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Math Help - Quotient or Factor Rings

  1. #1
    Super Member Bernhard's Avatar
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    Quotient or Factor Rings

    Can someone help with the following exercise from Beachy and Blair Ch 5 - Commutative Algebra

    Show that  Q[x]/ <x^2 - 2> \ \cong  \ Q[x] / <x^2 + 4x + 2>
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  2. #2
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    Re: Quotient or Factor Rings

    the ring on the left is better known as Q(√2) = {a + b√2: a,b in Q}. well, technically it's a "different" ring, but the two are ring-isomorphic, so it doesn't matter much.

    the ring on the right, is Q(u), where u is a root of x2 + 4x + 2, the good ol' "quadratic formula" tells us that u = -2 √2.

    it really doesn't matter "which" root we take to be u, so for convenience we'll take u = √2 - 2. now, informally, we see that u is in Q(√2), so Q(u) is contained in Q(√2).

    also √2 = u + 2, so Q(√2) is contained in Q(u).

    but...let's be formal and exhibit an isomorphism, which will definitely settle things.

    define φ:Q[x]/<x2-2> --> Q[x]/<x2+4x+2> by:

    φ((a+bx) + <x2-2>) = (a+2b)+bx + <x2+4x+2>

    the first thing we need to do is show that φ is well-defined.

    so suppose a'+b'x + <x2-2> = a+bx + <x2-2>.

    then (a-a') + (b-b')x is in <x2-2>, so since we have a linear polynomial that x2- 2 divides, it must be the 0-polynomial, we must have a = a', b = b', so that:

    φ((a+bx) + <x2-2>) = φ((a'+b'x) + <x2-2>) (i leave it to you to show that if deg(f(x)) > 1, f(x) = q(x)(x2 - 2) + a + bx for some polynomial a + bx...you might want to use the fact that Q[x] is euclidean).

    is φ a homomorphism?

    φ([(a+bx) + <x2-2> + (c+dx) + <x2-2>]) = φ((a+c) + (b+d)x + <x2-2>)

    = (a+c+2(b+d)) + (b+d)x + <x2+4x+2> = (a+2b) + bx + (c+2d) + dx + <x2+4x+2>

    = (a+2b) + bx + <x2+4x+2> + (c+2d) + dx + <x2+4x+2> = φ((a+bx) + <x2-2>) + φ((c+dx) + <x2-2>), so φ is an additive homomorphism.

    φ(((a+bx) + <x2-2>)((c+dx) + <x2-2>)) = φ((ac+(ad+bc)x+bdx2) + <x2-2>)

    = φ((ac+2bd)+(ad+bc)x+bd(x2-2) + <x2-2>) = φ((ac+2bd)+(ad+bc)x + <x2-2>)

    = (ac+2bd+2(ad+bc))+(ad+bc)x + <x2+4x+2> = (ac+2ad+2bc+2bd)+(ad+bc)x + <x2+4x+2>

    = (ac+2ad+4bc+2bd) + (ad+bc+4bd)x - 2bd - 4bdx + <x2+4x+2>

    = (ac+2ad+4bc+2bd) + (ad+bc+4bd)x + bdx2 + <x2+4x+2> (since bdx2 + <x2+4x+2> = -4bdx - 2bd + <x2+4x+2>)

    = ((a+2b) + bx + <x2+4x+2>)((c+2d) + dx + <x2+4x+2>) <--i urge you to work carefully through this step

    = φ((a+bx) + <x2-2>)*φ((c+dx) + <x2-2>), so φ is a ring homomorphism.

    given c+dx + <x2+4x+2> in Q[x]/<x2+4x+2>, note that:

    φ((c-2d) + dx + <x2-2>) = (c-2d+2d) + dx + <x2+4x+2> = c+dx + <x2+4x+2>, so φ is surjective.

    suppose that (a+bx + <x2-2>) is in ker(φ).

    this means a+2b + bx is in <x2+4x+2>, hence a+2b + bx is the 0-polynomial. thus a+2b = 0, and b = 0, whence a = 0, so the kernel of φ is simply <x2-2>,

    so φ is injective, and thus an isomorphism (egads that was awful).
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Re: Quotient or Factor Rings

    A much quicker, albeit high teck, solution is as follows. We know that there exists a unique ring map \mathbb{Q}[x]\to\mathbb{Q}[x] determined by x\mapsto x+2 (it's just p(x)\mapsto p(x+2)) call this ring map f. Note then that we can compose this with the canonical projection \mathbb{Q}[x]\to\mathbb{Q}[x]/(x^2+4x+2) to get a ring map \overline{f}:\mathbb{Q}[x]\to\mathbb{Q}[x]/(x^2+4x+2). Now, note that f(x^2-2)=(x+2)^2-2=x^2+4x+2 so that (x^2-2)\subseteq\ker\overline{f}. So, \overline{f} descends to a well-defined ring map g:\mathbb{Q}[x]/(x^2-2)\to\mathbb{Q}[x]/(x^2+4x+2). Now, note that this ring map is automatically injective since \mathbb{Q}[x]/(x^2-2) is a field (clearly x^2-2 is irreducible) and that this ring map is surjective because g is a \mathbb{Q}-algebra map and \mathbb{Q}[x]/(x^2-2) and \mathbb{Q}[x]/(x^2+4x+2) are both 2-dimensional \mathbb{Q}-spaces. So, g is an isomorphism.

    To reiterate, Deveno gave the better intuition for why this is true. Namely, \mathbb{Q}[x]/(x^2-2) is the "splitting field" for x^2-2 and similarly \mathbb{Q}[x]/(x^2+4x+2) is the splitting field for x^2+4x+2. Now, any two splitting fields are isomorphic, and you can show that \mathbb{Q}(\sqrt{2}) (literally the subfield of \mathbb{R} generated by \mathbb{Q} and \sqrt{2}) is also a splitting field for x^2-2 (over \mathbb{Q}) so that \mathbb{Q}[x]/(x^2-2)\cong\mathbb{Q}(\sqrt{2})\subseteq\mathbb{R}. But, using the same logic, one can prove that \mathbb{Q}(\sqrt{2}-2) is a splitting field for \mathbb{Q}[x]/(x^2+4x+2) and so \mathbb{Q}[x]/(x^2+4x+2)\cong\mathbb{Q}(\sqrt{2}-2)\subseteq\mathbb{R}. Since, by inspection, \mathbb{Q}(\sqrt{2})=\mathbb{Q}(\sqrt{2}-2) (since you are already including \mathbb{Q} in the generating set, there is no distinction between \sqrt{2} and \sqrt{2}-2) you can see that \mathbb{Q}[x]/(x^2-2)\cong\mathbb{Q}[x]/(x^2+4x+2).



    Note: The above actually goes to show that if [tex/]p(x)[/tex] and q(x) are rational (more generally of course) polynomials such that p(x+a)=q(x) for some rational a, then \mathbb{Q}[x]/(p(x))\cong\mathbb{Q}[x]/(q(x))--the same intuition though applies that
    \begin{aligned}\mathbb{Q}[x]/(p(x)) &\cong\mathbb{Q}(\text{roots of p})\\ &=\mathbb{Q}(\text{rationally perturbed roots of }q)\\ &=\mathbb{Q}(\text{roots of }q)\\ &\cong\mathbb{Q}[x]/(q(x))\end{aligned}.
    Last edited by Drexel28; December 17th 2012 at 02:56 PM.
    Thanks from Bernhard
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  4. #4
    Super Member Bernhard's Avatar
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    Re: Quotient or Factor Rings

    Thanks for the extra perspective on the problem - will work through this!

    Peter
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