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Let $K$ be the set of the following four-tuples of elements of $GF(3)$: $$K = \{(0,0,0,0),(1,2,1,1),(2,1,2,2),(1,0,0,1),(2,2,1,2),(2,0,0,2),(0,1,2,0),(0,2,1,0),(1,1,2,1)\}.$$ Define operations of addition, multiplication on $K$ to make $K$ into a field.

I have some misunderstanding. First, doesn't $GF(3)$ mean the same thing as $\mathbb{Z}/3$ which is the same thing as the set $\{0,1,2\}$? If that's the case, then how can we have tuples,vectors?, of length 4? Since I have $(0,0,0,0)$ already given, I have the additive identity element. I don't have $(1,1,1,1)$ which would have been the multiplicative identity element. Not sure where to go from here.

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$\,GF(q)\,$ usually denotes the field with q elements (q a power of a prime, of course), and it equals the ring $\,\Bbb Z/q\Bbb Z\,$ iff $\,q\,$ is a prime, so in this case yes: those two sets are the same. Now you form n-tuples with elements of this set (field), but not all the possible n-tuples, just as given. Use operations modulo 3 to try to make the given set with nine elements a field. – DonAntonio Nov 30 '12 at 3:53
Don, I don't quite follow the operations modulo 3. Can you elaborate further? – MathScratch Nov 30 '12 at 4:06
up vote 1 down vote accepted

Um. You would get a ring by taking all $81$ two by two matrices with entries in the field of three elements $\mathbb Z / 3 \mathbb Z. $ The only problem is that multiplication need not be commutative. Also, there would be zero divisors, that is not good either.

At the same time, the matrices with real entries of shape $$ \left( \begin{array}{cc} a & b \\ -b & a \end{array} \right) , $$ make a field, because the product of two matrices of this type is another, same for inverses, no zero divisors because all the determinants are nonzero. Oh, the big thing is that multiplication is commutative. You should check that. The result is a field, we call it the complex numbers.

What they have given you is just that, the nine matrices of that type with entries in the field of three elements. The reason the determinant is never zero for one of these (nonzero) matrices is now number theoretic. You need to confirm that as well.

To fit in a more familiar picture, take the field of three elements, call it $F.$ The ring of polynomials with those coefficients is written $F[x].$ Finally, the polynomial $$ x^2 + 1 $$ is irreducible in the ring, because there is simply no square root of $-1 \equiv 2$ in this field. So what i have described is the standard $$ F[x]/ (x^2 + 1), $$ where the matrix I display above is mapped to $$ a + b x $$ in this description of the nine-element field.

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Will, I do not quite understand where the 81 came from. Is it $3^4$ and why? and why is your matrix a 2 by 2? The given vectors were of length 4. – MathScratch Nov 30 '12 at 4:10
@Will, why do you think they gave him "matrices"? Of course, that may be part of the assignment: to llok at those 4-tuples as matrices and derive the operations from the ones on matrices. – DonAntonio Nov 30 '12 at 4:13
@DonAntonio, that was simply what I knew would work. Then I checked the 4-tuples he has, the first and fourth elements always agree, the second and third elements are additive inverses. – Will Jagy Nov 30 '12 at 4:31
@WillJagy, I know that. I just thought it would be nice the O)P would arrive by himself at the operations, with matrices orwithout them. Then somebody could tell him that in form of matrices the elements he's given..etc. – DonAntonio Nov 30 '12 at 4:34

Another approach is to say you have $9=3^2$ elements. There is precisely one field with $9$ elements. So just pick any bijection from your set with these $9$ elements and define addition and multiplication as the image of that bijection. This ignores the comment about $GF(3)$-you could equally biject them with some elements of the $16$ element field.

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Hint $\rm\ \ \Bbb F_3[{\it i}\,] \cong \Bbb F_3[x]/(x^2+1)$ is a field of $\,9\,$ elements, representable by the linear maps on its additive group $\rm\:\Bbb F_3\langle1,{\it i}\,\rangle,$ where the linear map of multiplication by $\rm\:a + b{\it i}\:$ has matrix

$$\begin{eqnarray}\rm (a+b{\it i})\, 1 &=&\rm \ \ \ \,a + b{\it i}\\ \rm (a+b{\it i})\ {\it i}\, &=&\rm\, -b+a{\it i}\end{eqnarray}\ =\ \left[\begin{array}{cc}\rm a &\rm b\\ \rm -b &\rm a\end{array}\right]\left[\begin{array}{} 1 \\ {\it i}\end{array}\right]$$

The given tuples are nothing but the flattened matrices $\rm\:(a,b,-b,a),\ $ for $\rm\: a,b\in\Bbb F_3$

Remark $\ $ Above is a special case of the fact that rings arise naturally as subrings of linear maps on their underlying additive groups (the left regular representation). This is a ring-theoretic analog of the Cayley represention of a group as subgroups of permutation, by acting on itself by left multiplication.

The problem is trivial as posed since one can always transport the structure of a field to any set of the same cardinality, by simply pulling back the field operations along any set-theoretic bijection of the elements. Presumably the author sought a solution respecting some further structure, e.g. vector space structure. Perhaps such a requirement was implicit in the ambient context of the source of the problem.

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See also this answer for the analogous crepresentation of $\,\Bbb C \cong \Bbb R[i].\ \ $ – Math Gems Feb 5 '13 at 15:45

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