Finite field adjunction - isomorphism that maps from the field into a polynomial ring factored over an ideal

Question

A field $(K, +, 0, *, 1)$ is given, and it is a finite field with $25$ elements. $P$ is the prime field over K. Further it was given that a polynomial $r(x):=x^2+x+1 \in P[x]$ is irreducible.

Now, the polynomial $y^2+y+1\in K[x]$ has two non-equal zeros in $K$, say $\alpha_1 \neq \alpha_2$.

The question is to describe all possible isomorphisms: $$ \varphi: K \rightarrow P[x]/(r) $$ by giving the values of $\varphi (\alpha_1)$ and $\varphi (\alpha_2)$.

There are a few things that I don't understand about this question.

First of all, what are the properties of $K[y]$ apart from it being a polynomial ring? I think that the additive group $(K, +, 0)$ should be isomorph to a cyclic group $C_5 \times C_5$ - can that help me in any way?

Then - when I try to find the $\alpha_1$ and $\alpha_2$ I do my calculations in the Modulo 25 arithmetic - is that right (because I failed to find the zeros)?

And since the $\varphi(k) = k + (r)$ is one of the possibilites to describe this isomorphism, my question is - are there any more of them? Because my textbook doesn't state that this isomorphism is unique.

Any kind of help - or a link to some other question is more than welcome.

Answer 1

The polynomial $r$ has two roots in $K$, $\alpha_1$ and $\alpha_2$. Their images must also be roots of the polynomial $r$ in $P[x]/(r)$. The notation here can be confusing since $x \in P[x]/(r)$ is an element of the field.

Let's compute the roots of $y^2+y+1$ in $P[x]/(r)$. They must be of the form $ax+b$ for $a,b\in P$, which is a root if and only if $a^2x^2 + 2abx + b^2 +ax + b + 1 = a^2x^2 +(2ab+a)x + (b^2+b+1)= 0$. This must be divisible by $x^2+x+1$, and $a$ is non-zero since $r$ is irreducible over $P$, so $a^2 = 2ab+a = b^2+b+1$. Hence $a =2b + 1$ and $(2b+1)^2 = b^2 +b+1$, so $b(b+1) = 0$. Hence the roots are $x$ and $-x-1$.

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This is somewhat ad-hoc. This can be clarified and generalized by thinking about homomorphisms $P[x]/(r) \to K$, which are determined by where $x$ goes. The image of $x$ must also be a root of $r$, so the only possibilities are if the image of $x$ is $\alpha_1$ or $\alpha_2$. These actually define homomorphisms since $r$ is irreducible, so the kernel of the evaluation map $P[x] \to K$, $x \mapsto \alpha_i$ is $(r)$.

Answer 2

This is not in any way an answer to your question.

Your elements of the field $K$ are expressions of the form $a+bx$ with $a,b\in P$. You add them in the obvious way, always treating $a$ and $b$ as things modulo $5$, and you multiply them by replacing $x^2$ by $-1-x=4+4x$ every time that you see an $x^2$.

Why don’t you take an element like $2+3x$ and write down a few of its first powers? I mean $(2+3x)^2$, $(2+3x)^3$, etc. If you’re enthusiastic, why don’t you write down all its powers, up to the twenty-fourth?

I promise you that by this exercise you will gain much more insight into the situation than by contemplating head-on the question put to you.