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The units of the ring F[x] are polynomials of degree 1. (i.e. nonzero constant polynomials )

I don't understand what this sentence means that. Can you give me proof or an example to understand?

thank you in advance.

lhf
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llee7 jkonn
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    "Degree $1$" is not correct. For example, $x$ is of degree $1$ but not a unit. What you write in the brackets, however, is correct. It is true that the units of $F[x]$ are precisely the nonzero constant polynomials. (Assuming $F$ is a field.) – Aryaman Maithani Jul 05 '21 at 16:36
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    You are also correct in noting that these are degree $\color{red}{0}$ polynomials and not $1$. (Assuming you are following a reasonable convention which takes care of the zero polynomial.) – Aryaman Maithani Jul 05 '21 at 16:37
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    thank you so much @aryaman maithani – llee7 jkonn Jul 05 '21 at 16:39
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    By the way, I noticed that you deleted the thoughts you had put. I'd suggest that you edit back and insert those since they showed the effort you put in (and you were correct as well!). – Aryaman Maithani Jul 05 '21 at 16:53
  • sorry I think that my effort is stupid after I saw the your answer. – llee7 jkonn Jul 05 '21 at 16:59

1 Answers1

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Claim. Let $K$ a field, and $K[X]$ its polynomial ring. Then the only units in $K[X]$ are precisely the constant, non-zero elements of $K$.

Proof. Let $f,g\in K[X]$ such that $f\cdot g = 1$. Then applying the degree map to both sides gives: $$\deg(fg) = \deg(f)+\deg(g)= \deg(1) = 0$$

So necessarily $\deg(f) = \deg(g) = 0$.

EDIT: A plus sign was missed when I was typing it. Oopsie.

Stefanos Van Dijk
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  • thank you so much for clear explanation – llee7 jkonn Jul 05 '21 at 16:42
  • What is the homomorphism that you mention here? (Homomorphism as what?) – Aryaman Maithani Jul 05 '21 at 16:43
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    So it should be $0=d(1)=d(fg)=d(f)+d(g)$, with $d(h)={\rm deg}(h)$. – Dietrich Burde Jul 05 '21 at 16:58
  • @lhf: I suspected that but what's the domain and codomain of $\deg$ here (and is it a homomorphism of what? groups? rings? monoids?)? Depending on that, one might want to consider $f, g \neq 0$. – Aryaman Maithani Jul 05 '21 at 17:10
  • @llee7jkonn: Every field is an integral domain (regardless of whether or not it is finite). – Aryaman Maithani Jul 05 '21 at 17:11
  • In a infinite field, the function $f: K \rightarrow K$ given by $f(x)=p(x)$ "describes entirely" the polynomial. In this terms, the equality $g=1$ means $g(x)=1$ for all $x \in K$, and it only happens for constant polinomal. That's the difference. Usually when we are learning polynomials in Calculus, we start with polynomial functions while in algebra we start with polynomial in one variable, which we labeled by x or capitalized x, I mean X. This conceps are entirely different -- while it can be compared in a infinite field. – R. W. Prado Jul 05 '21 at 17:25
  • To clear your doubts, I recommend that you build the set of polynomials using sequences, end then compare with the concept made in Calculus, i.e., polynomial functions . – R. W. Prado Jul 05 '21 at 17:32
  • @AryamanMaithani Apologies. Recently have been doing a lot of elliptic curves, where the degree homomorphism is a thing. It was a tongue slip. – Stefanos Van Dijk Jul 05 '21 at 17:43
  • The article https://en.wikipedia.org/wiki/Polynomial_ring gives the direction of the construction I have mentioned. – R. W. Prado Jul 05 '21 at 17:57
  • Please strive not to add more dupe answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Jul 05 '21 at 19:47