Why can't the Polynomial Ring be a Field? I'm

Thaddeus Nolan

Thaddeus Nolan

Answered question

2022-03-14

Why can't the Polynomial Ring be a Field?
I'm currently studying Polynomial Rings, but I can't figure out why they are Rings, not Fields. In the definition of a Field, a Set builds a Commutative Group with Addition and Multiplication. This implies an inverse multiple for every Element in the Set.
The book doesn't elaborate on this, however. I don't understand why a Polynomial Ring couldn't have an inverse multiplicative for every element (at least in the Whole numbers, and it's already given that it has a neutral element). Could somebody please explain why this can't be so?

Answer & Explanation

Jaydan Russell

Jaydan Russell

Beginner2022-03-15Added 5 answers

Hint rmrmx,f(x)=1, in R, by evaluating at x=0
Explanation:
This has a very instructive universal interpretation: if x is a unit in R[x] then so too is every R-algebra element r, as follows by evaluating xf(x)=1 at x=r. Therefore to present a counterexample it suffices to exhibit any nonunit in any R-algebra. A natural choice is the nonunit 0R, which yields the above proof.
Kaitlyn Yates

Kaitlyn Yates

Beginner2022-03-16Added 4 answers

For F[x] to be a field, you need to show there is an inverse for each element that isn't 0. Now xF[x], and clearly x0 (considered as a polynomial). But if you multiply x by any non-zero polynomial, the result will always contain x or higher powers, so it has no inverse.

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