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When does a polynomial in $GF$ have a multiplicative inverse?

Are there values of $n$ such that all polynomials in $GF(n)$ have multiplicative inverses?

EDIT: To address the comments, I mean:

  1. All coefficients are in $GF(n)$
  2. Addition and multiplication of polynomials is defined pointwise: $(f+g)(x) := f(x) + g(x)$ and $(f \cdot g)(x) := f(x) \cdot g(x)$.

I believe that definition solves the ambiguity raised. If it's not the standard terminology, please help me learn the right terms to use!

Fly by Night
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user61838
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    Do you mean elements of the finite field $\text{GF}(n)$ or do you mean elements of the polynomial ring $\text{GF}(n)[X]$? –  Feb 11 '13 at 16:35
  • I concur with @Hurkyl that this question needs to be rethought carefully, and then rewritten. I’m not even sure what $GF(n)$ is meant to be. – Lubin Feb 11 '13 at 16:41
  • @Lubin: I'm new to Galois theory. Can you help me understand what's unclear about the question? And what needs to be thought out better? – user61838 Feb 11 '13 at 16:53
  • @user61838: Polynomials are often used to represent elements of finite fields (in the same way that integers are often used to represent elements of the ring of integers modulo $n$). When you say "... in $\text{GF}(n)$", it sounds like you mean something that is an element of $\text{GF}(n)$. When "..." turns out to be "polynomial", on one hand it sounds like you really meant "... over $\text{GF}(n)$" , but on the other hand due to the aforementioned fact, it sounds like you really do mean elements of $\text{GF}(n)$, and that you're representing elements with polynomials. Thus ambiguity. –  Feb 11 '13 at 17:08
  • @Hurkyl - See post, edited appropriately – user61838 Feb 11 '13 at 17:49
  • @user61838: You do realize that this pointwise operation is not the definition of multiplication of polynomials. Are you sure you are not confusing polynomials and polynomial functions? As a lithmus test consider the polynomial $p(x)=x^2+x+1\in GF(2)[x]$. It gives rise to the same polynomial function from $GF(2)$ to itself as the constant polynomial $q(x)=1$, because $p(0)=1=p(1)$. Yet $p(x)$ and $q(x)$ are different polynomials! For example $p(x)$ is of degree two, and $q(x)$is of degree zero. A polynomial is a finite formal sum $\sum_{i}a_ix^i$. – Jyrki Lahtonen Feb 12 '13 at 20:16
  • (cont'd) and two such sums $\sum_i a_ix^i$ and $\sum_i a'_ix^i$ are the same polynomial, iff $a_i=a'_i$ for all $i$. You can plugg in something in places of $x$, but you cannot reliably test equality of polynomials as shown by the above example. – Jyrki Lahtonen Feb 12 '13 at 20:17

1 Answers1

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A non-constant polynomial can never have an inverse in the ring of polynomials over any field. To see this, note that if $f$ and $g$ are non-zero polynomials of degrees $n$ and $m$ then $fg$ is a polynomial of degree $n + m$.

Tobias Kildetoft
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