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:ain R},} where {[a]_sim={b:asim b}} for {sim} the equivalence relation defined by {asim b} iff {a-bin I.}

Since {sim} is an equivalence relation, we have that {[a]_sim=[b]_sim} if {asim b} and {[a]_simcap[b]_sim=emptyset} if {anotsim 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 {pin{mathbb F}[x],} i.e., given any polynomial {qin{mathbb F}[x],} {[q]_sim=0} iff {p|q} and, more generally, {[q]_sim=[r]_sim} (or, equivalently, {qsim r} or, equivalently, {rin[q]_sim}) iff {p|(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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