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Intuition for Quotient Ring in Polynomials

  1. Oct 23, 2012 #1
    I just had a discussion with someone who said he thought about quotient rings of polynomials as simply adjoining an element that is a root of the polynomial defining the ideal.

    For example, consider a field, F, and a polynomial, x-a, in F[x]. If we let (x-a) denote the ideal generated by x-a, then we can form the quotient ring F[x]/(x-a). He was saying we can think about this ring as F[a] (i.e. adjoining a (which is a root of x-a) to the original field F).

    This seems like a very useful intuition to have, but I am still struggling to see how it is true. Up until now, I've been thinking of the quotient ring simply in terms of a bunch of equivalence classes of polynomials, which is difficult to have an intuition for.
     
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  3. Oct 23, 2012 #2

    Erland

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    First, to choose a first degree polynomial x-a is not very interesting (although formally correct), since we only get F back: F[a]=F, since a in F (otherwise, x-a would not be a polynomial over F).

    Second, for this to work, we must use an irreducible polynomial over F, for otherwise, the quotient ring will not be a field.

    Suppose that we take the quotient ring F[x]/I, where I is the ideal generated by an irreducible polynomial p(x) over F with degree n.

    Then, two polynomials over F lie in the same equivalence class iff their difference is divisible by p(x). It follows that every equivalence class must contain a unique polynomial of degree < n. (Take any polynomial in the class and divide it with p(x). The remainder lies in the same class and has degree < n. If there are two such polynomials, their difference has degree < n and is divisble by p(x), so the difference is 0. Hence, this polynomial is unique.)

    So we can represent each equivalence class in a unique way by a polynomial over F of degree < n, and of course, every such polynomial lies in an equivalence class and is represented by this very class.

    The set of equivalence classes represented in this way by constant polynominals constitutes a subring of the quotient ring which is isomorphic to F, and this is therefore identified with F in a natural way. This means that we can substitute an element in the quotient ring for x in p(x) and get a new element in the quotient ring, so a polynomial over F can be considered as a polynomial over the quotient ring too. If we do this for the class [x], we see that p([x])=[p(x)])=[0]=0, so [x] ia zero pf p(x) over the quotient ring.
    If we rename [x] as a, we see that the quotient ring contains no proper subring that contains both F and a. Therefore, we can write the quitient ring as F(a).

    We must also prove that the quotient ring F(a) is a field. Let [q(x)] be a nonzero element in F(a). This means that the representative q(x) is not divisible by p(x) (otherwise [q(x)]=[0]=0). Since p(x) is irreducible over F (this is the only time we use this assumption) it follows from the Euclidean algorithm that 1 is a greatest common divisor of q(x) and p(x), and that there are polynomials b(x) and c(x) over F such that

    b(x)q(x)+c(x)p(x)=1, and hence, [b(x)][q(x)]=[1]=1, since [p(x)]=[0]=0.

    This means that [q(x)] has a multiplicative inverse in F(a): [q(x)]^(-1)=[b(x)].

    Since [q(x)] was an arbitrary nonzero element in F(a), F(a) is a field.

    It can also be proved that if K is any field which contains F, and an element b such that p(b)=0, then F(b), the smallest subfield of K which contains F and b, is isomorphic to the field F(a) constructed here, so in this sense, our extension is unique.
    (An isomorphism can easily be constructed by starting from the evaluation homomorphism for F[x] at b, and notice that this becomes well defined for the quotient field F(a) and indeed an isomorphism onto F(b).)
     
  4. Oct 24, 2012 #3

    mathwonk

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    by definition, forming a quotient ring means setting something that was previously non zero, equal to zero. hence if that something was a polynomial, then setting it equal to zero is the same as setting its variable equal to a root of that polynomial. yes??
     
  5. Oct 25, 2012 #4
    I followed you up to here:
    But then I lost you here:
    I don't understand what you mean by substituting an equivalence class (i.e. [x]) for x in p(x). I am thinking about the x's purely as formal variables. In my thinking, we could remove them and just use n-tuples with a special definition for multiplication. Could you elaborate on what is going on when you say p([x]) = [p(x)]. How do you get that?
     
  6. Oct 25, 2012 #5
    That makes sense. But I will need to think about this for a bit to see that it is equivalent to the definition in terms of cosets.
     
  7. Oct 26, 2012 #6

    Erland

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    Addition and multiplication in the quotient ring are defined by

    [p(x)]+[q(x)]=[(p+q)(x)] and [p(x)][q(x)]=[(pq)(x)] (*)

    It can be proved that this is well defined, that is, the sum and product of the equivalence classes do not depend upon which representatives we choose, and that the ring axioms are satisfied (so it is correct to call it "quotient ring").
    Since this ring contains the field F as a subring, every polynomial over F can also be considered as a polynomial over the quotient ring and can be evaluated at elements in this ring.

    You are right that variables in a polynomial can be considered as entirely formal, but polynomials over a ring R can also be evaluated at elements in R, formally we say that we apply the evaluation homomorhism at the element. This means that we substitute the element in R for x and compute the result. For example, if ##p(x)=x^2+4## over Z, then ##p(5)=5^2+4=29##.

    We then see that if we evaluate the polynomial x at the element [x], the result is [x] (for evaluation of x is the identity operator). If we instead evaluate the constant polynomial a, where a in F, at [x], then the result is a=[a], since it is a constant polynomial.

    If we now evaluate an arbitrary polynomial over F, say ##p(x)=a+bx+cx^2##, at [x], it follows from this and successive applications of the above definitions (*) that

    ##p([x])=a+b[x]+c[x]^2 = [a]+[x]+[c][x]^2=[a]+[bx]+[cx^2]=[a+bx+cx^2]=[p(x)]##.

    Thus p([x])=[p(x)], and this of course generalizes to all polynomials over F.
     
    Last edited: Oct 26, 2012
  8. Oct 26, 2012 #7

    mathwonk

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    some of the confusion may come from using the letter X for two different things.

    If we have any commutative ring R, and an element c of R, and X is a variable, there is an evaluation map

    R[X]-->R, sending f(X) to f(c). This is a ring map, i.e. takes sums to sums and products to products.


    Now suppose we choose a polynomial g(X) in R[X] and are looking to enlarge the ring R until it contains a root of g. We want g(X) to become zero when we replace X by the root.

    More generally if we have not just an "enlargement" of R, which is just an inclusion map of rings, but any ring map at all,

    R-->S, we get an induced map on polynomials R[X]-->S[X], by replacing the coefficients from R by their images in S.

    We want to find a ring map R-->S such that S contains a root of g. If we want to think of g as the same polynomial even though it is now in S[X], we will try to make the map R-->S injective. Then we can replace the images in S of the coefficients from R by the same coefficients from R, i.e. if R--S is injective, then R[X]-->S[X] is also injective.

    i.e. we want the chosen polynomial g(X) to go to zero under the composed map

    R[X]-->S[X]--S, where the right hand map is evaluation at c, some element of S.

    But the simplest map that takes g to zero, is the quotient map R[X]-->R[X]/(g).

    We want this to be the composed map above, i.e. we want S to be R[X]/(g).

    Now does that do everything we want? well, that IS an evaluation map, namely

    it is evaluation at X! i.e. the map H:R[X]-->R[X]/(g) takes X to the equivalence class {X},

    hence, it takes f(X) to {f(X)} which equals f({X}) since H is a ring map. Saying that f({X})

    = {f(X)} = H(f), says that the map H is evaluation at {X}.


    But {g(X)} = {0} = 0. by definition of S = R[X]/(g). so H(g) = g({X}) = 0. I.e. evaluation at {X} does take g to zero, so {X} is a root of g, in S.

    Now to make the map R-->S injective, it suffices that no element of R except zero, belong to the ideal (g), i.e. that no non zero element of R is a multiple of g. If we make R an integral domain, and take g to have degree one or more that will do it.


    Now one reason the argument may still be confusing is using X for the variable as well as for the equivalence class {X}.

    I.e. if we define S = R[X]/(g), then {X} is an element of S, hence not a variable over S.

    so strictly speaking g(X) is no longer a polynomial over S.

    Thus we really should use a new letter for the variable over S and write say S[T].

    Then {X} is an element of S = R[X]/(g) which is a root of the polynomial g(T) in S[T].

    But that is the same as saying it is a root of g(Z) where Z is any variable. I.e. being a root does not depend on the name of the variable.


    sorry this is so lengthy.
     
    Last edited: Oct 26, 2012
  9. Oct 26, 2012 #8

    mathwonk

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    shorter version: given g(X) of degree ≥ 1, in R[X] where R is a domain, instead of R[X] consider the isomorphic ring R[Y]. and

    let S = R[Y]/(g(Y)). Then we can consider R as a subring of S.

    hence R[X] is contained in S[X], and we can consider g(X) as a polynomial in S[X].

    Then the element {Y} of S is a root of g(X).

    I.e. g({Y}) in S, is the same element as {g(Y)}, because the map R[Y]-->S is a ring map.

    But {g(Y)} is zero by definition in S, and g({Y}) is the result of substituting {Y} into g(X) in place of X.

    Therefore {Y} is an element of S which is also a root of g(X).

    actually these are technical details, i already gave you the intuition you asked for in post #3.
     
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