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In our proof that every nonzero, nonunit, in a PID can be written as a product of irreducibles, there was a part of our proof that all of you took for granted. The following problem has you complete the details.
Problem 72 (Ideals Generated By A Reducible Are Properly Contained By Non Unit Factors)
Prove that if $d$ is reducible and $d=ab$ where $a$ and $b$ are non units, then $\left<d\right>$ is a proper subset of $\left<a\right>$.
Problem 74 (Existence Of A Splitting Field)
Let $F$ be a field and $f(x)\in F[x]$ have positive degree. We have already shown that there exists an extension field $E$ of $F$ in which $f(x)$ has a zero $a$, in other words we can write $f(x) = (x-a)g(x)$ where $a,g(x)\in E[x]$ (by the Factor Theorem). Prove that there exists an extension field $E'$ of $F$ in which $f(x)$ can be written as a product of linear factors, i.e. we can write $f(x) = (x-a_1)(x-a_2)\cdots (x-a_n)g$ where $a_1,a_2,\ldots, a_n,g\in E'$ and also $g$ is a unit.
Definition.Splitting Field of $f(x)$ over $F$
Look this on up in the text
Problem 73 (Unique Factorization In A PID)
Suppose that $D$ is a principal ideal domain. Suppose that $f=p_1p_2\cdots p_m = q_1q_2\cdots q_n,$ where each $p_i$ and $q_i$ is irreducible over $D$. Prove that $n=m$ and after rearranging, for each $i$ we have $p_i=u_iq_i$ for some unit $i$. In other words, prove that $D$ uniquely factors as a product of irreducibles.
We have now shown that in every principle ideal domain, every nonzero nonunit can be written as a product of irreducibles (we'll call this a factorization), and that any two factorization are essentially the same (after rearrangement and multiplying by units). Any integral domain that satisfies this property is called a unique factorization domain.
Definition (Unique Factorization Domain)
Let $D$ be an integral domain. We say that $D$ is a unique factorization domain if the following two properties hold.
- If $d\in D$ is not a unit, and not zero, then we can write $d$ as a product of irreducibles over $D$.
- If we have written $d$ as a product of irreducibles over $D$ in two ways, say $d=p_1p_2\cdots p_n$ and $d=q_1q_2\cdots q_m$, then $n=m$ and after rearranging we have for each $i$ that $p_i=u_iq_i$ for some unit $u_i$.
The next problem should follow immediately from a fact that we have already shown.
Problem 75 (Polynomial Rings Over A Field Are Unique Factorization Domains)
Prove that $F[x]$ is a unique factorization domain if $F$ is a field.
What was the key that made this all work? What is the key to proving that $F[x]$ is a principle ideal domain? The key is the Division Algorithm, which is also called the Euclidean Algorithm. If an integral domain has something similar to the division algorithm, then we'll call it a Euclidean domain.
Definition.Euclidean Domain
Please look this one up the book.
Take a look at the proof we used to show that $F[x]$ is a PID whenever $F$ is a field. The key to this proof was the division algorithm. You should be able to generalize this proof to show that any Euclidean domain is a principle ideal domain.
Problem 76 (Euclidean Domains Are PIDs)
Prove that a Euclidean domain is a principle ideal domain. Prove also that a Euclidean domain is a unique factorization domain.
Problem 77 (Both Z And FX Are Euclidean Domains)
Prove that $Z$ is a Euclidean domain with function $d(a)=|a|$. Prove that $F[x]$ is a Euclidean domain for a field $F$ with $d(f(x))=\deg f(x)$.
Problem 78 (The Gaussian Integers Is A Euclidean Domain)
Prove that $Z[i]$ is a Euclidean domain, using the function $d(a+bi)=a^2+b^2$.
For more problems, see AllProblems