Why is x not invertible in F[x]?

[bartieshaw]

Hey

First off, my terminology in the title may not be correct or what others use because when searching the forums for "polynomial ring" i had a lot of trouble finding anything that matched what i meant when i coined the term...


By polynomial ring i mean, for some ring R the polynomial ring over R is the ring R[x] = {a_0 + a_1x + ... + a_nx^n + ... ; a_i is in R} under the operations (+, .)

Now to my Question -

In the case that R (as above) is a Field, F, the i understand that F[x] is an integral domain, but not a field because of the simple example that my lecturers love to throw about, x in F[x] does not have an inverse (is not a unit)

Now i understand this argument and i don't dispute it, but i was wondering if any of you could tell me, or at least start me off (as it would probably help my understanding) on how to show / prove that x has no inverse in F[x]...

I have thought about this...but I am not sure the following argument is correct

suppose there was g(x) st. xg(x)=1

then xg(x)=x^0

hence g(x) = x^(0-1) = x^-1 which is not polynomial and hence not in F[x]

im thinking maybe a better argument would involve how i started but then somehow using the fact F[x] has no zero divisors would be more correct...but this is just guess work really.


sorry for the long post


cheers

Bart

 

[phoenixthoth]

Also, the degree of xg(x) is 1+deg(g) > 0=deg(1). Your argument seems circular to me because it rests on the premise that x itself doesn't have a polynomial inverse in R[x], which is what you're proving.

 

[bartieshaw]

Also, the degree of xg(x) is 1+deg(g) > 0=deg(1). Your argument seems circular to me because it rests on the premise that x itself doesn't have a polynomial inverse in R[x], which is what you're proving.

Yeah i know my attempt a a proof was pretty dismal...

Thankyou for your reply, i think that would be it. I am wondering does the degree of 1 (mult. identity) have to be 1? Maybe that is a stupid question, is it by definition deg 1 = 1...?

 

[phoenixthoth]

I think if f is a constant polynomial other than zero, deg(f) is defined to be zero and if f is the zero polynomial, then its degree is undefined or some people say negative infinity or something. I dunno, maybe some people define the degree of any constant polynomial to be zero...

 

[bartieshaw]

Yeah we had deg 0 = - infinity, but nothing defined about constants so maybe it was assumed...


anyway, thanks a bundle for sorting that out for me, now at least when I am in bed i won't be thinking about maths... : )

 

[phoenixthoth]

I'm surprised deg is defined for the zero polynomial but not other constant polynomials...perhaps it's embedded within the definition for other polynomials?

 

[bartieshaw]

hmm...ive looked it up in two textbooks now

The first one, Linear Algebra (LAY) - This is one was from 1st year - mentions non zero constant polynomials and says they have degree 0, but its talking about them in the context of the vector space P_n, so I am not sure if it would apply

My second textbook, A First Course in Abstract Algebra (FRALEIGH) - this one is from 2nd year - does discuss the degree of a polynomial in the context of the polynomial ring, but does not mention the degree of a non zero constant functin.

I will just ask my lecturer tomorrow, he comes around to everyone in the class individually before and after every lecture to see how we are going with his course so i shouldn't have any trouble getting to talk to him...


but for now, SLEEP, its 12.44 here and I am tired


THANK YOU again for your prompt and helpful responses


Bart

 

[Kummer]

In the case that R (as above) is a Field, F, the i understand that F[x] is an integral domain, but not a field because of the simple example that my lecturers love to throw about, x in F[x] does not have an inverse (is not a unit)

Yes, F[x] is an integral domain.

To show x has no inverse in F[x] we need to show x*f(x) ! = 1.

If, f(x) is a non-zero polynomial then deg f(x) = n where n>=0. And so deg (x*f(x)) >= n+1 >=1. Which means xf(x)!=1.

If, f(x)=0 then x*f(x) = 0 !=1. Unless this is a zero ring, i.e. 1=0. But by convention it is not considered a field. And so this is impossible.

 

[mathwonk]

the degree of a non zero constant is zero.