In this problem, we are dealing with the concept of units in the context of integral domains and polynomial rings. Specifically, we are given an integral domain RR with a unit element and are asked to prove that any unit in R[x]R[x] (the polynomial ring over RR) must also be a unit in RR.
Definitions
Integral Domain: An integral domain is a commutative ring RR with a multiplicative identity (unit element) such that the product of any two non-zero elements is non-zero.
Unit in a Ring: An element aa in a ring RR is called a unit if there exists an element bb in RR such that a*b=b*a=1a \cdot b = b \cdot a = 1, where 11 is the multiplicative identity in RR.
Polynomial Ring R[x]R[x]: The set of all polynomials with coefficients in RR.
Work/Calculations
Step 1: Assume f(x)f(x) is a Unit in R[x]R[x]
Let’s assume that f(x)f(x) is a unit in R[x]R[x]. This means there exists a polynomial g(x)g(x) in R[x]R[x] such that:
f(x)*g(x)=1f(x) \cdot g(x) = 1
Step 2: Examine the Degrees of f(x)f(x) and g(x)g(x)
The degree of the polynomial f(x)*g(x)f(x) \cdot g(x) is the sum of the degrees of f(x)f(x) and g(x)g(x). Since f(x)*g(x)=1f(x) \cdot g(x) = 1, a constant polynomial, the degree of f(x)*g(x)f(x) \cdot g(x) is 0.
Let “deg”(f(x))=m\text{deg}(f(x)) = m and “deg”(g(x))=n\text{deg}(g(x)) = n.
“deg”(f(x)*g(x))=m+n=0\text{deg}(f(x) \cdot g(x)) = m + n = 0
Step 3: Conclude m=n=0m = n = 0
Since m+n=0m + n = 0 and m,n >= 0m, n \geq 0, it must be the case that m=n=0m = n = 0.
Step 4: Show f(x)f(x) is a Unit in RR
Since m=0m = 0, f(x)f(x) is a constant polynomial, say f(x)=af(x) = a, where aa is in RR. Similarly, g(x)=bg(x) = b, where bb is in RR.
From f(x)*g(x)=1f(x) \cdot g(x) = 1, we have a*b=1a \cdot b = 1.
This shows that aa is a unit in RR, as a*b=1a \cdot b = 1.
Conclusion
We have shown that if f(x)f(x) is a unit in R[x]R[x], then it must be a constant polynomial and also a unit in RR. Therefore, any unit in R[x]R[x] is a unit in RR.