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Is the polynomial $t^6-3$ irreducible over $\mathbb{Z}/7\mathbb{Z}$? How would you prove that without doing all the computations? Is there a general method for this? Thank you.

Frank Zermelo
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1 Answers1

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A general method is based on identifying the cardinality of the field containing any of the roots. In the case of finite fields the multiplicative order of an element determines the field the element generates. Remember that the non-zero elements of the field $\mathbb{F}_q$ of $q$ elements have order that divides $q-1$.

This suggests the following approach:

  1. Show that $3$ is of order six, i.e. a primitive element in the field $\mathbb{F}_7=\mathbb{Z}/7\mathbb{Z}$.
  2. Show that any root $z$ of the polynomial $x^6-3$ (residing in some extension field of $\mathbb{F}_7$) is of order $36$. That is, $z^{36}=1$, but no exponent smaller than $36$ will do. As usual (Lagrange's theorem) ruling out proper divisors of $36$ suffices (and is not too difficult given step 1).
  3. Show that $n=6$ is the smallest exponent with the property $7^n\equiv1\pmod{36}$.
  4. Show that any root $z$ of your polynomial belongs to the field $\mathbb{F}_{7^6}$ but doesn't belong to any smaller field.
  5. Conclude that the minimal polynomial of any root of your polynomial has degree six, and hence your polynomial is irreducible.
Jyrki Lahtonen
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  • A catch in (a generalization of) step 2 is that it does not follow that the zeros of, for example, $x^5-3$ would all be of order 30. In fact one of the roots of that polynomial is of order 6, and thus that polynomial has a lineaar factor. Here the prime factors of 36 are already factors of 6, and things go smoothly. – Jyrki Lahtonen Apr 29 '13 at 05:41
  • Thank you, this was really helpful. – Frank Zermelo Apr 30 '13 at 00:12