# Diophantine Equations (Lec. 1, Thursday, January 14) ## Intro/Logistics See website for notes on books, intro to class. - Youtube Playlist: - Free copies of textbook: Paul's description of the course: > This course is an introduction to arithmetic beyond $\ZZ$, specifically arithmetic in the ring of integers in a finite extension of $\QQ$. > Among many other things, we’ll prove three important theorems about these rings: > > - Unique factorization into ideals. > - Finiteness of the group of ideal classes. > - Dirichlet’s theorem on the structure of the unit group. ## Motivation :::{.remark} The main motivation: solving **Diophantine equations**, i.e. polynomial equations over $\ZZ$. ::: :::{.example title="of a Diophantine equation"} Consider $y^2 = x^3 + x$. :::{.claim} $(x, y) = (0, 0)$ is the only solution. ::: To see this, write $y^2 = x(x^2+1)$, which are relatively prime, i.e. no $D\in \ZZ$ divides both of them. Why? If $d \divides x$ and $d \divides x+1$, then $d\divides (x^2+1) + (-x) = 1$. It's also the case that both $x^2+1$ and $x^2$ are squares (up to a unit), so $x^2, x^2 + 1$ are consecutive squares in $\ZZ$. But the gaps between squares are increasing: $1, 2, 4, 9, \cdots$. The only possibilities would be $x=0, y=1$, but in this case you can conclude $y=0$. ::: :::{.example title="Fermat"} Consider $y^2 = x^3-2$. :::{.claim} $(3, \pm 5)$ are the only solutions. ::: Rewrite \[ x^3 = y^2+2 &= (y+ \sqrt{-2})(y - \sqrt{-2}) \\ &\in \ZZ[\sqrt{-2}] \da \ts{a+b\sqrt{-2} \st a,b,\in \ZZ} \leq \CC .\] This is a subring of $\CC$, and thus at least an integral domain. We want to try the same argument: showing the two factors are relatively prime. A little theory will help here: :::{.definition title="Norm Map"} \[ N \alpha \da \alpha\bar \alpha && \text{for } \alpha\in \ZZ[ \sqrt{-2} ] .\] ::: :::{.lemma title="?"} Let \( \alpha, \beta \in \ZZ[\sqrt{-2}] \). Then 1. \( N(\alpha \beta) = N(\alpha) N(\beta) \) 2. $N( \alpha) \in \ZZ_{\geq 0}$ and $N(\alpha) = 0$ if and only if \( \alpha= 0 \). 3. \( N(\alpha) = 1 \iff \alpha\in R\units \) ::: :::{.proof title="?"} 1. Missing, see video (10:13 AM). 2. \( N(\alpha) = a^2 + 2b^2 \geq 0 \), so this equals zero if and only if \( \alpha= \beta= 0 \) 3. Write \( 1 = \alpha\bar \alpha \) if \( N(\alpha) = 1 \in R\units\). Conversely if \( \alpha\in R\units \) write \( \alpha \beta = 1 \), then \[ 1 = N(1) = N(\alpha \beta) = N(\alpha ) N(\beta ) \in \ZZ_{\geq 0} ,\] which forces both to be 1. ::: :::{.claim} The two factors \( y \pm \sqrt 2 \) are *coprime* in \( \ZZ[\sqrt{-2}] \), i.e. every common divisor is a unit. ::: :::{.proof title="?"} Suppose \( \delta\divides y\pm \sqrt{-2} \), then \( y + \sqrt{-2} = \delta \beta \) for some \( \beta\in \ZZ[\sqrt{-2}] \). Take norms to obtain \( y^2 + 2 = N \delta N \beta \), and in particular - \( N \delta y^2 +2 \) - \( \delta \divides (y+ \sqrt{-2} ) - (y - \sqrt{-2} ) = 2 \sqrt{-2} \) and thus \( N \delta \divides N(2 \sqrt{-2} ) = 8 \). In the original equation \( y^2 = x^3-2 \), if $y$ is even then $x$ is even, and $x^3 - 2 \equiv 0-2 \mod 4 \equiv 2$, and so $y^2 \equiv 2 \mod 4$. But this can't happen, so $y$ is odd, and we're done: we have \( N \delta\divides 8 \) which is even or 1, but \( N \delta\divides y^2 +2 \) which is odd, so \( N \delta = 1 \). ::: We can identify the units in this ring: \[ \ZZ[\sqrt{-2} ]\units = \ts{ a + b \sqrt{-2} \st a^2 + 2b^2 = 1} \] which forces $a^2 \leq 1, b^2 \leq 1$ and thus this set is $\ts{\pm 1}$. So we have \( x^3 = ab \) which are relatively primes, so $a,b$ should also be cubes. We don't have to worry about units here, since $\pm 1$ are both cubes. So e.g. we can write \[ y + \sqrt{-2} = (a + b \sqrt{-2} )^3 = (a^3-6ab^2) + (3a^2b -2b^3) \sqrt{-2} .\] Comparing coefficients of \( \sqrt{-2} \) yields \[ 1 = b(3a^2b - 2b^2) \in \ZZ \implies b \divides 1 ,\] and thus $b\in \ZZ\units$, i.e. $b\in \ts{\pm 1}$. By cases: - If $b=1$, then $1 = 3a^2 -2 \implies a^2 = 1 \implies a = \pm 1$. So \[ y = \sqrt{-2} = (\pm 1 + \sqrt{-2} )^3 = \pm 5 + \sqrt{-2} ,\] which forces $y=\pm 5$, the solution we already knew. - If $b = -1$, then \( 1 = -(3a^2 - 1) \) which forces $1=3a^2 \in \ZZ$, so there are no solutions. ::: ## Failure of Unique Factorization :::{.example title="where unique factorization fails"} Consider $y^2 = x^3 - 26$. Rewrite this as \[ x^3 = y^2 + 26 = (y + \sqrt{-26} )(y - \sqrt{-26} ) ,\] then the same lemma goes through with $2$ replaced by $26$ everywhere where the RHS factors are still coprime. Setting $y + \sqrt{-26} = (a + b \sqrt{-26} )^3$ and comparing coefficients, you'll find $b=1, a = \pm 3$. This yields $x=35, y=\pm 207$. But there are more solutions: $(x, y) = (3, \pm 1)$! The issue is that we used unique factorization when showing that $ab$ is a square implies $a$ or $b$ is a square (say by checking prime factorizations and seeing even exponents). In this ring, we can have $ab$ a cube with *neither* $a,b$ a cube, even up to a unit. ::: :::{.question} When does a ring admit unique factorization? Do you even *need* it? ::: :::{.remark} This will lead to a discussion of things like the **class number**, which measure the failure of unique factorization. In general, the above type of proof will work when the class number is 3! :::