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I am rather unskilled in algebra (have never done it to be precise), but today, I had to deal with some division rings (i.e. rings with multiplicative inverse) and I came to a point where I had to check whether

$$\{0,1,2,i,2i,i+1,i+2,2i+1,2i+2\}$$ With both addition and multiplication defined normally with the result modulo $3$ is a division ring. Now, I simply mechanically found inverses for all elements and checked the definition of a division ring. This was a rather lengthy process, so I wondered, can this be done faster? (Q1)

One idea I had, is that perhaps I could somehowsee that since $\{0,1,2,i,2i\}$ already is a division ring, the larger set is just created as all possible combinations of elements added together. But would that even be a true statement? (Q2)

Thanks!

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    $\begingroup$ Note that $\{0, 1, 2, i, 2i\}$ is not a ring because it isn't closed under addition. $\endgroup$
    – Thomas
    Commented Feb 11, 2014 at 0:09
  • $\begingroup$ Uhm, I meant once again with modulo $3$, or am I missing something? If not then I'll add that to the question. $\endgroup$
    – Dahn
    Commented Feb 11, 2014 at 0:09
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    $\begingroup$ The element $1+ i$ isn't an element in the set even when you do modulo $3$. $\endgroup$
    – Thomas
    Commented Feb 11, 2014 at 0:10
  • $\begingroup$ Ah, here's for the beginner's mistakes. Thanks. $\endgroup$
    – Dahn
    Commented Feb 11, 2014 at 0:10

1 Answer 1

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The question is whether $\mathbb{F}_3[x]/(x^2 + 1)$ is a division ring. Since it's commutative, the question is whether this is a field.

Proposition: Let $k$ be a field and $p(x) \in k[x]$ be a nonzero polynomial. Then $k[x]/p(x)$ is a field iff $p(x)$ is irreducible in $k[x]$.

This is a good exercise and I recommend it if you're planning on doing much with abstract algebra. In this case, $x^2 + 1$ is irreducible over $\mathbb{F}_3$ (since it has no roots and is a quadratic), so we're done. More explicitly,

$$\frac{1}{a + bi} = \frac{a - bi}{(a + bi)(a - bi)} = \frac{a - bi}{a^2 + b^2}$$

and $a^2 + b^2 \not\equiv 0 \bmod 3$ unless $a \equiv b \equiv 0 \bmod 3$, so this division works.

More generally, you might want to read up on finite fields. Here we've constructed the finite field $\mathbb{F}_9$.

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