Question

show that q /(x^2-2) is not isomorphic to q /(x^2-3)

Answer 1

q here being the set of rational numbers?

Answer 2

yes

Answer 3

Q[x] /(x^2-2) , Q[x] /(x^2-3), the congruence classes

Answer 4

Also, did you note that
\[\huge \frac{\mathbb Q}{<x^2-2>}\cong\mathbb{Q}(\sqrt2)\]?

Answer 5

Yes

Answer 6

oh and Q(sqr(3) is isomorphic to Q / x^2 - 3

Answer 7

And, also,
\[\huge \frac{\mathbb{Q}}{<x^2-3>}\cong\mathbb{Q}(\sqrt3)\]yes.

Answer 8

so i have all the pieces, i just dont know how to finish the puzzle

Answer 9

Well, here's one way to do it, suppose that there is an isomorphism from Q(sqrt2) to Q(sqrt3)

Answer 10

Let that isomorphism be
\[\huge \Phi:\mathbb Q(\sqrt2) \ \rightarrow \ \mathbb Q(\sqrt3)\]

Answer 11

So, what's \[\large \Phi(1)\]?

Answer 12

Unity must be mapped to unity, right?

Answer 13

yes

Answer 14

But I do not understand what \[\Phi(1) is\]

Answer 15

It's just the image of 1, under the isomorphism, phi. If you don't like phi, we can use f instead.
f(1) = 1
right?

Answer 16

yes

Answer 17

So, by extension, the image of any integer must be itself, since, for any integer n,
f(n) = f(1+1+1...+1) = f(1) + f(1) + f(1) ... + f(1) = 1+1+1+...+1 = n

Answer 18

I follow so far.

Answer 19

So, the image of any integer is itself, what does that make the image of any rational number?

Answer 20

itself?

Answer 21

Remember, since it's a field isomorphism,
\[\huge f\left(\frac{m}n\right)=\frac{f(m)}{f(n)}\]

Answer 22

but sqrt of 2 and 3 arent rational?

Answer 23

We'll get to those. So, we're clear that the image of any rational number must be itself, okay? :)

Answer 24

haha yes

Answer 25

Okay, so, what must be \[\huge f(\sqrt2)\]?

Answer 26

Relax, we don't know :)
But whatever it is, it must not be rational. Right?

Answer 27

Yes

Answer 28

so...
\[\huge f(\sqrt2)=a+b\sqrt3\]where a and b are rational, and b is not equal to zero.

Answer 29

Because Qsqrt 2 is iso to Qsqrt3

Answer 30

based on our supposition

Answer 31

Not that, yet, but only the fact that f maps Q(sqrt2) to Q(sqrt3)

Answer 32

oh okay

Answer 33

So, based on this, what now is
\[\huge f(\sqrt2^2)=f(2)\]?

Answer 34

We know that since 2 is rational, its image must also be 2, right?

Answer 35

yes

Answer 36

And since this is an isomorphism, it is also equal to
\[\huge (a+b\sqrt3)^2=a^2+2ab\sqrt3+3b^2\]

Answer 37

Yup

Answer 38

So,
\[\huge 2=a^2+2ab\sqrt3+3b^2 \]see anything wrong yet?

Answer 39

The left side is obviously rational, so it forces the right side to also be rational. So,
2ab must be 0, or ab must be zero.

Answer 40

but didnt we say b wasn't zero?

Answer 41

That's right, so a must be zero :)

Answer 42

And so,
\[\huge 2=3b^2\]

Answer 43

b = sqrt of 2/3

Answer 44

which is not rational.
Contradiction, anyone? :)

Answer 45

haha yes! because f(2) forces the other side to be rational?

Answer 46

Yes. So, satisfied? ^.^

Answer 47

Because assuming that Q(sqrt2) is isomorphic to Q(sqrt3)
leads us to
sqrt(2/3) is rational, a contradiction.

Answer 48

Make your presence felt, @urbanderivative :)

Answer 49

I was just reading

Answer 50

I get it now, thanks for all your help!

Answer 51

Okay, that's all I need to know :)
I need to go now, thanks for the nice question ^.^
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Terence out

Answer 52

I like to keep you sharp