Expert: Paul Klarreich - 10/1/2010

Question
Dear Sir,

I am unsure of the answer I have in polynomial in ring where I need to show that Let F be a field. Then f is a unit in F[x] if and only if f is a non-zero constant polynomial.

The solution is
Suppose f is a unit in F[x]. Then f
= 0F and there exists g
= 0F in F[x] such that
fg = 1F .( I know that it is due multiplicative inverse that is why fg=1F)

However this following part I do not understand why
deg (f) +deg (g) =deg{1}= 0 .
Since the degree of any element of F[x] is always a non-negative integer, we conclude that deg (f) = deg (g) =
0. So f must be a non-zero constant polynomial.

Can you help me to explain in simple term for me to understand it. Thanks.  

Answer
Questioner:   Raymond
Country:  Singapore
Category:  Advanced Math
Private:  No
 
Subject:  Abstract aglebra - Ring in polynomial
Question:  Dear Sir,

I am unsure of the answer I have in polynomial in ring where I need to show that Let F be a field. Then f is a unit in F[x] if and only if f is a non-zero constant polynomial.

The solution is
Suppose f is a unit in F[x]. Then f
= 0F and there exists g
= 0F in F[x] such that
fg = 1F .( I know that it is due multiplicative inverse that is why fg=1F)

However this following part I do not understand why
deg (f) +deg (g) =deg{1}= 0 .
Since the degree of any element of F[x] is always a non-negative integer, we conclude that deg (f) = deg (g) =
0. So f must be a non-zero constant polynomial.

Can you help me to explain in simple term for me to understand it. Thanks.
------------------------------------------------
I am not sure of the definitions, and it has been a long time, but:

For polynomials  p(x) and q(x),  deg(pq) = deg(p) + deg(q)

Now exactly what is your text's definition of a unit in a ring?  The one I found says that:

u is a unit in a ring R if there exists  v where  uv = 1.

(That means the units are the things that have inverses.)

For example, in the ring of integers,  1 and -1 are units.(each is its own inverse)

Now if p(x) is a unit,  there is an inverse q(x) such that pq = 1.

But  deg(1) = 0, and  deg(p(x)) >= 0

deg(q) >= 0   --->  - deg(q) <= 0

So :
       deg(p) + deg(q)   = 0
and             - deg(q) <= 0
--------------------------------    Adding:
       deg(p) <= 0
But   deg(p) >= 0
So  deg(p) = 0.  [If something   is  >= 0 and <= 0 at the same time....]

And the only polynomials of degree zero are the nonzero constants.
[Remember that  the degree of zero is undefined.]

Does that do it?