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# A question about fields and rational functions... watch

1. If k is a field, prove that , where k(x) is the field of rational functions.

Hint: Mimic a proof that is irrational.

The problem, in other words, so to show that there does not exist an element such that . This is certainly false if 2=0 in the field k, for then . So let's suppose in k; this implies that .

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 in k[x]. By unique factroization in k[x] we see that is divisible by and , so p is divisible by 1-x and 1+x.

I don't understand why in k[x] implies that p^2 is divisible by and . 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 ?

Since these irreducibles are distinct, and we write . Then , so q is also divisble by . This contradicts our assumption that p,q were relatively prime.

2. (Original post by Artus)
I don't understand why it's "certainly false if 2=0"
Well, it tells you. Suppose (think of ), so that

because .

(Original post by Artus)
I don't understand why in k[x] implies that p^2 is divisible by and . 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 ?
This should read: .. is divisible by and ..

Now, where and are clearly irreducible.

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

by primality.
3. (Original post by jack.hadamard)
Well, it tells you. Suppose (think of ), so that

because .
Thanks, but what's wrong when ?
4. (Original post by Artus)
Thanks, but what's wrong when ?
You would like to show that there is no element such that .
The fact that shows that .

In other words, if the field is has characteristic two, then has a square root.
5. (Original post by jack.hadamard)
You would like to show that there is no element such that .
The fact that shows that .

In other words, if the field is has characteristic two, then 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?
6. (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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