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A question about fields and rational functions...

If k is a field, prove that

\sqrt{1-x^2} \not\in k(x)

, where k(x) is the field of rational functions.

Hint: Mimic a proof that

\sqrt{2}

is irrational.

Answer from answer sheet:

The problem, in other words, so to show that there does not exist an element

f \in k(x)

such that

f^2 = 1 - x^2

. This is certainly false if 2=0 in the field k, for then

1-x^2 = 1+x^2 = (1-x)^2

. So let's suppose

2 \not= 0

in k; this implies that

1-x \not= 1+x

.

I don't understand why it's "certainly false if 2=0"

Suppose such an element f exists and write f=p/q with p,q in k[x] relatively prime. Then

p^2 = (1-x^2)q^2

in k[x]. By unique factroization in k[x] we see that

p^2

is divisible by

(1-x)^2

and

(1+x)^2

, so p is divisible by 1-x and 1+x.

I don't understand why

p^2 = (1-x^2)q^2

in k[x] implies that p^2 is divisible by

(1-x)^2

and

(1+x)^2

. Didn't the solution already assume taht 1-x is not equal to 1+x. Then how is p^2 divisible by both of them? Or are they just saying that

p_1 = q(1-x)^2

?

Since

2 \not= 0

these irreducibles are distinct, and we write

p=(1-x^2)p_1

. Then

p_1^2(1-x^2) = q^2

, so q is also divisble by

1-x^2

. This contradicts our assumption that p,q were relatively prime.

Thanks in advance

(edited 11 years ago)

Reply 1

jack.hadamard

Original post by Artus

I don't understand why it's "certainly false if 2=0"

Well, it tells you. Suppose

2 = 0

(think of

\mathbb{Z}/2\mathbb{Z}

), so that

(1 - x)^2 = 1 - 2x + x^2 = 1 + x^2 = 1 - x^2

because

-1 = 1

Original post by Artus

I don't understand why

p^2 = (1-x^2)q^2

in k[x] implies that p^2 is divisible by

(1-x)^2

and

(1+x)^2

. Didn't the solution already assume taht 1-x is not equal to 1+x. Then how is p^2 divisible by both of them? Or are they just saying that

p_1 = q(1-x)^2

This should read: ..

p^2

is divisible by

1 + x

and

1 - x

..

Now,

1 - x^2 = (1 - x)(1 + x)

where

1 - x

and

1 + x

are clearly irreducible.

Notice that

k[x]

is a PID, so these are in fact prime. Hence,

1 \pm x \mid p^2\ \Rightarrow\ 1 \pm x \mid p

by primality.

Reply 2

Artus

Original post by jack.hadamard

Well, it tells you. Suppose

2 = 0

(think of

\mathbb{Z}/2\mathbb{Z}

), so that

(1 - x)^2 = 1 - 2x + x^2 = 1 + x^2 = 1 - x^2

because

-1 = 1

Thanks, but what's wrong when

(1-x)^2 = 1-x^2

Reply 3

jack.hadamard

Original post by Artus

Thanks, but what's wrong when

(1-x)^2 = 1-x^2

You would like to show that there is no element

r \in k(x)

such that

r^2 = 1 - x^2

.
The fact that

(1 - x)^2 = 1 - x^2

shows that

1 - x = \sqrt{1 - x^2} \in k(x)

.

In other words, if the field is has characteristic two, then

1 - x^2

has a square root.

(edited 11 years ago)

Reply 4

Artus

Original post by jack.hadamard

You would like to show that there is no element

r \in k(x)

such that

r^2 = 1 - x^2

.
The fact that

(1 - x)^2 = 1 - x^2

shows that

1 - x = \sqrt{1 - x^2} \in k(x)

.

In other words, if the field is has characteristic two, then

1 - x^2

has a square root.

Oh, so the answer is telling us that the question is wrong when 2=0, so we just assume that 2 is not equal to 0?

Reply 5

jack.hadamard

Original post by Artus

Oh, so the answer is telling us that the question is wrong when 2=0, so we just assume that 2 is not equal to 0?

Yes. The statement is false for fields of characteristic two.

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