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I want to find a explicit morphism from $K_1$ to $K_2$, where

$$K_1=\mathbb F_2[x]/(x^3+x+1)\mbox{ and }K_2=\mathbb F_2[x]/(x^3+x^2+1).$$

I've found that it must exist because these polynomials are irreducible, hence these fields have $2^3$ and $2^3$ elements and $3\mid 3$.

But how could I find the explicit morphism?

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    You obviously mean $\mathbb{F}_2[x]$ then.2012-10-30
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    yes, I'm sorry.2012-10-30

4 Answers 4

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Since the multiplicative group $\mathbb F_8^*$ is the cyclic group of order $7$, any nontrivial element generates the full group. For example, in $K_1$, the group is $\{1,x,x^2, \ldots x^6\}$. Next, we know that the Frobenius map $x \mapsto x^2$ is a field automorphism, so that $x,x^2,x^4$ must have the same minimal polynomial, and likewise with $x^3,x^5,x^6$. Since the minimal polynomial of $x$ is $X^3+X+1$, this means that $x^3,x^5,x^6$ are the three roots of $X^3+X^2+1$ in $K_1$. Thus sending the $x$ of $K_2$ to any of those will work.

For example the map $K_2 \to K_1$ given by $x \mapsto x^{-1} (= x^6)$ is an isomorphism (and now we notice that the second polynomial was the first polynomial with the coefficients reversed)

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    +1: this is a very clear and beautiful explanation, mercio.2012-10-30
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    I don't understand the concept of minimal polynomial..2012-10-30
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    $X^3+X+1$ is an element of $K_1$, right? and $x^3$ is a root of $X^3+X^2+1$? How do you know that $x$ is a root?2012-10-30
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    @ Kits89, no $X^3+X+1$ is a polynomial with coefficients in $\mathbb F_2$, so it's an element of $\mathbb F_2[X]$. When we apply it to the class of $x$ in $K_1$, we obtain the class of $x^3+x+1$ in $K_1$, which happens to be the zero element of $K_1$. Hence, the class of $x$ in $K_1$ is a root of the polynomial $X^3+X+1$.2012-10-30
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    Okay! Then the class of $x$ is the algebraic element on $\mathbb F_2[x]$ of the field extension: $\mathbb F_2[x] \subseteq \mathbb F_2[x]/(x^3+x+1)$.2012-10-30
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    But using the same argument in your comment, the class of $x$ is also root of the other polynomial...2012-10-30
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    No, when you apply $X^3+X^2+1$ to the class of $x$ in $K_1$, you get the class of $x^3+x^2+1$, which is not the trivial class in $K_1$ because it's not a multiple of $x^3+x+1$. However, if you apply it to the class of $x$ in $K_2$, you get zero. When you build $\mathbb F_2[x]/P(x)$, this is designed to have a root (the class of $x$) of $P$ and this is called "adjoining a root of $P$ to $\mathbb F_2$"2012-10-30
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    I don't doubt that your answer is correct but I won't check it until I fully understand it ;)2012-10-30
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Hint: a finite field homomorphism is a fortiori a group homomorphism on its nonzero elements, which form a cyclic group. So, if you pick a primitive generator in each, you can map one generator to the other, and the rest you get for free.

Added: Lubin kindly reminded me to mention that this is necessary but not sufficient to get a field homomorphism. To make it sufficient, you additionally need to ensure the primitive roots you pick belong to the respective irreducibles you are making your fields from.

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    Sorry, this is not correct. You must map your first generator to another that is a *root of the same irreducible*. In OP’s example, call the indeterminate $y$ in $K_2$ instead of $x$. Then I'm guessing that he has to send $x$ to $y+1$. Check to see that this last is a root of the first irreducible and you’re done.2012-10-30
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    @Lubin It's a case of me saying more than I intended in a hint :)2012-10-30
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The map $K_2\rightarrow K_1$ is obtained by mapping $x\rightarrow x+1$.

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Hint: Both fields are isomorphic to $\mathbb{F}_{2^{3}}$ and explicitly the elements in both of them are of the form $[a+bx+cx^{2}]$ (by the Euclidean algorithm) (where $a,b,c\in\mathbb{F}_{2})$.

So any homomorphism $\mathbb{F}_{8}\to\mathbb{F}_{8}$ will be a good answer - given that you have explicitly isomorphisms between $\mathbb{F}_{8}$ to those fields. such isomorphism can be found by finding the multiplication table of both field and comparing it with the one of $\mathbb{F}_{8}$.

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    This is a funny answer: the question is «how to find an explicit isomorphism?» and you answer «well, any isomorphism will work, you just have to find it» :-)2012-10-30
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    @MarianoSuárez-Alvarez - but I did say that it can be done by writing the multiplication tables, its just hard work...after that its easy2012-10-30