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Let's consider the polynomial $x^d-1$. Theory tells us that it can have at most $d$ roots in (any extension of) a given field.

Here's my problem: let $A$ be the vector space spanned by $1,a,a^2,b,b^2,ab$ over $\mathbb{Z}_3$ (the integers mod $3$). $A$ is itself a field if we define $x^i*x^j=x^{i+j\pmod 3}$.

Choosing $d=2$ we should have that, at most, $2$ elements in $A$ should be equal to $1$...but I can count more than $2$ (at least $a^2,b^2,a^2b^2$). May somebody help me to point out the error in this reasoning?

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Manlio
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  • How do you define multiplication over these vectors? What is $a \cdot a^2$? – Ben Grossmann Mar 16 '14 at 16:49
  • $a\cdot a^2 = a^{3~mod~3} = a^0 = 1$ – Manlio Mar 16 '14 at 16:54
  • What are $;a,b;$ ? And if this is a field, what is the neutral element wrt addition...and what is the addition here? – DonAntonio Mar 16 '14 at 17:08
  • Addition can be defined as polynomial one, remembering that coefficients are in $\mathbb{Z}_3$, so $\lambda a + \mu a = ((\lambda + \mu)~mod~3) a$. $0$ is the neutral element wrt addition. $a,b$ are two generic elements, their sum remains $a+b$ – Manlio Mar 16 '14 at 17:19
  • @Saphrosit , where did you get this construction from? It's rather confusing (for me, of course. Perhaps not for others). Do you mean you have to take all the finite linear combinations of $;1,a,a^2,b...;$ with coefficients of $;\Bbb Z_3;$ ? – DonAntonio Mar 16 '14 at 17:27
  • @DonAntonio Sorry if it's confusing, it is just something I thought about while studying finite fields. Yes, I was thinking about all linear combinations with coefficients in $\mathbb{Z}_3$, sorry if it's not clear from my question. – Manlio Mar 16 '14 at 17:38

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This ring you have constructed is not a field, since we have (non-zero) zero-divisors. Note that $$ (a - 1)(a^2 + a + 1) = a^3 - 1 = 0 $$ The unique field (up to isomorphism) of size 27 is given by $$ F_{27} = \mathbb{Z}_3[t]/(t^3 + t^2 + t + 2) $$ The ring you have constructed can be thought of as $$ R = \mathbb{Z}_3[a,b]/((a^3-1)(b^3-1)) $$

Ben Grossmann
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  • According to what he wrote, perhaps he meant a field with $;3^6;$ elements, not $;3^3;$ ...who knows? – DonAntonio Mar 16 '14 at 17:28
  • @DonAntonio I figured the closest approximation for what he was attempting was the field over $\mathbb{Z}_3$ constructed by modding out a cubic polynomial – Ben Grossmann Mar 16 '14 at 17:43
  • @Omnomnomnom thank you very much, you're perfectly right. Can you please show me how did you get to the equivalent formulation for the ring I defined? – Manlio Mar 16 '14 at 18:05
  • @Saphrosit there's not much to it. $\mathbb{Z}_3[a,b]$ is the (commutative) ring of all polynomials on $a$ and $b$ with coefficients in $\mathbb{Z}_3$. In the quotient, I have the "defining relations" $a^3 - 1 = 0$ and $b^3 - 1 = 0$ (verify that $(ab)^3 = 1$ follows). As in group theory, a (polynomial) ring is completely determined by its generators and some set of relators. Here is a decent explanation. – Ben Grossmann Mar 16 '14 at 18:38