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

Show that $\displaystyle Q[x]/ <x^2 - 2> \ \cong \ Q[x] / <x^2 + 4x + 2> $

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- Dec 17th 2012, 12:24 AMBernhardQuotient or Factor Rings
Can someone help with the following exercise from Beachy and Blair Ch 5 - Commutative Algebra

Show that $\displaystyle Q[x]/ <x^2 - 2> \ \cong \ Q[x] / <x^2 + 4x + 2> $ - Dec 17th 2012, 03:17 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, 02:52 PMDrexel28Re: Quotient or Factor Rings
A much quicker, albeit high teck, solution is as follows. We know that there exists a unique ring map $\displaystyle \mathbb{Q}[x]\to\mathbb{Q}[x]$ determined by $\displaystyle x\mapsto x+2$ (it's just $\displaystyle p(x)\mapsto p(x+2)$) call this ring map $\displaystyle f$. Note then that we can compose this with the canonical projection $\displaystyle \mathbb{Q}[x]\to\mathbb{Q}[x]/(x^2+4x+2)$ to get a ring map $\displaystyle \overline{f}:\mathbb{Q}[x]\to\mathbb{Q}[x]/(x^2+4x+2)$. Now, note that $\displaystyle f(x^2-2)=(x+2)^2-2=x^2+4x+2$ so that $\displaystyle (x^2-2)\subseteq\ker\overline{f}$. So, $\displaystyle \overline{f}$ descends to a well-defined ring map $\displaystyle g:\mathbb{Q}[x]/(x^2-2)\to\mathbb{Q}[x]/(x^2+4x+2)$. Now, note that this ring map is automatically injective since $\displaystyle \mathbb{Q}[x]/(x^2-2)$ is a field (clearly $\displaystyle x^2-2$ is irreducible) and that this ring map is surjective because $\displaystyle g$ is a $\displaystyle \mathbb{Q}$-algebra map and $\displaystyle \mathbb{Q}[x]/(x^2-2)$ and $\displaystyle \mathbb{Q}[x]/(x^2+4x+2)$ are both $\displaystyle 2$-dimensional $\displaystyle \mathbb{Q}$-spaces. So, $\displaystyle g$ is an isomorphism.

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

Note: The above actually goes to show that if [tex/]p(x)[/tex] and $\displaystyle q(x)$ are rational (more generally of course) polynomials such that $\displaystyle p(x+a)=q(x)$ for some rational $\displaystyle a$, then $\displaystyle \mathbb{Q}[x]/(p(x))\cong\mathbb{Q}[x]/(q(x))$--the same intuition though applies that

$\displaystyle \begin{aligned}\mathbb{Q}[x]/(p(x)) &\cong\mathbb{Q}(\text{roots of p})\\ &=\mathbb{Q}(\text{rationally perturbed roots of }q)\\ &=\mathbb{Q}(\text{roots of }q)\\ &\cong\mathbb{Q}[x]/(q(x))\end{aligned}$. - Dec 17th 2012, 03:07 PMBernhardRe: Quotient or Factor Rings
Thanks for the extra perspective on the problem - will work through this!

Peter