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Section Day 19
This is an outline of the topics we covered in the nineteenth day of class. The skeleton notes are in a handout, which can be printed out using the printer icon at the top right of its section of the page for filling in during class. Filled notes for each day will be posted after class to Canvas.
Handout Thursday 7/26
Objectives: Advanced Learning Outcomes
During our class meeting, we will work on learning the following. Fluency with these is not expected or required before class.
State and instantiate the definition of: unique factorization domain
State the following mathematical results: Unique Factorization in Z[x], PID Implies Irreducible Equals Prime, PID Implies UFD,
Theorem 196 . \(\Z[x]\) Has Unique Factorization.
Every polynomial in \(\Z[x]\) that is not zero and not a unit can be written as
\begin{equation*}
b_1 b_2 \dots b_s p_1 p_2 \dots p_m\text{,}
\end{equation*}
where the \(b_i\) are irreducible constant polynomials and the \(p_j\) are irreducible non-constant polynomials. If
\begin{equation*}
b_1 b_2 \dots b_s p_1 p_2 \dots p_m = c_1 c_2 \dots c_t q_1 q_2 \dots q_n
\end{equation*}
then \(s=t, m=n\text{,}\) and after reordering \(b_i=\pm c_i, p_i=\pm q_i\text{.}\)
Theorem 197 .
In a principal ideal domain, an element is irreducible if and only if it is prime.
Proof.
Note: Most of our examples have unique factorization, but not all are PIDs
Definition 198 . Unique Factorization Domain.
An integral domain \(D\) is a unique factorization domain (UFD) if
every nonzero non-unit element of
\(D\) can be written as a product of irreducible elements of
\(D\)
this factorization is unique up to associates and order of the factors.
Note: Not all domains are UFDs! You’ll show this.
Lemma 199 . Ascending Chain Condition for Principal Ideal Domains.
In a PID, any strictly increasing chain of ideals
\(I_1 \subset I_2 \subset I_3 \subset \dots\) must be finite in length.
Proof.
Theorem 200 . PID Implies UFD.
Every principal ideal domain is a unique factorization domain.
Proof.
