Results 1 to 10 of 10

Math Help - Reducible Polynomials

  1. #1
    Member
    Joined
    Dec 2009
    Posts
    225

    Reducible Polynomials

    Let p be a prime. How do we show that the number of reducible polynomials over \mathbb{Z}_p of the form x^2+ax+b is p(p+1)/2?

    I just know that \mathbb{Z}_p is a field, and a polynomial of the form x^2+ax+b is of degree 2, so there is a theorem which says a polynomial of degree 2 or 3 is reducible over a field if and only if it has a root in that field.

    So how can I show the total number of polynomials x^2+ax+b which have a zero in \mathbb{Z}_p is p(p+1)/2? Any help would be appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by demode View Post
    Let p be a prime. How do we show that the number of reducible polynomials over \mathbb{Z}_p of the form x^2+ax+b is p(p+1)/2?

    I just know that \mathbb{Z}_p is a field, and a polynomial of the form x^2+ax+b is of degree 2, so there is a theorem which says a polynomial of degree 2 or 3 is reducible over a field if and only if it has a root in that field.

    So how can I show the total number of polynomials x^2+ax+b which have a zero in \mathbb{Z}_p is p(p+1)/2? Any help would be appreciated.

    A polynomial x^2+ax+b is reducible iff x^2+ax+b=(x-\alpha)^2\,\,or\,\,x^2+ax+b=(x-\alpha)(x-\beta)\,,\,\alpha\neq \beta .

    Of the first kind are there p different pol's, and of the second kind are there \frac{p(p-1)}{2} different pol's (why do we divide by 2?)

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2009
    Posts
    225
    Quote Originally Posted by tonio View Post
    A polynomial x^2+ax+b is reducible iff x^2+ax+b=(x-\alpha)^2\,\,or\,\,x^2+ax+b=(x-\alpha)(x-\beta)\,,\,\alpha\neq \beta .

    Of the first kind are there p different pol's, and of the second kind are there \frac{p(p-1)}{2} different pol's (why do we divide by 2?)

    Tonio
    Could you please explain why you wrote that expression for the kind (x-\alpha)(x-\beta)? For the first kind there are p different polynomials because it is in \mathbb{Z}_p... but in for the second one I don't know why you wrote that expression and divided by 2...

    but it is correct because p + \frac{p(p-1)}{2} = \frac{p(p+1)}{2}.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by demode View Post
    Could you please explain why you wrote that expression for the kind (x-\alpha)(x-\beta)? For the first kind there are p different polynomials because it is in \mathbb{Z}_p... but in for the second one I don't know why you wrote that expression and divided by 2...

    but it is correct because p + \frac{p(p-1)}{2} = \frac{p(p+1)}{2}.
    I wrote the expression for the second kind as I did because a quadratic pol. may have two different roots, and in the calculations I divided by two since (x-\alpha)(x-\beta)=(x-\beta)(x-\alpha) ...

    Tonio
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Dec 2009
    Posts
    225
    The question further asks: "Determine the number of reducible quadratic polynomials over \mathbb{Z}_p".

    I think again we need to consider the number of distinct expressions of the form (x-c)(x-d).

    I know that quadratic polynomials have the form ax^2 + bx + c. So this time we need to use the quadratic formula for the roots:

    \frac{-b \pm \sqrt {b^2-4ac}}{2a}

    but how could I find the number of reducible polynomials from this formula?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by demode View Post
    The question further asks: "Determine the number of reducible quadratic polynomials over \mathbb{Z}_p".

    I think again we need to consider the number of distinct expressions of the form (x-c)(x-d).

    I know that quadratic polynomials have the form ax^2 + bx + c. So this time we need to use the quadratic formula for the roots:

    \frac{-b \pm \sqrt {b^2-4ac}}{2a}

    but how could I find the number of reducible polynomials from this formula?


    You don't need that formula. Again, if a quadratic is red. then ax^2+bx+c=a(x-\alpha)^2\,\,or\,\, =a(x-\alpha)(x-\beta) , and you make calculations very simmilar to the ones in the monic case.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Dec 2009
    Posts
    225
    Quote Originally Posted by tonio View Post
    You don't need that formula. Again, if a quadratic is red. then ax^2+bx+c=a(x-\alpha)^2\,\,or\,\, =a(x-\alpha)(x-\beta) , and you make calculations very simmilar to the ones in the monic case.

    Tonio
    So for the kind ax^2+bx+c=a(x-\alpha)^2 there are p different polynomials. Is this correct? or do I need to add another p to this?

    and for the kind a(x-\alpha)(x-\beta) there are \frac{p(p-1)}{2} + p = \frac{p(p+1)}{2} polynomials. I added the p because "a" can be a root.

    So in total \frac{p(p+1)}{2} + p polynomials?

    P.S. Isn't it also possible for the "a" in the first type to be different from the "a" in the second type?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Dec 2009
    Posts
    225
    a(x-\alpha)^2 or a(x-\alpha)(x-\beta).

    of the first kind there are p+p=2p polynomials. And of the second kind there are \frac{p(p-1)}{2} + p = \frac{p(p+1)}{2} polynomials.

    So there are 2p + \frac{p(p+1)}{2} = \frac{p(p+5)}{2} in total.

    Is this alright?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by demode View Post
    a(x-\alpha)^2 or a(x-\alpha)(x-\beta).

    of the first kind there are p+p=2p polynomials.


    No...the quadratic coefficient a\in\mathbb{Z}_p cannot be zero, so for any choice of \alpha\in\mathbb{Z}_p there are p-2 choices for a

    (since 0 cannot be and you ALREADY counted a=1!).

    Take it from here.

    Tonio


    And of the second kind there are \frac{p(p-1)}{2} + p = \frac{p(p+1)}{2} polynomials.

    So there are 2p + \frac{p(p+1)}{2} = \frac{p(p+5)}{2} in total.

    Is this alright?
    .
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Dec 2009
    Posts
    225
    Oops, yes the quadratic coefficient can't be zero!

    Previously we have shown that the number of reducible polynomials over \mathbb{Z}_p of the form x^2+ax+b are p(p+1)/2. So can we just make use of this result and deduce that the number of reducible quadratic polynomials ax^2+bx+c is (p-2) \frac{p(p+1)}{2}?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Reducible
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: November 10th 2010, 07:35 PM
  2. [SOLVED] Reducible Equations
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: July 14th 2010, 04:56 AM
  3. Reducible problem. Help!
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: April 18th 2010, 03:33 AM
  4. reducible and irreducible
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 21st 2010, 10:33 PM
  5. R[x] - Reducible?
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: October 18th 2009, 09:27 PM

Search Tags


/mathhelpforum @mathhelpforum