March 14


  • Questions
  • The last class ended up focusing on the constructability of regular polygons and the search for Fermat primes, so this one is for algebraic number theory and the fundamental theorem of arithmetic - the end of our work on the Fermat conjecture.

Student Questions:

The discussion page can be found here, for observations and remarks.

Over the spring break, please send your ideas for your term paper topics to Professor Bolker.


Cyclotomic Integers

Build the cyclotomic integers $Z[\omega]$, where $\omega$ is a primitive $p^{th}$ root of unity. Factor $x^p + y^p = z^p$ there in hopes of proving the Fermat conjecture.

  • Unique factorization fails for $p \ge 23$. (Needs good definitions of "prime" and "irreducible", which may differ!).
  • Note that the cyclotomic integers (Eisenstein's) for $p=3$ live in a quadratic extension.
  • Look at some quadratic examples. build the Gaussian integers by adjoining $\sqrt{-1}$ to the integers. The resulting ring enjoys unique factorization.
  • Unique factorization fails for the adjunction of $\sqrt{-5}$. The norm is the tool of choice for proving this.
  • The _Eisenstein integers_ do enjoy unique factorization. Since $-3 \equiv 1 (\mod 4)$ those weird halves come into play. Factoring 4 there is interesting - it answers Matt's question at the end of the last class, but in a way that may feel unsatisfying.

If time: units in real quadratic extensions, Pell's equation.

Gaussian Integers

Numbers of the form $a+ib : a,b \in \mathbb{Z}$.

The norm of a Gaussian Integer: $N(a+ib)=a^2+b^2$.
Generally: $N(\alpha \cdot \beta) = N(\alpha) \cdot N(\beta)$.

  • Question: What GIs have norm 1?
  • Answer: $\pm 1$ and $\pm i$.
  • Question: What GIs have reciprocals? i.e., which GIs are units?
  • Answer: If $\alpha$ is a unit, then $\frac{1}{\alpha}$ is a GI. We know that $\frac{1}{\alpha} = \frac{\overline{\alpha}}{N(\alpha)}$. Then for $\frac{1}{\alpha}$ to be a GI, $N(\alpha)= 1$. Then the only GIs with reciprocals are $\pm 1$ and $\pm i$.

$\alpha$ is irreducible when $\alpha = \beta \cdot \gamma$ implies that either $\beta$ or $\gamma$ are units.

$\alpha$ is prime when $\alpha | (\beta \cdot \gamma)$ implies either $\alpha | \beta$ or $\alpha | \gamma$.

In certain rings (sets), a number can be irreducible but not prime.

Consider the ring $\{a+b\sqrt{-5} : a, b \in \mathbb{Z} \}$.

2 is irreducible.

Suppose that $2=\alpha \cdot \beta$. Then $N(2)=4=N(\alpha) \cdot N(\beta)$.
Then $N(\alpha)$ is either 1,2 or 4. Let $\alpha=x+y\sqrt{-5}$. If $N(\alpha)=2$, then $x^2+5y^2=2$. This is clearly impossible, since $x^2$ and $y^2$ are positive integers. Then $N(\alpha)$ is either 4 or 1. If $N(\alpha)=1$, then $\alpha$ is a unit, and we are done. If $N(\alpha)=4$, $N(\beta)=1$, and we are done. Q.E.D

2 is not prime.

Note that in this particular ring, we can write: $2 \cdot 3=6 = (1+\sqrt{-5})(1-\sqrt{-5})$. Then 2 divides $6=(1+\sqrt{-5})(1-\sqrt{-5})$, but 2 doesn't divide either $(1+\sqrt{-5})$ or $(1-\sqrt{-5})$.

In this ring, 7 is irreducible and a prime. Proof left as exercise.

In GIs, every irreducible is also prime. This is equivalent to saying that GIs have unique factorization.

The proof depends on the division algorithm for GIs.

Notes: Shira
Formatting: Rob

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