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

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- Dec 17th 2012, 01:24 AMBernhardQuotient or Factor Rings
Can someone help with the following exercise from Beachy and Blair Ch 5 - Commutative Algebra

Show that - Dec 17th 2012, 04:17 AMDevenoRe: 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 x^{2}+ 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]/<x^{2}-2> --> Q[x]/<x^{2}+4x+2> by:

φ((a+bx) + <x^{2}-2>) = (a+2b)+bx + <x^{2}+4x+2>

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

so suppose a'+b'x + <x^{2}-2> = a+bx + <x^{2}-2>.

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

φ((a+bx) + <x^{2}-2>) = φ((a'+b'x) + <x^{2}-2>) (i leave it to you to show that if deg(f(x)) > 1, f(x) = q(x)(x^{2}- 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) + <x^{2}-2> + (c+dx) + <x^{2}-2>]) = φ((a+c) + (b+d)x + <x^{2}-2>)

= (a+c+2(b+d)) + (b+d)x + <x^{2}+4x+2> = (a+2b) + bx + (c+2d) + dx + <x^{2}+4x+2>

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

φ(((a+bx) + <x^{2}-2>)((c+dx) + <x^{2}-2>)) = φ((ac+(ad+bc)x+bdx^{2}) + <x^{2}-2>)

= φ((ac+2bd)+(ad+bc)x+bd(x^{2}-2) + <x^{2}-2>) = φ((ac+2bd)+(ad+bc)x + <x^{2}-2>)

= (ac+2bd+2(ad+bc))+(ad+bc)x + <x^{2}+4x+2> = (ac+2ad+2bc+2bd)+(ad+bc)x + <x^{2}+4x+2>

= (ac+2ad+4bc+2bd) + (ad+bc+4bd)x - 2bd - 4bdx + <x^{2}+4x+2>

= (ac+2ad+4bc+2bd) + (ad+bc+4bd)x + bdx^{2}+ <x^{2}+4x+2> (since bdx^{2}+ <x^{2}+4x+2> = -4bdx - 2bd + <x^{2}+4x+2>)

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

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

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

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

suppose that (a+bx + <x^{2}-2>) is in ker(φ).

this means a+2b + bx is in <x^{2}+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 <x^{2}-2>,

so φ is injective, and thus an isomorphism (egads that was awful). - Dec 17th 2012, 03:52 PMDrexel28Re: Quotient or Factor Rings
A much quicker, albeit high teck, solution is as follows. We know that there exists a unique ring map determined by (it's just ) call this ring map . Note then that we can compose this with the canonical projection to get a ring map . Now, note that so that . So, descends to a well-defined ring map . Now, note that this ring map is automatically injective since is a field (clearly is irreducible) and that this ring map is surjective because is a -algebra map and and are both -dimensional -spaces. So, is an isomorphism.

To reiterate,**Deveno**gave the better intuition for why this is true. Namely, is the "splitting field" for and similarly is the splitting field for . Now, any two splitting fields are isomorphic, and you can show that (literally the subfield of generated by and ) is also a splitting field for (over ) so that . But, using the same logic, one can prove that is a splitting field for and so . Since, by inspection, (since you are already including in the generating set, there is no distinction between and ) you can see that .

Note: The above actually goes to show that if [tex/]p(x)[/tex] and are rational (more generally of course) polynomials such that for some rational , then --the same intuition though applies that

. - Dec 17th 2012, 04:07 PMBernhardRe: Quotient or Factor Rings
Thanks for the extra perspective on the problem - will work through this!

Peter