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Let $F = \Bbb Z_2$ and let $f (x) = x^3 + x + 1 \in F[x]$. Suppose that $a$ is a zero of $f (x)$ in some extension of $F$. How many elements does $F(a)$ have? Express each member of $F(a)$ in terms of $a$.

I can't think of a single extension that this would have a root in.

Adam Hughes
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Oliver G
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  • It doesn't really matter what $a$ is, other than that it is a zero in some extension. They are just asking about the structure of $F(a)$. – user332239 Sep 20 '16 at 13:32
  • I guess your difficulty is related to the fact that you need to construct something new here - the field you end up with is not a subset of any of the fields you are likely to have met earlier in your studies. That is precisely why the courses in abstract algebra describe the tool of quotients of polynomial rings for constructing those. First read Adam Hughes' answer (+1). For a little bit more about another way how multiplication/addition work in this particular field you can later take a look at this. – Jyrki Lahtonen Sep 21 '16 at 05:33

1 Answers1

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Since the degree of the irreducible--no roots and low degree--polynomial, $f$, is $3$ we know that the degree of the extension it generates will be $3$. This means that as a vector space, the dimension of $F(a)$ over $\Bbb Z_2$ is $3$, so it is isomoprhic to $\Bbb Z_2 \times\Bbb Z_2\times\Bbb Z_2$ which has $8$ elements. By the first ring isomorphism theorems,

$$F(a)\cong \Bbb Z_2[x]/(x^3+x+1)$$

by the map $\varphi: \Bbb Z_2[x]\to F(a)$ where $p(x)$ maps to $p(a)$, so for example $a+1, a^2+a, a^2+a+1$ et cetera are all elements of $F(a)$.

Adam Hughes
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