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Math Help - show that a quotient ring is in integral domain

  1. #1
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    show that a quotient ring is in integral domain

    Let A be an integral domain, and a,b are elements in A. Let B=A[x]/(ax+b), where (ax+b) is ideal generated by ax+b. Suppose (a)\cap(b)=(ab), show that B is an integral domain.

    I think it suffices to show that (ax+b) is prime under the assumption, and hence need to show that if f(x)g(x) is in (ax+b), then either f(x) or g(x) is in (ax+b).

    thank you for help!
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  2. #2
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    Quote Originally Posted by frankmelody View Post
    Let A be an integral domain, and a,b are elements in A. Let B=A[x]/(ax+b), where (ax+b) is ideal generated by ax+b. Suppose (a)\cap(b)=(ab), show that B is an integral domain.

    I think it suffices to show that (ax+b) is prime under the assumption, and hence need to show that if f(x)g(x) is in (ax+b), then either f(x) or g(x) is in (ax+b).

    thank you for help!
    first of all we also need to assume that a \neq 0, \ b \neq 0. i've been trying for a couple of hours now to find a simpler proof for this problem but i haven't been able to do so! so, i'm just going to give

    you my "not very simple" proof. we need a lemma:


    Lemma: suppose d \mid a^n c and d \mid b^n c, for some c,d \in A and integer n \geq 0. then d \mid c.

    Proof: by induction on n. if n=0 or c=0, there is nothing to prove. so suppose the claim is true for n-1 and d \mid a^nc, \ d \mid b^n c, where c \neq 0. so we have a^n c = rd, \ b^n c = sd, for some r,s \in A,

    which gives us a^ns = b^nr \in <a> \cap <b> = <ab>. hence a^ns=abu, \ b^nr=abv, for some u,v \in A. therefore a^{n-1}s=bu and b^{n-1}r=av. hence by the induction hypothesis b \mid s and a \mid r.

    so s = bt, \ r = az, for some t,z \in A. but then a^n c = rd = azd, \ b^n c = sd = btd, and so a^{n-1}c = zd, \ b^{n-1}c = td, and we're done by the induction hypothesis. \Box


    now suppose f(x)g(x) \in <ax + b>, for some f(x),g(x) \in A[x]. we want to prove that either f(x) \in <ax + b> or g(x) \in <ax+b>. obviously we may assume that both f(x),g(x) are non-

    zero. let Q be the field of fractions of A. clearly in Q we have f(-b/a)g(-b/a)=0. so we may assume that f(-b/a)=0. so there exists h(x) \in Q[x] such that f(x)=(ax+b)h(x), which we

    will call it (1). let f(x)=c_0x^n + \cdots + c_n and h(x)=\frac{d_0}{d}x^{n-1} + \cdots + \frac{d_{n-1}}{d}, where c_j, d_j , d \in A and d \neq 0. so (1) becomes: dc_0x^n + \cdots + dc_n = (ax+b)(d_0x^{n-1} + \cdots + d_{n-1}). call this (2).

    the claim is that d \mid d_j, for all j, which will complete the solution because then h(x) \in A[x] and thus by (1): f(x) \in <ax+b>. here's the proof of this claim:

    from (2) we have: dc_0=ad_0, \ dc_n = bd_{n-1}, \ dc_j = ad_j + b d_{j-1}, \ 1 \leq j \leq n-1. call this one (3). now, ignoring dc_n = bd_{n-1} in (3), the rest of the relations can be written as CX=dY, where:

    C is an n \times n lower triangular matrix with a on the main diagonal, X an n \times 1 vector with the entries d_0, \cdots , d_{n-1}, and Y an n \times 1 vector with the entries c_0, \cdots , c_{n-1}. multiplying CX=dY

    from the left by \text{adj}(C) gives us a^n X = \det(C) X = d \cdot \text{adj}(C)Y and so d \mid a^n d_j, for all j. similarly if we ignore dc_0=ad_0 in (3) and write the remaining relations in terms of matrices, we will get

    d \mid b^n d_j, for all j. therefore d \mid d_j, for all j, by the lemma. \Box
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  3. #3
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    hi, thank you for your help! I think most of your solution is correct, only the last part of your proof has some problems. Since A is only an integral domain, adj(C) need not to exist in A. But still, I can get d|d_j under your idea, just by considering each equation one by one. I can solve the problem now, thank you so much!
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  4. #4
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    Quote Originally Posted by frankmelody View Post

    ... Since A is only an integral domain, adj(C) need not to exist in A.
    that is not true! actually for the adjoint matrix of a square matrix C over a ring A to exist, we even don't need A to be integral domain!! we only need A to be commutative with identity.

    i suggest you take a look at a standard textbook in linear algebra or just google it!
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  5. #5
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    oh~yep! sorry for my mistake! we do not need division in the denifition of the adjoint matrix, thank you!
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