime $p$. If the difference of an infinite arithmetic progression $(a)$ is non congruent to $0$ modulo $p$, then $(a)$ contains all the residues mod $p$. Recall that $n>3$ is a prime. In the first Lemma you can generate the sequence $a_t=t(k-j)+j$, because $a_0=j$, $a_1=k$ and $(a_t,a_{t+1})\to a_{t+2}$ for $t\ge0$. In the second lemma WLOG let $k\neq0$. You can generate $a_t=2tk+j$ in the following way: $a_0=j$, $(k,j)\to 2k+j=a_1$ and $(k,a_t)\to a_{t+1}$ for $t\ge0$. $\endgroup$ – B.B. May 16 '18 at 14:04
  • $\begingroup$ Thanks a lot for preparing such a detailed summary. Very helpful! $\endgroup$ – UnrealVillager Dec 30 '20 at 0:27
  • $\begingroup$ Just curious, aren't all z_k actually units in Z[r]? And if so, Lamé's entire argument is not valid even regardless of the unique factorization. z_k can be reduced to 1 + r^2k. But isn't that one of those so-called "cyclotomic units"? Maybe I'm missing something... $\endgroup$ – UnrealVillager Jan 4 at 0:43

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