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How to define operations $+ $and $\cdot$ if a set $F$ has $3$ elements {j, k,l } rather than the usual $2$ elements using the table so that it is a field? And more broadly, how can I define operation on a set that has more than $2$ elements and how can I intuitively understand?

My working: \begin{array}{c|ccc}+&j&k&l\\\hline j&j&k&l\\k&k&l&j\\l&l&j&k\end{array}

\begin{array}{c|ccc}\cdot&j&k&l\\\hline j&j&j&j\\k&j&k&l\\l&j&l&\color{red}{?}\end{array}

CountDOOKU
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    Rename the elements to ${0,1,2}$, and define addition and multiplication as for the quotient ring $F=\Bbb Z/3\Bbb Z$. Then $(F,+,\cdot)$ is a finite field with $3$ elements. – Dietrich Burde Mar 04 '21 at 13:23
  • We could tell you the answer of how they all interact... but I think it is more important for you to actually go through the effort of doing this one yourself via trial and error. Just remember that you'll necessarily have one of the elements be "zero" and be the additive identity and there be additive inverses. That should, once you choose which of the elements plays the role of zero force the entire addition table. Next, you'll focus on the multiplication properties... choosing one of the nonzero elements to play the role of "One", the multiplicative identity. – JMoravitz Mar 04 '21 at 13:24
  • Just make yourself an addition table and a times table and fill it in to organize your thoughts. I'll get you started... if $j$ plays the role of zero, then we have $\begin{array}{c|ccc}+&j&k&l\\hline j&j&k&l\k&k\l&l&&\color{red}{?}\end{array}$ where we filled these in since the additive identity plus an element returns that same element, so $j+k=k$ for instance. What I marked with the red $?$ you will fill in what $l+l$ is equal to, continue... – JMoravitz Mar 04 '21 at 13:28
  • Writing down $+$ and $\cdot$ tables for ${0,1,2}$ (where $0,1$ are as specified in the definition of "field", and $3$ is just some arbitrary additional object) should be easy using the properties that a field should have. Proving the "associative" laws for your result, however, can be tedious. That is when @DietrichBurde comment is useful. You get associative laws for your system, assuming you already know associativity for $\mathbb Z$. Have you already proved that multiplication and addition are associative for $\mathbb Z$? – GEdgar Mar 04 '21 at 13:28
  • @JMoravitz Of course! Of course! I have tried and tried and I am not sure about my answer so I thought I would ask. Thanks! – CountDOOKU Mar 04 '21 at 13:31
  • So, you have an attempt at an answer? You should share that. You should always share that in every post regardless of how sure you are of it – JMoravitz Mar 04 '21 at 13:32
  • @JMoravitz Hello, thanks for trying to help! :) Really embrassed about it but I have updated my working. I have no idea whats $k \cdot k$ equal to – CountDOOKU Mar 04 '21 at 13:56
  • @GEdgar Thanks for commenting! But we haven't learnt about the quotient ring stuff so I am confused. I have updated to include my answer – CountDOOKU Mar 04 '21 at 13:58
  • What is $l \cdot l$? Since $l \ne 0$, we know $l \cdot l \ne l \cdot j$ and $l \cdot l \ne l\cdot k$. So... – GEdgar Mar 04 '21 at 16:47
  • @GEdgar why can't $l \cdot l = l\cdot k$, is it because of $l \ne 1$ and then $l \cdot l = k$? Also is the addition operation defined correctly? Thanks you so much! – CountDOOKU Mar 04 '21 at 23:34
  • In any field, if $l \ne 0$, then from $l\cdot l = l\cdot k$ we get $l\cdot l - l\cdot k = 0$ and then $l\cdot(l-k) = 0$, since $l \ne 0$ conclude $l-k = 0$ so $l=k$. – GEdgar Mar 04 '21 at 23:54
  • @GEdgar so $ l \cdot l = k \cdot k = k$? – CountDOOKU Mar 05 '21 at 01:07
  • $l=k$ is a contradiction. So conclude $l\cdot l \ne l\cdot k$. – GEdgar Mar 05 '21 at 01:51
  • You wanted this to be a field, yes? So... there is a "$0$" and there is a "$1$". From your work above, it is clear that you chose $j$ to be your "$0$" and $k$ to be your "$1$". Now, as it is a field, every nonzero element must have a multiplicative inverse. Since $l\cdot 0 = 0$ and since $l\cdot 1 = l$, it follows that the only choice for $l^{-1}$ is $l$ itself... in other words $l\cdot l = 1$... or using your variable names, $l\cdot l = k$ – JMoravitz Mar 05 '21 at 03:23
  • Compare to $\Bbb F_3$... you have $2\cdot 2 = 1$ (noting that in $\Bbb Z$ you have $2\cdot 2 = 4 = 3+1$) – JMoravitz Mar 05 '21 at 03:24
  • @JMoravitz Thanks for helping me! I have a final question. if I want to verify the field axioms do I just go through each and every one of it? (ie the associativity) but can I somehow cut this down? – CountDOOKU Mar 08 '21 at 13:12
  • @GEdgar Back to proving the associative laws, when you said tedious did you mean that as in I probably have to create tables and prove them one by one? is there a shortcut that I can understand? I don't understand the ring stuff :( – CountDOOKU Mar 08 '21 at 13:21
  • Check associativity from the table of an operation? See https://math.stackexchange.com/q/168663/442 The answer by boumol notes a (slightly) simpler method when the table is commutative – GEdgar Mar 08 '21 at 16:56
  • However in our case: to check $a(bc)=(ab)c$ ... if any of $a,b,c$ is $1$, easy; if any of $a,b,c$ is $0$, easy. So we only have to check $2(22) = (22)2$, done. For $+$ if any is $0$, easy, so that leaves $8$ cases to check. – GEdgar Mar 08 '21 at 17:06
  • @GEdgar So I don't have to prove commutativity because the table is symmetrical. I can check $a + 0 = a$ for all $a$ for additive identity. So my question is why do I have to prove for repeating element (like you have done there with $2(22)=(22)2$? So I have to prove, for example: $(1 + 2) + 2 = 1 + ( 2 + 2)$, $1 + ( 1 + 1) = (1 + 1)+ 1$ and so on? – CountDOOKU Mar 09 '21 at 13:24

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