I am wondering if someone can help me to look over the following;
It is a question consisting of a few different but related parts. I was asked to consider the finite field with 7 elements, and to list all its squares and cubes in a chart. I did so and I came up with that they are
$\mathbb{Z}/7\mathbb{Z}:$$$(0 , 1 , 2 , 3 , 4 , 5 , 6)$$
Squares: $$(0 , 1, 4, ,2 ,2, 4, 1)$$
Cubes:$$(0, 1, 1 , 6 , 1, , 6 ,6)$$
respectively.
Then I am asked to construct a field with 49 elements.
So I thought, well $x^{2}-3$ has a root of $3$, and is a quadratic , so it will be reducible if and only if it has a root, but from our list of squares we can see 3 is not an option, hence,
$$\mathbb{L}=F_{7}[X] / (x^{2}-3)$$ is a field with $7^{2}=49$ elements
Then I was asked to explain/show and find the solutions to the polynomials $x^{2}-3$ and $x^{2}+x-1$ in this field.
Well I know $x^{2}-3$ certainly can be solved, as this is similar to how we proved that every polynomial can be solved in some field, we construct that field with the polynomial the ideal, which in this case is what we have. Correct me if im wrong , but we could also word it as showing the polynomials $t^{2}-3$ and $t^{2}+t-1$ can be solved, to possibly avoid confusion, because t is really congruence classes, not necessarialy in our first list of numbers $0$ to $6$?
My attempt for the first polynomial is , $(a+bx)^{2}-3=0 $
$a^{2}+2abx+b^{2}x^{2}=a^{2}+2abx+b^{2}(3)$
giving a system, $ (a^{2}+3b^{2}-3)+2(abx)=0$
$a^{2}+3b^{2}-3=0$
$2abx=0$
if $b=0$ then this would imply $a=\sqrt{3}$ which we showed cannot be solved in this field, also since a field is an integral domain this implies that $a=$0 as $2abx=0$ so this leads us to $3b^{2}=3$ or $b^{2}=1$ , looking at our list this corresponds to $b=1$ or $b=6$,
so the solutions (my guess) would be $x$ and $6x$.
But this is part of where I am stuck, i dont know how to go about solving these polynomials?
Final, it asks to construct a field with $7^{3}$ elements and show that $x^{2}-3$ cannot be solved in it. For this I chose $$\mathbb{K}=F_{7}[X]/(x^{3}-3)$$ as it is irreducible so it holds. In terms of showing the polynomial cant be solved, I am stumped on that here as well.
In summary, I understand some of the question, but am having trouble understanding how to solve the polynomials in it, what the elements are etc.
I also know that in $\mathbb{L}$ every element can be written as $ax+b$ where $a,b \in F_{7}$ and similar for $\mathbb{K}$ but $ax^{2}+bx+c$
So I am looking for some help to understand this , etc.
If time is an issue, my main focus is on understanding how to solve the polynomials in $\mathbb{L}$ , the other stuff would be great to, but that is my main priority. Thank you!
For the sake of clarity, let's use $\gamma$ to construct the 49 element field:
$$\mathbb{L}=F_{7}[\gamma] / (\gamma^{2}-3)$$
So now, $\mathbb{L}$ is a field with 49 elements, each of the form $a\gamma + b$, where $a,b\in F_7$.
Now, we're trying to solve polynomials in the ring $\mathbb{L}[x]$, and $\mathbb{L}$ is our coefficient field.
$x^2-3$ has the obvious solutions $\pm\gamma$.
How about $x^2+x+1$? Let's set $x=a\gamma+b$ and expand:
$$(a^2 \gamma^2 + 2ab\gamma + b^2) + (a\gamma+b) + 1 = 0$$
$$(3a^2 + 2ab\gamma + b^2) + (a\gamma+b) + 1 = 0$$
$$(3a^2 + b^2 + b +1) + (2ab + a)\gamma = 0$$
Both components must be zero, so either $a=0$ or $b=3$ ($\gamma$ component). Setting $b=3$ in the other component gives $3a^2 + 6=0$, so $a^2 = -2 = 5$, which has no solution in $F_7$. Setting $a=0$ yields $b^2+b+1=0$, which has solutions 2 and 4 in $F_7$, so these are also our solutions in $\mathbb{L}$. A little anti-climatic, but there it is.
Does this help?