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If $f(x)\in F[x]$ is irreducible, $F$ is a finite field, then the quotient $F[x]/(f(x))$ is a field.

Is $x + (f(x))$ always a generator of this field? Meaning powers of it always generate the whole field except $0$. Does it hold or not?

James Lee
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    Welcome to MSE. Please read this text about how to ask a good question. – José Carlos Santos May 31 '22 at 08:46
  • I have an impression that x powers can generate the multiplicative group of field for smaller sizes, but not sure if it holds for arbitrary sizes. – James Lee May 31 '22 at 08:49
  • Do you mean, say, F is a finite field? If $F=\mathbb R$ then the quotient field has the cardinality of continuum, while powers can be at most countable. – lisyarus May 31 '22 at 08:50
  • Yes, only consider finite field for this question. – James Lee May 31 '22 at 08:50
  • Because in crypto only finite field is used. That's why I tagged with "finite-fields". – James Lee May 31 '22 at 08:52
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    @JamesLee Your problem can be restated as, whether a generator $\alpha$ of a finite field extension $K/F$ (an element such that $K=F(\alpha)$) is a generator of the multiplicative group of $K$. This is not true in general, but the converse holds. –  May 31 '22 at 15:22

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This is not always true. Consider the quotient field $$ \Bbb F_3[x]/(x^2+1). $$ Here $x$ is not a generator of the nonzero elements in this field.

Dietrich Burde
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  • Is it always true for finite fields with characteristic 2 ? – James Lee May 31 '22 at 09:01
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    @JamesLee No, but then the smallest example is the quartic $f(x)=x^4+x^3+x^2+x+1$. With that choice the powers of the coset $x+\langle f(x)\rangle$ start repeating after the fifth. – Jyrki Lahtonen May 31 '22 at 16:08