abstract algebra questions
kazu
test
1)first question
Let R be the field of real numbers and R[x] the ring of polynomials in
x with coefficients in R. Which of the following subsets of R[x] is a
subring of R[x]?
I. ALl polynomials whose coefficient of x is zero
II. All polynomials whose degree is an even integer, together with the
zero polynomial
III. all polynomials whose coefficients are rational numbers
(A) I only
(B) II only
(C) I and III only
(D)II and III only
(E)I,II, and III
2)second question
A cyclic group of order 15 has an element x such that the set
{x^3,x^5,x^9} has exactly two elements. The number of elements in the
set {x^(13n): n is a positive integer} is
(A) 3
(B) 5
(C) 8
(D) 15
(E) infinite
Stephen J. Herschkorn
>
>1)first question
>Let R be the field of real numbers and R[x] the ring of polynomials in
>x with coefficients in R. Which of the following subsets of R[x] is a
>subring of R[x]?
>
>I. ALl polynomials whose coefficient of x is zero
>II. All polynomials whose degree is an even integer, together with the
>zero polynomial
>III. all polynomials whose coefficients are rational numbers
>
>(A) I only
>(B) II only
>(C) I and III only
>(D)II and III only
>(E)I,II, and III
>
(E). Think about the definition of a ring.
>
>2)second question
>A cyclic group of order 15 has an element x such that the set
>{x3,x5,x9} has exactly two elements. The number of elements in the
>set {x^(13n): n is a positive integer} is
>(A) 3
>(B) 5
>(C) 8
>(D) 15
>(E) infinite
>
(A). Just work in Z_15. Unless I made a mistake, x = 5 there.
--
Stephen J. Herschkorn hers...@rutcor.rutgers.edu
Ignacio Larrosa Cañestro
mensaje|n3F523BF...@rutcor.rutgers.edu:
>> 1)first question
>> Let R be the field of real numbers and R[x] the ring of polynomials
>> in x with coefficients in R. Which of the following subsets of R[x]
>> is a subring of R[x]?
>>
>> I. ALl polynomials whose coefficient of x is zero
>> II. All polynomials whose degree is an even integer, together with
>> the zero polynomial
>> III. all polynomials whose coefficients are rational numbers
>>
>> (A) I only
>> (B) II only
>> (C) I and III only
>> (D)II and III only
>> (E)I,II, and III
>>
>
> (E). Think about the definition of a ring.
>
>>
The difference of two polynomials of degree k evenwith the same doefficient
of x^k an distinct coeeficients of x^(k-1) is a polynomial of odd degree,
among other cases.
The correct answer is (C).
--
Best regards,
Ignacio Larrosa Cañestro
A Coruña (España)
ilarrosaQUIT...@mundo-r.com
Peter L. Montgomery
>Can you solve the following questions? they are from Math GRE subject
>test
>
>1)first question
>Let R be the field of real numbers and R[x] the ring of polynomials in
>x with coefficients in R. Which of the following subsets of R[x] is a
>subring of R[x]?
>
>I. ALl polynomials whose coefficient of x is zero
>II. All polynomials whose degree is an even integer, together with the
>zero polynomial
>III. all polynomials whose coefficients are rational numbers
>
>(A) I only
>(B) II only
>(C) I and III only
>(D)II and III only
>(E)I,II, and III
>
are closed under addition and multiplication. In I, for example, if you
add the polynomials 3*x^3 + 7*x^2 - pi and sqrt(3)*x^4 - 6.456*x^2 + 1
(both of which has a zero coefficient of x^1), will the sum have
a zero coefficient of x^1? What about the product?
>2)second question
>A cyclic group of order 15 has an element x such that the set
>{x^3,x^5,x^9} has exactly two elements. The number of elements in the
>set {x^(13n): n is a positive integer} is
>(A) 3
>(B) 5
>(C) 8
>(D) 15
>(E) infinite
Since the set has exactly two elements, either x^3 = x^5 or
x^3 = x^9 or x^5 = x^9 holds, but not all three are equal.
In each case figure out the order of x and the order of x^13.
Or you can start by noting that the order of x must divide 15,
hence be 1, 3, 5, or 15. Which of these orders is/are
consistent with the hypothesis? Ask yourself how many
different values {x^13, x^26, x^39, ...} will have.
For these test questions, where there is no answer
"(F) Cannot be determined from the hypothesis", once you
find one order for x consistent with the hypothesis
you can count the corresponding {x^(13n)} and select
your answer, not worrying about other values for the order of x.
--
Wanted: Experts at choosing the best of 100+ applicants for a position.
Register as a California voter by September 22, and vote on October 7.
Peter-Lawren...@cwi.nl Home: San Rafael, California
Microsoft Research and CWI
Stephen J. Herschkorn
>
>Oh thanks a lot. if you have time, can you let me know
>how to solve this question?
>
>What is the greatest integer that divides p^4-1 for
>every prime number p greater than 5?
>(A) 12 (B) 30 (C) 48 (D) 120 (E) 240
>
>
I'm a little rusty on this stuff - in fact, others have pointed out that
my answer to the first question below is incorrect; the correct answer
is C, since the difference of two polynomials of even degree may have
odd degree.
I would approach your question above this way, but there may be a faster
way:
p^4 - 1 = (p-1) (p^3 + p^2 + p +1). p-1 is even. By doing the
arithmetic mod 4, one can determine that the second factor is divisible
by 4. Hence, 8 always divides the number, ruling out answers (A) and (B).
7^4 -1 = 6 x 500, which is not divisible by 16, hence (E) is ruled
out. If p = 1 mod 5, then 5 | (p-1); else, you may check that 5
divides the second factor. Hence, I get (D) to be the answer. Is this
correct?
I have posted this to sci.math, in case others have a faster way.
>--- "Stephen J. Herschkorn"
><hers...@rutcor.rutgers.edu> wrote:
>
>
>>> >
>>> >
>>> >1)first question
>>> >Let R be the field of real numbers and R[x] the
>>
>>
>> ring of polynomials in
>
>
>>> >x with coefficients in R. Which of the following
>>
>>
>> subsets of R[x] is a
>
>
>>> >subring of R[x]?
>>> >
>>> >I. ALl polynomials whose coefficient of x is zero
>>> >II. All polynomials whose degree is an even
>>
>>
>> integer, together with the
>
>
>>> >zero polynomial
>>> >III. all polynomials whose coefficients are
>>
>>
>> rational numbers
>
>
>>> >
>>> >(A) I only
>>> >(B) II only
>>> >(C) I and III only
>>> >(D)II and III only
>>> >(E)I,II, and III
>>> >
>>
>>
>>
>> (E). Think about the definition of a ring.
>
--
Stephen J. Herschkorn hers...@rutcor.rutgers.edu
db
> >
> >
> >Oh thanks a lot. if you have time, can you let me know
> >how to solve this question?
> >
> >What is the greatest integer that divides p^4-1 for
> >every prime number p greater than 5?
> >(A) 12 (B) 30 (C) 48 (D) 120 (E) 240
> >
> >
> I'm a little rusty on this stuff - in fact, others have pointed out that
> my answer to the first question below is incorrect; the correct answer
> is C, since the difference of two polynomials of even degree may have
> odd degree.
>
> I would approach your question above this way, but there may be a faster
> way:
> p^4 - 1 = (p-1) (p^3 + p^2 + p +1). p-1 is even. By doing the
> arithmetic mod 4, one can determine that the second factor is divisible
> by 4. Hence, 8 always divides the number, ruling out answers (A) and (B).
> 7^4 -1 = 6 x 500, which is not divisible by 16, hence (E) is ruled
> out. If p = 1 mod 5, then 5 | (p-1); else, you may check that 5
> divides the second factor. Hence, I get (D) to be the answer. Is this
> correct?
>
I'd factor p^4-1 as (p^2-1)(p^2+1) = (p-1)(p+1)(p^2+1)
Then you can see that all 3 factors are divisible by 2,
plus one of (p-1) or (p+1) must be divisible by 4, so
p^4-1 must be divisible by 16.
Also, one of (p-1) or (p+1) must be divisible by 3.
By letting p = 1,2,3,4 (mod 5) you can see that one of
the 3 factors must be divisible by 5.
I believe the answer is 16 * 3 * 5 = 240 which is (E).
David
Bill Dubuque
>kazu...@yahoo.com (kazu)
>>
>> What is the greatest integer that divides p^4-1 for
>> every prime number p greater than 5?
>> (A) 12 (B) 30 (C) 48 (D) 120 (E) 240
>
>
> Then you can see that all 3 factors are divisible by 2,
> plus one of (p-1) or (p+1) must be divisible by 4, so
> p^4-1 must be divisible by 16.
>
> Also, one of (p-1) or (p+1) must be divisible by 3.
>
> By letting p = 1,2,3,4 (mod 5) you can see that one of
> the 3 factors must be divisible by 5.
>
> I believe the answer is 16 * 3 * 5 = 240 which is (E).
More generally use the Carmichael lambda function (see below).
y(240) = y(2^4 3 5) = lcm(2^2,2,4) = 4
=> x^4 = 1 (mod 240) for x coprime to 2,3,5
------------
Chris Nolen <cmn...@ualr.edu> wrote:
:
: In my number theory homework I showed that if n is an integer
: not divisible by 2 or 3 then n^2 + 23 must be divisible by 24.
: Is there a general theory that relates to this?
You showed that a^2 = 1 (mod 24) if (a,24) = 1. More generally,
k = y(m) => a^k = 1 (mod m) if (a, m) = 1
where y(m) is the Carmichael lambda function, defined as is the
Euler phi function on prime powers, but combined via lcm vs. times:
y(1) = y(2) = 1, y(4) = 2,
y(2^e) = 2^(e-2) for e > 2
y(p^e) = p^(e-1) (p-1) for odd prime p
y(p^i q^j ...) = lcm( y(p^i), y(q^j), ...)
In your case y(24) = y(2^3*3) = lcm(2,2) = 2, as you showed,
and it's easy to see that 24 is the largest m with y(m) = 2.
In group theoretic language one says the y(m) is the exponent
of the unit group of Z/(m), i.e. the smallest k such that
x^k = 1 for all elts x of the group.
= http://groups.google.com/groups?threadm=y8zhekq6s2d.fsf%40nestle.ai.mit.edu
-Bill Dubuque