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Math Help - primitive, Gauss Lemma

  1. #1
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    primitive, Gauss Lemma

    Let A be a commutative ring and let A[x] be the ring of polynomials in an indeterminate x, with coefficients in A. Let f = a_0 + a_1x + \cdots + a_nx^n \in A[x]. f is said to be primitive if (a_0, a_1, \cdots, a_n)=(1). Prove that if f, g \in A[x], then fg is primitive iff f and g are primitive.
    [This is Atiyah-Macdonald #1.2d]
    I am particularly confused with \Leftarrow which I assume uses Gauss Lemma.

    For the \Rightarrow, I have:
    Suppose that fg is primitive. Now suppose that f were not primitive.
    Then (a_0, a_1, \ldots, a_n) = (1) so there is a common factor to all of the terms. But then fg would have a common factor as well. Thus f is primitive, and so must g be.
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  2. #2
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    Quote Originally Posted by xianghu21 View Post
    Let A be a commutative ring and let A[x] be the ring of polynomials in an indeterminate x, with coefficients in A. Let f = a_0 + a_1x + \cdots + a_nx^n \in A[x]. f is said to be primitive if (a_0, a_1, \cdots, a_n)=(1). Prove that if f, g \in A[x], then fg is primitive iff f and g are primitive.
    [This is Atiyah-Macdonald #1.2d]
    I am particularly confused with \Leftarrow which I assume uses Gauss Lemma.

    For the \Rightarrow, I have:
    Suppose that fg is primitive. Now suppose that f were not primitive.
    Then (a_0, a_1, \ldots, a_n) = (1) so there is a common factor to all of the terms. But then fg would have a common factor as well. Thus f is primitive, and so must g be.
    Let f(x) = a_nx^n + ... + a_1x + a_0 and let g(x) = b_mx^m+...+b_1x+b_0. Define h(x) = f(x)g(x). Let \pi be an irreducible in R. Now if a_p is smallest coefficient not divisible by \pi and b_q is smallest coefficient not dividible by \pi then c_{p+q} is smallest coefficient (in h(x) = c_Nx^N + ... + c_0 ) not divisible by \pi.

    Prove the result above. After that, Gauss lemma's follows.
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  3. #3
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    Quote Originally Posted by xianghu21 View Post
    Let A be a commutative ring and let A[x] be the ring of polynomials in an indeterminate x, with coefficients in A. Let f = a_0 + a_1x + \cdots + a_nx^n \in A[x]. f is said to be primitive if (a_0, a_1, \cdots, a_n)=(1). Prove that if f, g \in A[x], then fg is primitive iff f and g are primitive.
    [This is Atiyah-Macdonald #1.2d]
    I am particularly confused with \Leftarrow which I assume uses Gauss Lemma.

    For the \Rightarrow, I have:
    Suppose that fg is primitive. Now suppose that f were not primitive.
    Then (a_0, a_1, \ldots, a_n) = (1) so there is a common factor to all of the terms. But then fg would have a common factor as well. Thus f is primitive, and so must g be.
    let f(x)=\sum_{i=0}^na_ix^i, and g(x)=\sum_{j=0}^mb_jx^j. let f(x)g(x)=\sum_{k=0}^{m+n}c_kx^k. first see that by the definition of the coefficients c_k if t_k \in R, \; 0 \leq k \leq m+n, then \sum_{k=0}^{m+n}t_kc_k=\sum_{i=0}^nr_ia_i=\sum_{j=  0}^ms_jb_j, for

    some r_i, s_j \in R. so if fg is primitive, clearly both f and g must be primitive too.

    conversely, suppose f and g are primitive but fg is not primitive. so \sum_{k=0}^{m+n}Rc_k \subseteq \mathfrak{m}, for some maximal ideal \mathfrak{m} of R. now for any z \in R let z+\mathfrak{m}=\overline{z} \in \frac{R}{\mathfrak{m}}. so \overline{c_k}=0, \; 0 \leq k \leq m+n. thus:

    \overline{f}(x)\overline{g}(x)=(\sum_{i=0}^{n}\ove  rline{a_i}x^i)(\sum_{j=0}^{m}\overline{b_j}x^j)=\s  um_{k=0}^{m+n}\overline{c_k}x^k=0. but \frac{R}{\mathfrak{m}}[x] is an integral domain. thus either \overline{f}(x)=0 or \overline{g}(x)=0, which means either \sum_{i=0}^nRa_i \subseteq \mathfrak{m} or \sum_{j=0}^mRb_j \subseteq \mathfrak{m}. contradiction!
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post

    Let f(x) = a_nx^n + ... + a_1x + a_0 and let g(x) = b_mx^m+...+b_1x+b_0. Define h(x) = f(x)g(x). Let \pi be an irreducible in R. Now if a_p is smallest coefficient not divisible by \pi and b_q is smallest coefficient not dividible by \pi then c_{p+q} is smallest coefficient (in h(x) = c_Nx^N + ... + c_0 ) not divisible by \pi.

    Prove the result above. After that, Gauss lemma's follows.
    be careful! the ring is not even an integral domain! the problem is not that straightforward.
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  5. #5
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    Quote Originally Posted by NonCommAlg View Post
    be careful! the ring is not even an integral domain! the problem is not that straightforward.
    Yes, you are correct. I usually only deal with unique factorization domains.
    I guess this proof can work for special types of domains.
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  6. #6
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    Alternatively:

    => contrapositive. wlog suppose f is not primitive. let I be the proper ideal generated by the coefficients of f. It is easily seen that all the coefficients of fg lie in I so that fg is not primitive.

    Other direction requires a rigorous argument like above.
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