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As I understand a finite field of order $q$ exists if and only if the order $q$ is a prime power $ p^k $ (where $p$ is a prime number and $k$ is a positive integer). So when taking $ p=3 $ and $ k=2 $ this should result in a finite field with $ q=9 $ (Modulo 9) and thus multiplication, addition, subtraction and division are defined.

Maybe I didn't understand this correctly but it seems to me that 3 and 6 (and 9) have no multiplicative inverses? I used this website to calculate them based on the extended euclidian algorithm (http://planetcalc.com/3311/).

But if they have no multiplicative inverses, how can it be a field?

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  • $\begingroup$ Aside: there are a few ways one can reasonably say that the inverse of $3 \bmod 9$ is $\frac{1}{3} \bmod 1$. $\endgroup$ – user14972 Sep 26 '15 at 22:48
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The field $\Bbb F_9$ of order $9$ is (as a ring) not isomorphic to the ring $\Bbb Z / 9 \Bbb Z$ of integers modulo $9$. (In fact, even the underlying additive groups of the two rings are nonisomorphic: $\Bbb Z / 9 \Bbb Z$ has elements of order $9$ under addition, but all nonzero elements of $\Bbb F_9$ have order $3$ under addition.)

To construct $\Bbb F_9$, it's enough to pick a polynomial $p$ of degree $2$ over the prime field $\Bbb F_3$, say, $x^2 + 1$, and form the quotient ring $\Bbb F_3[x] / \langle p(x) \rangle$. Since the polynomial is irreducible, the ideal $\langle p(x) \rangle$ is maximal, and hence the quotient ring is a field. On the other hand, every equivalence class in the quotient ring has exactly one linear polynomial representative in $\Bbb F_3[x]$, and there are $3 \cdot 3$ such polynomials, so, as desired, this quotient is the field with $9$ elements.

On the other hand, like you say, $\Bbb Z / 9 \Bbb Z$ has zero divisors (as, e.g., $3 \cdot 3 \equiv 0 \pmod 9$), so this ring is not a field.

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  • $\begingroup$ Thank you Travis! I'm still trying to understand this subjects and I hope I did understand you correctly. So, just to clarify: taking any prime number power does not automatically result in a field and it is in deed true that 3 and 6 have no multiplicative inverses in the above case? $\endgroup$ – holistic Sep 26 '15 at 17:29
  • $\begingroup$ You're welcome. The point is that the ring of integers $\Bbb Z / n \Bbb Z$ modulo $n$ is only a field for prime $n$. (For $n$ not prime, say, $n = ab$ for $a, b > 1$, then $a$ and $b$ are both zero divisors and hence the ring is not a field.) On the other hand, and as you say, there are fields of order $p^k$ for any prime $p$ and positive integer $k$, but for $k > 1$ these are not respectively isomorphic to $\Bbb Z/p^k \Bbb Z$ (which we just established cannot be fields). It is in particular true that $3, 6$ are zero divisors in $\Bbb Z/9\Bbb Z$ and hence do not have multiplicative inverses. $\endgroup$ – Travis Willse Sep 26 '15 at 17:42
  • $\begingroup$ You're welcome! $\endgroup$ – Travis Willse Sep 27 '15 at 17:13
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Indeed $\mathbb Z/9\mathbb Z$ is not a field. That some field with $9$ elements exists does not mean that this is it.

Exercise: Show that in every finite field $\displaystyle\min\{n\ge2 : \underbrace{1+\cdots+1}_n = 0\}$ is prime.

You've already done most of this exercise if you've figured out why $3$ and $6$ cannot have multiplicative inverses modulo $9$.

From this exercise it can be seen to follow that in a field with $p^k$ elements, one must have $\displaystyle\underbrace{1+\cdots+1}_p = 0$.

Hence in a field with $9$ elements, one has $1+1+1=0$.

Notice that in such a field, $1^2=1$ and $(1+1)^2=1$, so $1+1$ has no square root within the set $\{0,1,1+1\}$. Thus the polynomial $x^2-2$ cannot be factored within that field with three elements. Now look at the quotient ring $F[x]/(x^2-2)$, where $F$ is the field whose elements are $0,1,1+1$, and show that it's a field in which $x^2-2$ factors, so $2$ has a square root. That is a field with $9$ elements.

PS: That every finite field $F$ is a subfield of some larger finite field can be seen as follows: let $$ f(x) = 1 + \prod_{a\in F} (x-a). $$ This polynomial has no zeros in $F$; it therefore has no first-degree factors with coefficients in $F$. From that it can be shown to follow that if $g(x)$ is any factor of $f(x)$ that is irreducible over $F$, then $F[x]/(g(x))$ is a field, and it has $F$ as a proper subfield.

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  • $\begingroup$ Why can't the polynomial $f$ in the postscript be factored as a product of two (or more) irreducible polynomials, each of degree $> 1$? $\endgroup$ – Travis Willse Sep 26 '15 at 19:01
  • $\begingroup$ @Travis : oh .... I wasn't thinking of that, but I think what I've got is enough to show there's a bigger finite field. ${}\qquad{}$ $\endgroup$ – Michael Hardy Sep 26 '15 at 19:30
  • $\begingroup$ @Travis : ok, I've done some emendations. $\endgroup$ – Michael Hardy Sep 26 '15 at 19:32
  • $\begingroup$ Oh, I see, yes, that's sufficient, I agree. Cheers! $\endgroup$ – Travis Willse Sep 27 '15 at 1:57
  • $\begingroup$ Thank you as well Michael, it's much clearer to me now although I need to read a little bit more to understand all of it :). $\endgroup$ – holistic Sep 27 '15 at 13:15

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