# Finite Fields of Order 3

I have been trying to learn Real Analysis, though I'm having trouble with a problem.

1. Show that there exists one and (essentially) only one field with three elements.

Any help will be appreciated.

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Maybe you can construct an isomorphism between any two such. – Andrew Nov 9 '12 at 22:42
How did this question appear in real analysis? As for your question every field is an additive group. How many groups do you know with 3 elements? (Upto isomorphism) – fretty Nov 9 '12 at 22:52

When you see a claim in mathematics that says "There exists a unique $X$ up to isomorphism", or "There is one, and (essentially) only one $X$", this means that you are required to prove two things:

1. There is some $X$ with the wanted properties and structure.
2. If $Y$ also has the wanted properties then there is an isomorphism between $X$ and $Y$.

In the case of a field with three elements, you really have little choice of how to define everything. Show that you can do that on the three elements $\{0,1,2\}$, where $0$ is the additive neutral and $1$ is the multiplicative neutral (those have to exist anyway). Now given a field of three elements $\{a,b,c\}$ you have to send the additive neutral to zero, the multiplicative neutral to $1$ and the third element to...?

Now you have to show that this mapping preserves addition and multiplication, and the rest of the field structure. In the case of three elements this is quite a simple checking indeed.

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