I have a question:
Write down the two definition for prime elements in $\Bbb Z[\sqrt 2 ]$, as well as the Unique Factorization Theorem for $\Bbb Z[\sqrt 2 ]$.
How would I do this?
Definition for Prime Elements in $\Bbb Z[\sqrt 2 ]$
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elementary-number-theory
1 Answers
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From Wikipedia: An element $p$ of a commutative ring $R$ is said to be prime if it is not zero or a unit and whenever $p$ divides $ab$ for some $a$ and $b$ in $R$, then $p$ divides $a$ or $p$ divides $b$. Equivalently, an element $p$ is prime if, and only if, the principal ideal $(p)$ generated by $p$ is a nonzero prime ideal.
You can see that the funciotn $N: \mathbb Z[\sqrt 2] \to\mathbb N$ such that for $\alpha=a + b\sqrt 2$ is defined as $$N(\alpha)= |a^2-2b^2|$$ makes our ring into an Euclidean Domain. It is a general fact that every ED is a PID, and that every PID is a UFD. Let's try to see these two facts. Suppose we have an ED, we want to show it is also a PID. Take an ideal $I$, we want to show it is principal, ie it exists an element $a$ such that $I=(a)$. Pick $a$ as an element with the minimum norm in the $ED$, you can do that because the norm takes values in the natural numbers. Now pick any other element $i$ in $I$. You can perform the euclidean division since you are working in an ED $$i=aq+r$$ and this tells us that $i \in (a)$. Thus $I=(a)$ and our ED is a PID. Now let's prove that every PID is a UFD. Suppose that in a given PID there is an element that can be written in two different ways as a product of irreducible elements with no repetitions ($p_i \not = p_j$ if $i \not =j$ and the same fare the $q_i$). $$p_1^{a_1}\cdot p_2^{a_2}\cdot \dots\cdotp_r^{a_s}=q_1^{b_1}\cdot q_2^{b_2}\cdot \dots \cdot q_m^{b_m}$$ These are prime elements also, because in a PID every irreducible element is prime (and every prime element is irreducible). Since $p_1$ is prime and it divides the right hand side of the equation it must divide one of the $q_i$, let's say by reordering the product that it divides $q_1$, but $q_1$ is irreducible, thus $p_1=cq_1$ where $c$ is a unit, ie an invertible element. Moreover of course $a_1=b_1$. Now repeat this argument for $p_2$ etc until you finish the factors. We have shown that in a PID there is a unique way to write an element as a product of irreducible elements up to a unit and up to order.