Re: Infinite integral domain and zero polynomial
- From: quasi <quasi@xxxxxxxx>
- Date: Sun, 25 Nov 2007 14:17:51 -0500
On Sun, 25 Nov 2007 13:36:22 EST, jane <jane1806@xxxxxxxxxx> wrote:
On Sun, 25 Nov 2007 11:15:00 -0500, quasi
<quasi@xxxxxxxx> wrote:
<jane1806@xxxxxxxxxx> wrote:
On Sun, 25 Nov 2007 08:11:30 EST, jane
the elements of D.
Let D be an integral domain and f in D[x].
How can i prove that
(f = 0 <=> f(a) = 0 for all a in D)
if and only if D is infinite integral domain?
First suppose D is finite. Let {a_1, ..., a_n} be
Then the polynomial f(x) = (x - a_1)* ... *(x - a_n)is a nonzero
element of D[x] (the leading term is x^n), but ftakes the value 0 for
all elements of D.of D[x]. Then
Next suppose D is infinite.
We need a preliminary lemma ...
lemma [the Factor Theorem]:
Let D be a commutative ring and let f be an element
D[x]
f(a) = 0 iff f(x) = (x - a)*g(x) for some g in
typically found in a
The proof is essentially the same as the proof
precalculus text. Note, as indicated above, theresult holds for any
commutative ring D, even if D is not an integraldomain. You should
work out the proof -- it's simple and short.infinitely many
Back to the proof ...
Let f be a nonzero polynomial of least degree with
roots in D. Suppose a in D is root of f. Then f(a) =0, hence, by the
Factor Theorem, f(x) = (x - a)*g(x) for some g inD[x] But every root
of f, except possibly for a, is also a root of g,since if b is a root
of f, b not equal to a, thenzero divisors.
0 = f(b) = (b - a)*g(b) => g(b) = 0
Note -- this last claim uses the fact that D has no
Everywhere else in the proof, D was regarded simplyas a commutative
ring.D[x], and g also
Continuing with the proof, g is a nonzero element of
has infinitely many roots in D, contradiction, sincethe degree of g
is less than the degree of f.infinitely many roots in
It follows that the only element of D[x] with
D is the zero polynomial, as was to be shown.
To jane:
You really should try problems before just asking for
help.
The only reason I gave a solution is because another
poster had
already provided a full solution (and an awkward one
at that). Had a
solution not already been posted, I would have given
this very simple
hint, which should have been sufficient to get you
started ...
Hint: The Factor Theorem.
But really, you should try problems -- it seems you
are getting a
false education by proxy, posting textbook exercises,
and getting the
solutions spoon fed.
How do I know you didn't try this problem? That's
obvious. To see
that, make believe you were just starting on the
problem, ok? Let's
suppose that after a few minutes of thought, you had
no idea how to
prove the claim. Fine -- but why would you not
consider trying special
cases.
For example D = Z_2. With such a small ring, you
would surely find the
polynomial x*(x-1) which is nonzero in D[x] but
equals zero for all
values of D. Progressing to Z_3, you would then find
x*(x-1)*(x-2). At
that point, the finite case resolves, right?
Next try some infinite D's. How about D = Z? Or
perhaps D = Q. Could a
nonzero univariate polynomial with rational
coefficients have
infinitely many roots? While you know in your heart
that infinitely
many roots shouldn't be possible, you might have some
trouble making
that rigorous. Of course, the Factor Theorem would
clinch it, and
thus, that's my would-be hint.
It's obvious you have talent. Why let yourself be
spoon fed?
Instead, if you get stuck, at least show your
progress on a problem,
and then specifically ask for hints, not solutions.
Now, in some cases, a problem is not solvable by
hints. For example,
perhaps there's some obscure trick which either you
see it or you
don't. In that case, the replier can make the
judgement that giving
hints is a waste of time -- both yours and theirs,
and just show you
the solution. But if the textbook is any good, most
problems have some
kernel of insight which is only truly internalized
when you solve the
problem yourself or at least struggle with it.
Thank you very much for the solution and advices, i really appreciate that.
What concerns my progress on the problem : i solved before the "only if" part and knew how to do the "if" part if you are working over the field. The only missing part is that i didn't come up with the idea that our integral domain can be just embedded into a field of fractions and that's it.
Ok, but if you look at my alternative proof, there no mention of
"field".
Maybe i should have written that in my 1'st post, but i'm not sure whether people need to read and figure out with my thoughts, which might be wrong and it might be somehow spoiling of the time.
Yes, in some cases, the work is too unfinished to post -- I understand
that. But at least state what parts you have proved and where you seem
to be stuck.
This is a coin with two sides in fact : i understand that some people think that i haven't even tried the problem myself as you probably did for example.
Yes, sorry -- I couldn't understand why you wouldn't have tried a few
special rings, just to get a feel for the problem.
I'll try to write my work, especially when i have some doubts about my ideas, whether they are wrong or not.
Not necessarily -- use your judgement. But rather than ask for full
solutions, specifically demand hints (and hope the hint providers
don't give away too much). If a hint doesn't work, ask for a different
hint, or an additional one. If you can't resolve it even with multiple
hints, perhaps ask for a simpler problem of the same kind. If at some
point, you really decide to give up, fine, the problem beat you (that
time), but at least you put up a good fight.
quasi
.
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