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Math Help - polynomial ring

  1. #1
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    polynomial ring

    Let R be a ring, suppose that 2 non-zero functions f and g with:
    f = a_0 + a_1x +...+ a_mx^m belongs to R[x] , a_m not equal 0 and deg(f) = m
    g = b_0 + b_1x +....+ b_nx^n belongs to R[x], b_n not equal 0 and deg(g) = n

    (a) Let R be and Integral Domain (ID). Show that f.g is nonzero and f.g has leading term is (a_m)(b_n)x^(m+n). Deduce that R[x] is a ID and deg(f.g) = deg(f) +deg(g). If R[x] is a ID, show that R is also a ID.
    (b) Let R be a ID. Show that U(R[x]) = U(R)
    (c) Let a in R and a^n = 0 for n>1. Show that (1-a) belongs to U(R). What is (1-a)^-1?
    (d) Show that U(Z_4) is strictly not subset of U(Z_4[x]). Why doesn't this contradict part (b)?
    Note: Z_4 is set of integer mod 4

    Ok, Part (a) is easy and straightforwards.
    f.g is nonzero sine both f and g are nonzero.
    Also, the highest power of f.g is (a_m)(b_n)x^(m+n) obtained when u multiplied 2 functions.
    R[x] is an ID because either f = 0 or g = 0 (not both equal o at the same time)
    deg(f.g) = m + n = deg(f) +deg(g)

    Part (b), (c) and (d) I dont know how to begin with

    Can someone show me how to do part b, c and d please?

    Thank you
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  2. #2
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    I don’t think you’ve even done part (a) properly.

    Recall that the coefficient of x^k in fg is

    \sum_{i\,=\,0}^ka_ib_{k-i}

    When k=m+n, the sum is just a_mb_n because a_i=0 for all i>m and b_{m+n-i}=0 for all i<m. Hence the coefficient of x^{m+n} in fg is a_mb_n. And since a_m\ne0,\ b_n\ne0 and R is an integral domain, a_mb_n\ne0 and so fg is nonzero.

    Do you understand this so far?
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  3. #3
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    Quote Originally Posted by knguyen2005 View Post
    (a) .....
    If R[x] is a ID, show that R is also a ID.
    Let f(x)=a_0 + a_1x + \cdots +a_mx^m and g(x)=b_0 + b_1x + \cdots + b_nx^n be two nonzero polynomials in R[x]. Since R[x] is an integral domain, f(x)g(x) is nonzero and we have a non-zero leading term a_kb_lx^{k+l}. This follows that a_k \neq 0 and b_l \neq 0. Since R[x] is an integral domain, we obtain  a_kb_l \neq 0. Thus R is an integral domain.
    Think about this argument ((a_k \neq 0 \wedge b_l \neq 0) \Rightarrow a_kb_l \neq 0) \Leftrightarrow (a_kb_l = 0 \Rightarrow (a_k = 0 \vee b_l=0)).


    (b) Let R be a ID. Show that U(R[x]) = U(R)
    A polynomial f(x) of degree m times a polynomial g(x) of degree n is a polynomial f(x)g(x) of degree m+n in an integral domain R[x]. Suppose f(x) is a polynomial whose degree is 1 or greater. Then no polynomial g(x) exists such that f(x)g(x)=1, which is a polynomial of degree 0. This forces that units of R[x] only exists with a degree zero. Thus U(R[x])=U(R).

    (c) Let a in R and a^n = 0 for n>1. Show that (1-a) belongs to U(R). What is (1-a)^-1?
    Think about the Maclaurin series of 1/(1-a) and use a nilpotency of a.

    (1-a)(1+a+ \cdots +a^{n-1}) = 1, where a^n=0

    (d) Show that U(Z_4) is strictly (not?) subset of U(Z_4[x]). Why doesn't this contradict part (b)?
    Note: Z_4 is set of integer mod 4
    U(\mathbb{Z}_4) = \{1,3\}. They are coprime to 4. Meanwhile U(\mathbb{Z}_4[x]) has additional units including U(\mathbb{Z}_4). For instance, 1 + [2]x is a unit in U(\mathbb{Z}_4[x]). We know that \mathbb{Z}_4 is not an integral domain. This implies that if R is not an integral domain, U(R) \subset U(R[x]).
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  4. #4
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    Thank you both proscientia and aliceinwonderland
    I already understood and know how to do it now

    Thanks again for spending your time on this question
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