# Field extension, extension field, simple field

Discussion in 'Undergraduate Math' started by winshum, Feb 7, 2007.

1. ### winshumGuest

I just start learning these objects. I have read Hungerford and Fraleigh. It seems that they have different approaching introducing these concepts
and this make me confusing. Here i want to ask some questions. Any hints or underlying principles are enough. Thanks very much

1. C/R is a field extension, This is not a quotient ring, but what does it "look like"?

2. i know Q(i)={a+bi}, Is Q(pi) also be the set {a+bpi} ?
why Q(3+i)=Q(1-i) ? C=R(a+bi) ?

3. why is Q(pi) isomorphic to Q(x) ?
Is Q[pi] isomorphic to Q[x] ?
4. Is Q an extension field of Z_2 ?
5. C is simple extension of R, Q(i) is simple extension of Q, are there any more examples ?

Thanks every much.

winshum, Feb 7, 2007

2. ### Arturo MagidinGuest

Suggestion: Mathforum screws up the line breaks (doesn't provide
any). Your lines run off the screen. Use a hard carriage return rather
than relying on their on-screen editor.
Huh? R is a field. C is a bigger field that contains R. Looks like a
field with a smaller field inside it.
Please use your shift key, especially when you have square roots of -1
floating around, when using the first person pronoun.
This is extremely poor and sloppy notation. You mean

Q(i) = {a + bi | a and b are rationals}

(otherwise, you are saying that Q(i) is a set that contains one and
only one element).

What is "pi"? Is it p times the square root of -1, or is it pi, the
transcendental number that is the ratio of circumpherence of a circle
to its diameter?

If the latter, NO, it is not the set of all expressions of the form
a+b*(pi) with a and b rationals. In the case of Q(i), you can express
any powers of i greater than 1 in terms of rationals and i, but you
cannot do so with pi. pi^2 cannot be expressed as a+b*pi with a and b
rationals; if it could you would have that pi is the root of the
polynomial x^2 - bx -a, which would mean that pi is algebraic; we know
it is transcendnetal.
Q(3+i) = Q(1-i) because both fields are equal to Q(i), the smallest
subfield of the rational numbers that contains Q and i. Alternatively,
1-i is certainly in Q(3+i), as 4 - 1*(3+i), so Q(1-i) is contained in
Q(3+i). And 3+i is in Q(1-i), since 3+i = 4 - 1*(1-i), giving the
reverse inclusion.

As for why the complex numbers are equal to R(a+bi) whenever a and b
are real numbers AND b is not zero, same reason as above. Notice that
C = R(i).

The underlying reason is that it is because pi is transcendental. If
you map Q(x) to R by mapping the rationals to themselves and x to pi,
the only thing that maps to 0 is 0, and the image is called Q(pi).
Yes. Same reason: map Q[x] to R by mapping Q to itself identically,
and x to pi.
Not just no, but "hell no." If you cannot figure it out, you are in
much deeper trouble than even your questions above suggest.
Loads of them. Pick your favorite real number that is not a
rational. Call it r. Then Q(r) is a simple extension of Q.

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================

Arturo Magidin
magidin-at-member-ams-org

Arturo Magidin, Feb 7, 2007