305 -Extension fields revisited (3)

April 15, 2009

1. Isomorphisms

We return here to the quotient ring construction. Recall that if {R} is a commutative ring with identity and {I} is an ideal of {R,} then {R/I} is also a commutative ring with identity. Here, {R/I=\{[a]_\sim:a\in R\},} where {[a]_\sim=\{b:a\sim b\}} for {\sim} the equivalence relation defined by {a\sim b} iff {a-b\in I.}

Since {\sim} is an equivalence relation, we have that {[a]_\sim=[b]_\sim} if {a\sim b} and {[a]_\sim\cap[b]_\sim=\emptyset} if {a\not\sim b.} In particular, any two classes are either the same or else they are disjoint.

In case {R={\mathbb F}[x]} for some field {{\mathbb F},} then {I} is principal, so {I=(p)} for some {p\in{\mathbb F}[x],} i.e., given any polynomial {q\in{\mathbb F}[x],} {[q]_\sim=0} iff {p\mid q} and, more generally, {[q]_\sim=[r]_\sim} (or, equivalently, {q\sim r} or, equivalently, {r\in[q]_\sim}) iff {p\mid (q-r).}

In this case, {{\mathbb F}[x]/(p)} contains zero divisors if {p} is nonconstant but not irreducible.

If {p} is 0, {{\mathbb F}[x]/(p)\cong{\mathbb F}.}

If {p} is constant but nonzero, then {{\mathbb F}[x]/(p)\cong{0}.}

Finally, we want to examine what happens when {p} is irreducible. From now on suppose that this is the case.

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305 -Extension fields revisited (2)

April 15, 2009

Most of our work from now on depends on the following simple, but very useful observation:

Theorem 1 Let {{\mathbb F}:{\mathbb K}} be a field extension. Then {{\mathbb F}} with its usual addition is a vector space over {{\mathbb K},} where multiplication of elements of {{\mathbb F}} by elements of {{\mathbb K}} is just the usual product of {{\mathbb F}}.

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