Detached mathematics Polynomials

 25 June 05:19   In this area we attending at the polynomial in some capricious ring with identity. What is absorbing is that belief polynomials over some capricious ring with character acts actual abundant like numbers; the aforementioned rules generally are obeyed by both.

    A polynomial over some capricious ring with character R is an announcement in the form

    : $sum\_^n\; a\_jx^j;\; forall\; a\_j\; in\; R,\; a\_n$
ot = 0

    and n ∈ N, and x is some general (not a variable)

    Given the first nonzero appellation in the polynomial, ie the appellation a_nxⁿ above:

    In the above, if a_i=0 for all i, the polynomial is the aught polynomial.

    Let R[x] be the set of all polynomials of all degrees. Acutely R is bankrupt beneath accession and multiplication (although in a non-straightforward way), and appropriately we accept that R[x] is itself a capricious ring with identity.

    Assume now R is a acreage F; we do this so we can ascertain some advantageous accomplishments on polynomials

    Firstly anamnesis the analysis algorithm for numbers, that anniversary amount can be addle into the form

    : n = qk+r

    where q is the caliber and r the butt and r<n.

    Now, back we accept that F is a field, we can do something agnate with the polynomials over F, F[x].

    If f(x), g(x) ∈ F[x], with g(x) nonzero:

    : f(x) = q(x)g(x)+r(x)

    Again, q(x) is accepted as the caliber polynomial and r(x) the butt polynomial. Furthermore, we accept the amount of r(x) ≤ amount of f(x)

    We accomplish capacity by polynomial continued division. For brevity we omit the x^k terms. Heres an example. We bisect x³+x+2 by x-1. First write:

    :$$

    egin

     & & & & & & & 1 & -1 & | & 1 & 0 & 1 & 2 & \end

    

    Note we abode a 0 in any polynomial appellation not present. Now x³/x = x², so we abode a 1 in the additional cavalcade to get

    :$$

    egin

     & & & & 1 & & & 1 & -1 & | & 1 & 0 & 1 & 2 & \end

    

    Multiply x² throughout the divisor x-1 to get x³-x², which is (1 -1), so address this beneath like the following:

    :$$

    egin

     & & & & 1 & & & 1 & -1 & | & 1 & 0 & 1 & 2 & & & | & 1 &-1 & & & \end

    

    Now decrease (1 0) and (1 -1), bead the third 1 to get:

    :$$

    egin

     & & & & 1 & & & 1 & -1 & | & 1 & 0 & 1 & 2 & & & | & 1 & -1 & & & & & | & & 1& 1 & & \end

    

    Now repeat, but bisect by x² now (since we accept subtracted and gotten (1 1) - x² + x), and abide in the aforementioned fashion, to get:

    :$$

    egin

     & & & & 1 & 1 & 2 & 1 & -1 & | & 1 & 0 & 1 & 2 & & & | & 1 & -1& & & & & | & & 1 & 1 & & & & | & & 1 & -1 & & & & | & & & 2 & 2 & & & | & & & 2 & -2 & & & | & & & & 4 & \end

    

    So the caliber is x²+x+2, and the butt is 4.

    Now we accept a alive analysis algorithm for polynomials, the Euclidean algorithm, and appropriately the gcd of two polynomials can readily be found.

    Lets use a agnate archetype above: what is gcd(x³+x+2, x-1)?

    Weve apparent already aloft that

    x³+x+2=(x²+x+2)(x-1) + 4

    Proceeding in the accustomed appearance in the Euclidean algorithm

    :x³+x+2=(x²+x+2)(x-1) + 4

    :gcd(x³+x+2, x-1) = gcd(x-1,4)

    and the greatest accepted divisor of any monomial and an accumulation is acutely 1, so x³+x+2 and x-1 are coprime.

    For a additional example, accede gcd(x²-1,x²+2x+1)

    :x²+2x+1=(x²-1):x²-1=(2x+2):2x+2=-(x+1)

    Since factors of -1 create no difference, gcd(x²-1,x²+2x+1) is -(x+1)

    Weve apparent that x³+x+2 and x-1 are coprime; they accept no factors in common. So, are we able to actuate prime polynomials? Absolutely we can - depending on the acreage that the polynomial lies in. We alarm these irreducibles instead of primes.

    Take p(x)=x³ + x² + 2 over Z₃. Now we can agency this polynomial if it has a basis - from the agency assumption (which aswell holds for polynomials over any capricious ring with identity) p(k)=0 agency k is a root. So, lets attending at the following:

    Since were in Z₃, we luckily alone charge to analysis three values

    p(0)=2

    p(1)=1

    p(2)=2

    So we accept p(x) accepting no roots - it is irreducible (prime).

    Now beam an absorbing fact. Yield the exact aforementioned polynomial but instead over Z₂. The polynomial then is agnate to

    : x³+x²+0

    and appropriately has basis p(0)=0 and appropriately is reducible but over Z₂

    So the reducibility of the polynomial depends on the acreage it is in.

    The accepted action to appearance a polynomial is irreducible is:

    effectively a affidavit by cases.

    For example, accede the polynomial q(x)=x⁴+x+2 in Z₃. To prove it is irreducible, beam that q(x) could be factorized in the afterward ways:

    # linear, irreducible cubic

    # linear, linear, irreducible quadratic

    # linear, linear, linear, linear

    # irreducible quadratic, irreducible quadratic

    So we can prove 1, 2, 3 by assuming it has no beeline factors. 4 is a little added difficult. Let us advance to appearance it has no beeline factors:

    Observe

    : q(0)=2

    : q(1)=1

    : q(2)=2

    So q has no beeline factors. Now, we charge to appearance that q is not the artefact of two irreducible quadratics.

    In Z₃, we accept the quadratics

    :

    We can analyze the irreducible quadratics calmly by inspection. We then obtain

    :

    If we can appearance that neither of these polynomials bisect q(x)=x⁴+x+2, we accept apparent q(x) is irreducible.

    Let us try x²+1 first.

    :

     & & & & & & 1 & 0 & 2 1 & 0 & 1 & | & 1 & 0 & 0 & 1 & 2 & & & & 1 & 0 & 1 & & & & & & & & 2 & 1 & 2 & & & & & & 2 & 0 & 2 & & & & & & & 1 & 0 \end

    We accept a remainder, so x²+1 doesnt bisect q.

    On adding the additional polynomials, we all get a remainder. (Verify this for yourself as practice).

    So q(x) is irreducible in Z₃.

    Since we accept a alive polynomial analysis and agency theorem, and that polynomials arise to actor the behaviour of the integers - can we analytic ascertain some array of modular addition with polynomials?

    We can indeed. If we accept a acreage Z_p[x] and we ambition to acquisition all the remainders (remember, these remainders are polynomials) on adding by some polynomial m(x), we can do so by polynomial continued division.

    If m(x) is irreducible, then the set of remainders as aloft forms a field.