A field is an integral domain
In Dummit and Foote Chapter 13: Field Theory, the authors give several examples of field extensions on page 515 - see attached.
In example (3) we read (see attached)
" (3) Take and , irreducible over by Eisenstein's Criterion, for example"
Now Eisenstein's Criterion (see other attachment - Proposition 13 and Corollary14) require the polynomial to be in R[x] where R s an integral domain.
In example (3) on page 515 of D&F we are dealing with a field, specifically .
My problem is, then, how does Eisenstein's Criterion apply?
Can anyone please clarify this situation for me?
Peter
Hi Idea,
Post by Deveno is post #8 at http:///linear-abstract-algebra-14/f...19-a-6555.html
Part of Deveno's post reads as follows:
1. When applying Eisenstein to rational polynomials, we typically turn them into integral polynomials by multiplying by the lcm of the denominators of the coefficients. This turns our polynomial into one over an integral domain, the integers. In the integers, we HAVE prime ideals, namely the ideals generated by a prime integer.
Peter
If R is a ring, then $R[x]$ , the ring of polynomials in x with coefficients in R, consists of all formal sums $\displaystyle \sum_{i=0}^\infty a_i x^i$ , where $a_i = 0$ for all but finitely many values of i.
If $\displaystyle \sum_{i=0}^\infty a_i x^i$ is a nonzero polynomial, the degree is the largest $n \ge 0$ such that $a_n \ne 0$ . The zero polynomial has degree $-\infty$ .
$R[x]$ becomes a ring with the usual operations of polynomial addition and multiplication.
If F is a field, the units in $F[x]$ are exactly the nonzero elements of F.
If $F[x]$ is a field, $f(x), g(x) \in F[x]$ , and $g(x) \ne 0$ , there are unique polynomials $q(x), r(x) \in F[x]$ such that
$$f(x) = q(x)\cdot g(x) + r(x), \quad\hbox{where}\quad \deg r < \deg g.$$ \item{$\bullet$} Let F be a field and let $f(x) \in F[x]$. c is a root of $f(x)$ in F if and only if $x - c \mid f(x)$.
Let R be an integral domain. An element $x \in R$ is irreducible if $x \ne 0$ , x is not a unit, and if $x = yz$ implies either y is a unit or z is a unit.
Let R be an integral domain. An element $x \in R$ is prime if $x \ne 0$ , x is not a unit, and $x \mid yz$ implies $x \mid y$ or $x \mid z$ .
In an integral domain, primes are irreducible.
Let F be a field. If $f(x), g(x) \in F[x]$ are not both zero, then $f(x)$ and $g(x)$ have a greatest common divisor which is unique up to multiplication by units (elements of F).
Let F be a field, let $f(x), g(x) \in F[x]$ , and let $d(x)$ be a greatest common divisor of $f(x)$ and $g(x)$ . There are polynomials $u(x), v(x) \in F[x]$ such that
$$d(x) = u(x)f(x) + v(x)g(x).$$
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