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IrreduciblesPolynomialsBehaveLikePrimeNumbersOverTheIntegers

We know that if we were in $\mathbb{Q}$ instead of $\mathbb{Z}$, then the problems we have worked on before make this trivial. So what connection is there between reducibility over $\mathbb{Z}$ and $\mathbb{Q}$. We've work on problems that give exactly what we need.
Definition (Associates Irreducibles Primes)

Let $D$ be an integral domain. All elements below are elements of $D$.

  • We say that two elements $a$ and $b$ are associates if $a=ub$ for some unit $u$.
  • If $a$ is nonzero and not a unit, then we say $a$ is an irreducible if whenever $a=bc$, then either $b$ or $c$ is a unit.
  • If $a$ is nonzero and not a unit, then we say $a$ is a prime if $a\mid bc$ implies $a\mid b$ or $a\mid c$.

Problem 62 (Prime Implies Irreducible)

In an integral domain, prove that if an element $a$ is prime, then $a$ must be irreducible.

Problem 72 (Irreducible Implies Prime in a PID)

We've already shown that every prime is irreducible. The converse is not always true. However, suppose that $D$ is a principle ideal domain. Prove that if an element $a$ is irreducible then it is prime.

Definition (Associates Irreducibles Primes)

Let $D$ be an integral domain. All elements below are elements of $D$.

  • We say that two elements $a$ and $b$ are associates if $a=ub$ for some unit $u$.
  • If $a$ is nonzero and not a unit, then we say $a$ is an irreducible if whenever $a=bc$, then either $b$ or $c$ is a unit.
  • If $a$ is nonzero and not a unit, then we say $a$ is a prime if $a\mid bc$ implies $a\mid b$ or $a\mid c$.

Problem 63(Irreducible Polynomials Have A Zero In Some Extension Field)

Let $F$ be a field and suppose that $p(x)$ is an irreducible polynomial over $F$.

  1. Prove that $E=F[x]/\left<p(x)\right>$ is an extension field of $F$. (Note: An extension field of $F$ is a field $E$ such that $E$ contains a subfield isomorphic to $F$.)
  2. Show that the element $x+\left<p(x)\right>\in E$ is a zero of $p(x)$ in $E$.

For more problems, see AllProblems