# Construct an isomorphism between fields

The first part of my question asked : State all the irreducible Polynomials in $\mathbb{Z}_2[x]$ of order 3.

I was able to do this and get the following polynomials :

$x^3 + x^2 + x + 1 \Rightarrow$ reducible
$x^3 + x^2 + 1 \Rightarrow$ irreducible
$x^3 + x + 1 \Rightarrow$ irreducible
$x^3 + 1 \Rightarrow$ reducible

and the other polynomials in $\mathbb{Z}_2[x]$ are trivially reducible.
I set $f=x^3 + x^2 + 1$, $g=x^3 + x + 1$.
I then have to take $\mathbb{Z}_2[x]/f$ and $\mathbb{Z}_2[x]/g$ and construct an isomorphism between the two.

My real problem is understanding what these two fields $\mathbb{Z}_2[x]/f$ and $\mathbb{Z}_2[x]/g$ really are- and how I can build an isomorphism between them.

If anyone could help I'd be very grateful :)

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First of all, you can treat $\mathbb{F}_2[x]/(f)$ just as a ring. As you know, a quotient of a ring is a set of cosets with the obvious operations of addition and multiplication inherited from the big ring. It just so happens that quotients of integral domains by maximal ideals are also fields, but you needn't be too worried about that at the moment. Now, to write down an isomorphism between quotient rings, the most naive way of doing it would be to write down an isomorphism between the big rings (i.e. in this case between $\mathbb{F}_2[x]$ and itself) that takes the one ideal bijectively to the other, i.e. $(f(x))$ to $(g(x))$. Again, the most naive way of accomplishing that would be to just send one generator to the other. See if you can do that and convince yourself that this procedure really does always give you an isomorphism of quotient rings.

Another Exercise: if two isomorphic rings turn out to be fields, why is an isomorphism of rings also an isomorphism of fields?

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rtel thanks very much :) can i just check, when you say one generator to the other do you mean my function f to my function g? as in a bijection from x^3+x^2+1 to x^3+x+1? – Lucy Marshall Jan 24 '11 at 9:28
@Lucy Exactly. You want to send the element $f$ of $\mathbb{F}_2[x]$ to the element $g$, then you will also bijectively map the respective ideals they generate one to the other. – Alex B. Jan 24 '11 at 10:55

The first field looks like this: \begin{align*} \mathbb{Z}_{2}[x]/\langle f(x)\rangle &= \{ p(x) + \langle f(x)\rangle\mid p(x)\in\mathbb{Z}_2[x],\ \deg(p)\lt\deg(f)\}\\ &= \left\{ \begin{array}{l} 0 + \langle f(x)\rangle,\quad 1 + \langle f(x)\rangle,\quad x + \langle f(x)\rangle, \quad x^2 + \langle f(x)\rangle,\quad (1+x) + \langle f(x)\rangle,\\ (1+x^2) + \langle f(x)\rangle,\quad (x + x^2) + \langle f(x)\rangle, \quad (1+x+x^2) + \langle f(x)\rangle \end{array}\right\} \end{align*}

You know that you have found all elements by counting them as follows:

You have $|\mathbb{Z}_{2}| = |{ 0, 1}| = 2$ choices for any coefficient.

In any polynomial, you can have terms of degree $0$, $1$ or $2$, each such term can have any of the two possible coefficients.

So your resulting field has $2 \times 2 \times 2 = 2^3 = 8$ elements.

Regarding notation: $\langle f(x)\rangle$ in the above means the ideal generated by $f$. This means, the set of all elements $p(x) \in \mathbb{Z}_{2}[x]$ that are multiples of $f$.

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Regarding factor (or quotient) rings in general: If $F$ is a field and $I$ an ideal of $F$, then $F/I$ is only a field, if $I$ is a maximal ideal in $F$. In the case where $I=<p(x)>$ for some polynomial, $<p(x)>$ is maximal if and only if $p(x)$ is irreducible. – Rudy the Reindeer Jan 24 '11 at 16:27
A very good and easy to read book about undergrad algebra is J.A.Gallian's "Contemporary Abstract Algebra". This recommendation just in case you haven't got a book already. – Rudy the Reindeer Jan 24 '11 at 16:30
You have got your arithmetic wrong, as you can see from general algebraic facts: a finite field has order a prime power, so certainly not 6. – Alex B. Jan 24 '11 at 17:21
Thanks, you're right, I'm sorry. Very embarrassing. – Rudy the Reindeer Jan 24 '11 at 17:24
Matt, you confusingly typed $F$ instead of $F[x]$ three times in your comment . Please, correct that. – Georges Elencwajg Jan 25 '11 at 21:27