Finding the subfields of a field

[Mr Davis 97]

Homework Statement

Suppose that $p(x)$ is an irreducible polynomial of prime degree in $F[x]$. What are the subfields of $K=F[x] / \langle p(x) \rangle$?

Homework Equations

The Attempt at a Solution

For some reason I am not seeing how to approach this problem. Some pointers would be helpful.

 

[fresh_42]

What is $[\,K\, : \,F\,]$ and what does the degree formula tells you, which you have read some minutes ago in the other thread?

 

[Mr Davis 97]

What is $[\,K\, : \,F\,]$ and what does the degree formula tells you, which you have read some minutes ago in the other thread?

Is $[\,K\, : \,F\,] = q$, where $q$ is the degree of $p(x)$? Does the degree formula show that the only subfields are $K$ and the trivial subfield because $q$ is prime and can't be decomposed into factors?

 

[fresh_42]

Yes. Although I wouldn't call $F$ the trivial subfield. And you need the irreducibility of $p(x)$, too, for otherwise the degree of the field extension would be less than $q$.

 

[Mr Davis 97]

Yes. Although I wouldn't call $F$ the trivial subfield. And you need the irreducibility of $p(x)$, too, for otherwise the degree of the field extension would be less than $q$.

One more slightly related thing, that I don't really want to make another thread for. I know that $\mathbb{Z} / \langle x^2 + 2x + 2\rangle$ is isomorphic to $\mathbb{Z} / \langle x^2 + x + 2\rangle$, because they are finite fields with order as a power of a prime, since they both have order $3^2 = 9$. I'm just curious, is there a standard way that I could find an explicit isomorphism?

 

[fresh_42]

One more slightly related thing, that I don't really want to make another thread for. I know that $\mathbb{Z} / \langle x^2 + 2x + 2\rangle$ is isomorphic to $\mathbb{Z} / \langle x^2 + x + 2\rangle$, because they are finite fields with order as a power of a prime, since they both have order $3^2 = 9$. I'm just curious, is there a standard way that I could find an explicit isomorphism?

Let me guess: you meant $\mathbb{Z}_3[x] / \langle x^2 + 2x + 2\rangle$ and $\mathbb{Z}_3[x] / \langle x^2 + x + 2\rangle$?
I would determine the roots of the polynomials and map root to root. Say $\xi^2+2\xi +2 = 0$ and $\eta^2+\eta +2 =0$ then $\varphi(\xi):=\eta$ (and $\varphi(a)=a$ for $a\in \mathbb{Z}_3$) should do.

 

[Mr Davis 97]

Let me guess: you meant $\mathbb{Z}_3[x] / \langle x^2 + 2x + 2\rangle$ and $\mathbb{Z}_3[x] / \langle x^2 + x + 2\rangle$?
I would determine the roots of the polynomials and map root to root. Say $\xi^2+2\xi +2 = 0$ and $\eta^2+\eta +2 =0$ then $\varphi(\xi):=\eta$ (and $\varphi(a)=a$ for $a\in \mathbb{Z}_3$) should do.

But aren't the polynomials irreducible?

 

[fresh_42]

Irreducibility depends on where. So, yes, they are irreducible in $\mathbb{Z}_3[x]$, but they are not in $\mathbb{Z}_3[\xi][x]$, resp. $\mathbb{Z}_3[\eta][x]$. Of course, $\xi$ and $\eta$ are artificial numbers, but so is, e.g. $\sqrt{2}$ which also is only an abbreviation of a number that satisfies a polynomial equation, in this case $x^2-2=0$.

 

[Mr Davis 97]

Irreducibility depends on where. So, yes, they are irreducible in $\mathbb{Z}_3[x]$, but they are not in $\mathbb{Z}_3[\xi][x]$, resp. $\mathbb{Z}_3[\eta][x]$. Of course, $\xi$ and $\eta$ are artificial numbers, but so is, e.g. $\sqrt{2}$ which also is only an abbreviation of a number that satisfies a polynomial equation, in this case $x^2-2=0$.

I see. I am little confused about the function definition though. For example, where would it send the element $(ax+b )+ \langle x^2 + 2x +2 \rangle$ for example?

 

[fresh_42]

I see. I am little confused about the function definition though. For example, where would it send the element $(ax+b )+ \langle x^2 + 2x +2 \rangle$ for example?

Let's see:

$(ax+b )+ \langle x^2 + 2x +2 \rangle$ corresponds to the element $a\xi +b \in \mathbb{Z}_3[x]/\langle x^2+2x+2 \rangle$.
So $\varphi(a\xi +b) = \varphi(a)\varphi(\xi)+\varphi(b)=a\eta +b \in \mathbb{Z}_3[x]/\langle x^2+x+2 \rangle$ which corresponds to the polynomial class $(ax+b )+ \langle x^2 + x +2 \rangle$.

To be sure, one would have to check, whether $\varphi((ax+b) \cdot (cx+d))=\varphi(ax+b)\cdot \varphi(cx+d)$ and $\varphi((ax+b) + (cx+d))=\varphi(ax+b) + \varphi(cx+d)$ hold. However, it is better to operate with $\xi$ and $\eta$, because then it is clear, that they satisfy two different equations, whereas the difference in the multiplications of the polynomials isn't as obvious.
So better to check $\varphi((a\xi +b) \cdot (c\xi + d))=(a\eta +b)\cdot (c\eta +d)$ and $\varphi((a\xi +b) + (c\xi +d))=(a\eta +b) + (c\eta+d)$.

Things become more difficult in extensions of higher degrees. But in the end, it's only a basis change of $\mathbb{Z}_3$ vector spaces. Since $\varphi(a)=a$ for all $a \in \mathbb{Z}_3$, there is not much choice left rather than to map $\xi \mapsto \eta$ or on its conjugate, which would be the other root (that can be found by the division $(x^2+x+2)\, : \,(x-\eta)\, = \, \ldots\;\;$ Here's an example of how to do this with abstract numbers: https://www.physicsforums.com/threa...r-a-polynomial-over-z-z3.889140/#post-5595083).