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So far I've been working with (simple) extensions of the rationals, so this is the first time I've come across a problem of trying to extend another field. I've come across the notion that $\mathbb{Q}(k) = \{a + bk \mid a, b \in \mathbb{Q} \}$ where $k \not \in \mathbb{Q}$ (otherwise you're not extending anything), and think it'd be the same for any other type of simple extension.

So, specifically, if I want to find $[\mathbb{Z}_3(\frac{1}{2}(1 + i\sqrt{7})): \mathbb{Z}_3]$ (a simple extension), i'd want to look at the basis for the set $\{ a + b\frac{1}{2}(1 + i\sqrt{7}) \mid a,b \in \mathbb{Z}_3 \} = \{a + b + i\frac{b}{2}\sqrt{7} \mid a,b \in \mathbb{Z}_3 \} = \{c + i\frac{b}{2}\sqrt{7} \mid c,b \in \mathbb{Z}_3\}$? Which would be $\{1, \sqrt{7}\}$? Is this a way to characterize all simple extensions?

Edit: This is the full problem that I'm working with:

Calculate the degree of $[\mathbb{Z}_3(\alpha) : \mathbb{Z}_3]$ where $\alpha$ is the root of the polynomial $x^3 + x + 2 \in \mathbb{Z}_3[x]$. (I've determined that the 3 roots are $-1$ and $\frac{1}{2}(1 \pm i\sqrt{7})$.

Thanks for reading!

Bill Dubuque
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Struggles
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    Do you mean to work in $,\Bbb Z_3[x]/(x^2+1)\ $ with $, i := x =\sqrt{-1}?\ $ Else it makes no sense. – Bill Dubuque Feb 07 '19 at 00:34
  • Not that I'm aware of, unless I secretly am without knowing it. I've written out the full statement of the problem I'm trying to solve near the bottom in an edit. – Struggles Feb 07 '19 at 00:49
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    The quadratic factor is $,x^2-x-1,,$ Do you intend to adjoin a root of that? (the polynomial must be irreducible to obtain a field). It seems you may be jumping ahead before you have mastered some of the basics. What textbook are you using? – Bill Dubuque Feb 07 '19 at 01:03
  • This was an given-in-class exercise to be done where the class isn't following any particular textbook. I do have a copy of Dummit & Foote on hand that I've been going through whenever I feel like I need a different explanation. Yes I do intend to adjoin its roots: $\frac{1}{2}(1 \pm i\sqrt{7})$, sorry if I'm being unclear. – Struggles Feb 07 '19 at 01:23
  • Could you please give the exact statement of the problem. Please don't rephrase it in any way. – Bill Dubuque Feb 07 '19 at 01:26
  • Calculate the degree: $[\mathbb{Z}_3(\alpha) : \mathbb{Z}_3]$ where $\alpha$ is the root of the polynomial $x^3 + x + 2 \in \mathbb{Z}_3[x]$. <---- the exact statement – Struggles Feb 07 '19 at 01:34
  • Double check the degree, possibly it is $2$ vs. $3$ (or a typo and should be $2).,$ Then it makes sense as written. Or possibly it was meant to be an irreducible cubic and there is a typo in one of the coef's – Bill Dubuque Feb 07 '19 at 01:37
  • The polynomial is $x^3 + x + 2$ (quintuple-checked to make sure). I'm not sure I understand why there's a possible issue with the given polynomial. Is there something invalid about factoring it to $(x+1)(x^2 - x - 1)$ and adjoining a root of the quadratic to $\mathbb{Z}_3$? That's the stage I'm at in thinking about the question. – Struggles Feb 07 '19 at 01:51
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    The problem is that the field obtained by adjoining "the root" of a reducible polynomial is not well-defined. Are you perchance working with splitting fields? – Bill Dubuque Feb 07 '19 at 01:55
  • Ah I see. In that case, adjoining a root of $x^2 - x - 1 \in \mathbb{Z}_3[x]$ would make sense as an action? Splitting fields were not introduced as a concept when this exercise was given, though have been introduced recently. – Struggles Feb 07 '19 at 02:04
  • Yes, but that's not what the statement says, so something is fishy. You might find helpful this thread Why can't the notation $\mathbb{Q}( \sqrt[4]{4})$ be used? – Bill Dubuque Feb 07 '19 at 02:27
  • Gotcha. I'll have a read. Thanks for sharing your wisdom! – Struggles Feb 07 '19 at 02:35

2 Answers2

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over $\Bbb Z_3$ one root lies in the field, and we have the factorization. $$ x^3 + x + 2 = (x+1)(x^2 + 2x + 2) $$

the quadratic factor is $(x+1)^2 +1^2$ which is irreducible because $3$ is a prime of the form $4n+3$ (Fermat).

if $\alpha \ne 2$ is one root of the quadratic then the other root is $1-\alpha$ since the sum of roots is $1 $ .

hence the extension is $\Bbb Z_3[\alpha]$ of degree 2.

David Holden
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  • Wait a minute, doesn't $x^3 + x + 2 = (x+2)(x^2 - x - 2)?$ I do follow the method of confirming irreducibility, though I don't follow why $\alpha \neq 2$ is necessary, and the necessity of the other root being $1-\alpha$ due to the sum of the roots being $-2$. Do you mind elaborating? – Struggles Feb 07 '19 at 01:47
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    $(x+2)(x^2-x-2)=x^3+x^2-4x-4 \not \equiv x^3+x+2 $ over $ \mathbb Z_3$ but $(x+1)(x^2+2x+2)=x^3+3x^2+4x+2 \equiv x^3+x+2 $ over $ \mathbb Z_3$ – J. W. Tanner Feb 07 '19 at 03:00
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    i've posted a few comments as a separate answer. as Bill remarked, you may be mixing up real arithmetic with the arithmetic of $\Bbb Z_3$ and its algebraic extensions. you may use $i$ for the square root if you wish, but the established significance of $i$ is so central to math, that this may be more of a hindrance than a help to your developing understanding. – David Holden Feb 07 '19 at 03:02
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    Oh my goodness... It appears I need to revisit first semester algebra with my habit of forgetting about modulo arithmetic.

    So the method of thinking is: Factoring the polynomial under $\mathbb{Z}_3$ gives us one root in the field, and an irreducible quadratic. Then extending $\mathbb{Z}_3$ by one of these roots would mean it's a degree 2 extension since the minimum polynomial of that root is a quadratic? Also, that the sum of the roots of a quadratic is equal to $\frac{b}{a}$ for a quadratic $ax^2 + bx + c$?

     – Struggles Feb 07 '19 at 03:26
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    precisely! btw like you, i tend to forget, but the upside is that often, when you look again at some rusty topic you find your insight goes a bit deeper. some of these ideas took decades to develop in the first place – David Holden Feb 07 '19 at 04:24
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in $\Bbb Z_3$ we have $-1 \equiv 2$ so apparently different forms of an equation can be equivalent over any finite field. note also that $-7 \equiv -4 \equiv -1$ so if you wish to solve using the quadratic formula, then the roots are:

$$ \frac{1 \pm "\sqrt{2}"}2 $$

the quotes indicate that we are not dealing with the $\sqrt{2}$ of real arithmetic, but a root over $\Bbb Z_3$ of the equation $$ x^2 - 2 = 0 $$

David Holden
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