See here. The question is as follows.

How do we see that there do not exist nonconstant, relatively prime, polynomials $a(t)$, $b(t)$, and $c(t) \in \mathbb{C}[t]$ such that$$a(t)^3 + b(t)^3 = c(t)^3?$$

There is the following answer, purportedly elementary.

We give a completely elementary proof simply working in the ring $\mathbb{C}[t]$ and using that it is a UFD. Suppose there are some solutions of$$a(t)^3 + b(t)^3 = c(t)^3$$in $\mathbb{C}[t]$. Choose a solution $(a(t), b(t), c(t))$ such that the maximum $m > 0$ of the degrees of $a$, $b$, $c$ is minimal possible among all solutions. Clearly, this choice ensures that $a(t)$, $b(t)$, $c(t)$ are coprime. Then we have$$a(t)^3 = c(t)^3 - b(t)^3 = (c(t) - b(t))(c(t) - \omega b(t)) (c(t) - \omega^2 b(t)),$$where $\omega$ is a third primitive root of unity. Now, we look at the factors $c(t) - b(t)$, $c(t) - \omega b(t)$. Suppose that they have a common factor $q(t)$. Considering their sum and difference, we conclude that $c(t)$ and $b(t)$ have a common factor too. Moreover, $q(t)$ is a factor of $a(t)$. Thus, $a$, $b$, $c$ are not relatively prime, which is a contradiction. Repeating the same game with other pairs of factors, we see that all three factors $c(t) - b(t)$, $c(t) - \omega b(t)$, $c(t) - \omega^2 b(t)$ are pairwise coprime. Therefore,$$c(t) - b(t) = d_1(t)^3,\text{ }c(t) - \omega b(t) = d_2(t)^3,\text{ }c(t) - \omega^2b(t) = d_3(t)^3,\text{ where }d_1,\,d_2,\,d_3 \in \mathbb{C}[t].$$Note that$$\omega^2 + \omega + 1 = 0.$$Multiplying the second equation by $\omega$ and the third equation by $\omega^2$ and adding all three, we arrive at $$d_1(t)^3 + \omega d_2(t)^3 + \omega^2d_3(t)^3 = 0.$$Choosing $\eta_1$ and $\eta_2$ as any third roots of $-\omega$ and $-\omega^2$, respectively, and letting$$a_1 = d_1^3,\text{ }b_1 = \eta_1d_2^3,\text{ }c_1 = \eta_2d_2^3,$$we get$$a_1^3 = b_1^3 + c_1^3.$$By construction, at least one of $a_1$, $b_1$, $c_1$ is a nonconstant polynomial, and the maximum of their degrees is smaller than that of $a$, $b$, $c$. This is a contradiction to the choice of $a$, $b$, $c$.

Unfortunately, I have never seen any abstract algebra before (I am only a student who has taken calculus), and therefore, do not really understand what is going on, aside from the fact infinite descent is being invoked. Can anyone explain the motivation/what is going on/the key steps/help me understand the proof? Thanks.

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    $\begingroup$ This proof is quite long: you should tell us what are the specific steps you don't understand (for example, do you know what $\omega$ is? Do you know what is an UFD?), in order to better help you. $\endgroup$ – Crostul Dec 7 '15 at 8:13
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    $\begingroup$ Yes, the answer is elementary. If you do not understand it, then you need to learn abstract algebra. It's not a big deal, many people on Earth don't know anything about algebra -- but you still need it to understand this proof. There's no way around it. You wouldn't expect someone who's never run in their life to be able to run a marathon, or someone who's never studied medicine to successfully perform surgery, right? $\endgroup$ – Najib Idrissi Dec 7 '15 at 20:27
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    $\begingroup$ I'm really confused by the positive attention this question has received. The OP has verbatim copied a long answer from another question and asked us to explain it to them. There is no evidence of any effort (eg we don't even know where the OP first gets stuck); all we know is that they have no background in abstract algebra. Without more information, I don't think it is possible to answer this question adequately. $\endgroup$ – Mathmo123 Dec 9 '15 at 9:47
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    $\begingroup$ This argument just uses simple arithmetic of polynomials (long division etc.) that is essentially high school algebra. The assertion that $\mathbb C[t]$ is a UFD appears, but this is just short hand for saying that polynomials have unique factorization into linear polynomials, which is again a piece of high school algebra. Did you try actually working through it? $\endgroup$ – tracing Dec 9 '15 at 12:21
  • $\begingroup$ Just answered to your question with lots of details about the formal framework behind it and on it as well. $\endgroup$ – Olórin Dec 15 '15 at 22:18

The formal framework behind your concrete problem is the following : you want to show that "something sizeable" does not exist. "Something that does not exist" is coded by a set $X$ that we want to show to be empty. "Being sizeable" means that you have a "size function" $s : X \rightarrow \mathbf{N}$. If $X$ is not empty, as the set $s(X)$ is a non-empty part of $\mathbf{N}$, it has a smallest element $d_0$ reached for some $x_0\in X$ such that $s(x_0)=d_0$. Now, if by some argument, some "recipe", from $x_0$ you arrive to construct another element $x_1$ from $X$ such that $s(x_1) = d_1 < d_0$ then you will have contradicted the fact that $s(x_0)$ is supposed minimal, and show that $X$ is empty. The descent name of the argument comes from the fact that from $x_1$ you could, with the same recipe construct an $x_2$ such that $s(x_2) < s(x_1)$, and then an $x_3$ etc etc : by induction you construct a sequence $(x_n)_n$ such that the sequence $(s(x_n))_n$ strictly decreases : it "descends infinitely", hence the "descente infinie" french term given by Fermat himself to such kind of proofs.

Now, in the case of your concrete problem, can you find the right set $X$, and the right "size" function $s : X \rightarrow \mathbf{N}$ ? Can you then find the right recipe to get $x_1$ from $x_0$ ? (Note that the recipe is just a method giving you a $y\in X$ such that $s(y) < s(x)$, starting from a given $x\in X$.) I will answer to these question through the details I give below.

First, no need to know what a UFD (unique factorization domain) is : we are in the $\mathbf{C}[T]$ case and the only thing we will use is this : for any non constant polynomial $P\in\mathbf{C}[T]$ you have a unique (modulo the order of the factors) factorization $$P = a_d (T-\lambda_1^{\mu_{\lambda_1}})\ldots(T-\lambda_d^{\mu_{\lambda_d}}) = z \prod_{\lambda\in S} (T - \lambda^{\mu_{\lambda}})$$ where $S$ is the set of roots of $P$ and $\mu_{\lambda}$ is the multiplicity of the root $\lambda\in S$ (this factorization is called nice factorization from now on) where the $\lambda_i\mathbf{C}$ are the distinct roots of $P$, $a_d\in\mathbf{C}$ is $P$'s leading coefficient and $d$ is the degree of $P$. You already know this : it is called that the field $\mathbf{C}$ is algebraically closed.

I fixed once for all an integer $n\geq 3$. Now we note $X$ the set of triples $(a(T),b(T),c(T))\in\left(\mathbf{C}[T]\right)^3$ such that

  • $a,b$ and $c$ are non constant and relatively prime : this means that you cannot find a $\lambda\in\mathbf{C}$ such that the polynomial $T-\lambda$ divides $a, b$ and $c$, that is, you cannot write $a(T) = (T-\lambda) a_0(T), b(T) = (T-\lambda) b_0(T)$ and $c(T) = (T-\lambda) c_0(T)$ for a triple $(a_0(T),b_0(T),c_0(T))\in\left(\mathbf{C}[T]\right)^3$ : that is the polynomials $a,b$ and $c$ have no common factor
  • $a(t)^n + b(t)^n = c(t)^n$ (note that I replaced $3$ by my $n$, I will prove something more general than what you want to prove, and I think that the greater generality will make you understand more easily what is formal behind the proof and what is not.)

and we want to show that $X = \varnothing$. Note that the degree of the zero polynomial is $-\infty$ let's say by convention, and that for a triple $(a(T),b(T),c(T))\in X$, at least one polynomial is non-zero, as if all were zero, $a,b$ and $c$ wouldn't be relatively prime. All this to say that for a triple $(a,b,c)\in X$ we have $s((a,b,c)) := \max\left( \textrm{deg}(a), \textrm{deg}(b), \textrm{deg}(c) \right) \in \mathbf{N}$ : we have our size function !

We want to show that $X = \varnothing$, and as I wrote at the beginning, we suppose that $X \not= \varnothing$ and we therefore take a $(a,b,c)\in X$, and we saw at the beginning that we have the right to choose $(a,b,c)$ such that $s((a,b,c))$ is minimal.

Remark. If $(a,b,c)$ is a solution in $\mathbf{C}[T]^3$ such that $s((a,b,c))$ is minimal, then $a,b,c$ are de facto relatively prime : would they have a common non trivial (that is, of degree $>0$) factor $Q$, we could write $a = Q a_0$, $b=Q b_0$ and $c = Q c_0$ and plugging these relations in the equation $a^n + b^n = c^n$ and factoring out the $Q^n$ shows that $a_0^n + b_0^n = c_0^n$, so that $(a_0,b_0,c_0)$ is also a solution but with $s((a_0,b_0,c_0)) < s((a,b,c))$ which contradicts the minimality of $s((a,b,c))$ and shows that $a,b,c$ are relatively prime indeed

Let's proceed further. We rewrite $a(t)^n + b(t)^n = c(t)^n$ as $a(t)^n = c(t)^n - b(t)^n$. Now remark that would we have $n=2$, you would know how to factor $c(t)^n - b(t)^n$, as we have the remarkable identity $c^2 - b^2 = (c-b)(c+b)$. We look for such an identity for $c(t)^n - b(t)^n$ in the general case $n\geq 3$. I am sure you heard about $n$-th roots of unity : these are complex numbers $\omega$'s such that $\omega^n = 1$. You have to know that there are so-called primitive $n$-th roots of unity : those are $n$-th roots of unity $\omega$'s such that the list $1,\omega,\ldots,\omega^{n-1}$ give you all $n$-th roots of unity. (Note that there are indeed $n$ $n$-th roots of unity : can you see why ?) Fix such a primitive $n$-th roots of unity $\omega$. As the $\omega^i$ for $0\leq i < n$ are all roots of unity, that is, roots of the polynomial $T^n - 1$, we can write : $$T^n - 1 = \prod_{i=0}^{n-1} (T - \omega^i ).$$ This is a formal identity, so that you can replace $T$ by $\frac{c(t)}{b(t)}$ in it and you will then see that you have $$c(t)^n - b(t)^n = \prod_{i=0}^{n-1} (c(t) - \omega^i b(t))$$ which is the remarkable identity we were looking for.

Now, we can rewrite the equation $a(t)^n = c(t)^n - b(t)^n$ as $$a(t)^n = \prod_{i=0}^{n-1} (c(t) - \omega^i b(t)).$$ What do we see here ? A product (in the right-hand side) that is equal to the $n$-th power of something (of $a$, in the left-hand side.) We would like to be able to say that this implies that each term of the product is the $n$-th power of something. We are allowed to make this implication when the factors of the product is constituted by relatively prime polynomials. (This is by the way the same situation as in $\mathbf{Z}$.) I will detail this :

(1) The $c(t) - \omega^i b(t)$ form relatively prime polynomials. Indeed : suppose they aren't and let $Q$ a common non trivial (that is, of degree $>0$) factor $Q$. We have $c - b = Q R_1$ and $c - \omega b = Q R_2$ for polynomials $R_1$ and $R_2$. Solving the system of equations constituted by two previous equations will show that $Q$ divides $b$ and $c$, and as $a^n = c^n - b^n$, $Q$ divides also $a^n$, and therefore divides also $a$ (you see why ?). So $Q$ divides $a$, $b$ and $c$ : contradiction as $a$, $b$ and $c$ are relatively prime. So, we shown that $c(t) - \omega^i b(t)$ form relatively prime polynomials.

(2) Previous (1) and the equation $c(t)^n - b(t)^n = \prod_{i=0}^{n-1} (c(t) - \omega^i b(t))$ imply that all $c(t) - \omega^i b(t)$ are $n$-th powers. To see this, simply write nice factorizations each of the $c(t) - \omega^i b(t)$'s : $$c(t) - \omega^i b(t) = z_i \prod_{\lambda\in S_i} (T-\lambda^{\mu_{\lambda}})$$ where $S_i \subseteq \mathbf{C}$ for all $i$ such that $0\leq i < n$. As the $c(t) - \omega^i b(t)$'s are relatively prime we have $S_i \cap S_j$ for all $i,j$ such that $0\leq i,j <n$ and $i\not=j$. Now if you you note $a = z\prod_{\lambda\in S} (T-\lambda^{\mu_{\lambda}})$ the nice factorization of $a$ (where $S\subseteq \mathbf{C}$), we must have $ S = \cup_{i=0}^{n-1} S_i$. Plugging all linear factorization described in this paragraph into the equation $a(t)^n = \prod_{i=0}^{n-1} (c(t) - \omega^i b(t))$ and regrouping in the left-hand side linear factors according to if they belong to the same $S_i$ and comparing with the right-hand side shows indeed that each $c(t) - \omega^i b(t)$ is the $n$-th power of some polynomial $D_i(T)$ : $c(t) - \omega^i b(t) = D_i(T)^n$ for $0\leq i < n$.

From now on I will diverge slightly from the proof you quote. As the $c(t) - \omega^i b(t)$'s are relatively prime, so are the $D_i (T)$'s (exercise : prove it). Looking at leading coefficients from $c$ and $b$ we see that at most one $D_i (T)$ can be constant. Now choose any triplet $(x,y,z)$ of distinct elements among the $D_i (T)$'s. (This is possible as $n\geq 3$.) As $x^n,y^n,z^n$ are in the sub-vector space of $\mathbf{C}[T]$ generated by $b$ and $c$ (remember what the definition of the $D_i (T)$'s is !) which as dimension (over $\mathbf{C}$) $\leq 2$, we have a non trivial relation $$ \alpha x^n + \beta y^n = \gamma z^n$$ where $(\alpha,\beta,\gamma)\not=(0,0,0)$. But (as $\mathbf{C}$ is algebraically closed) as each complex number is a $n$-th power of some other complex number, we can write $\alpha = \alpha_O^n$, $\beta = \beta_0^n$ and $\gamma = \gamma_0^n$ and then write $x_0^n + y_0^n = z_0^n$ where $x_0 = \alpha_0 x$, $y_0 = \beta_0 y$ and $z_0 = \gamma_0 z$.

Now, $x_0, y_0, z_0)$ are relatively prime non constant polynomial satisfying the equation, and such that $s(x_0,y_0,z_0)) < s((a,b,c))$, which is the contradiction with the minimality of $s((a,b,c))$ we were looking for. ;-)

Remark. All of this is based on the fact that $\mathbf{C}[T]$ is an UFD. You could mimic the same proof over $\mathbf{Z}$ instead of $\mathbf{C}[T]$ and conclude falsely that you get a proof of Fermat's great theorem, as it is believed Fermat did himself. Why falsely ? Because there would be and error in the "over $\mathbf{Z}$" analogue of the point (2). First over $\mathbf{Z}$, this argument would have to take place inside the sub-ring $\mathbf{Z}[\omega]:=\{a+b\omega\;|\;a,b\in\mathbf{Z}\}$ of $\mathbf{C}$. Second, for the argument to be true, you would need $\mathbf{Z}[\omega]$ to be a UFD, which is false. It is believed that Fermat falsely thought the $\mathbf{Z}[\omega]$'s to be a UFD for $\omega$ primitive $n$-th root of $1$. I think you can find historical details on this in the historical notes from Algèbre, Bourbaki.


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