Isomorphisms between finite fields.

Question

Let $P(X) \in \mathbb{F_{5}}[X]$ be an irreducible monic polynomial of degree $2$.

Prove that the quotient $\frac{\mathbb{F_{5}}[X]}{(P(X))}$ is isomorphic to the field $\mathbb{F_{25}}$ and that $P$ has two roots in $\mathbb{F_{25}}$.

I have no idea how to start this. I figure it has to do with the cardinality of $\frac{\mathbb{F_{5}}[X]}{(P(X))}$ being $5^{2} = 25$, but I don't know whar arguments to use.

Answer 1

Since $P$ is irreducible of degree 2, $k=F_5[X]/(P)$ is an extension of degree 2 of $F_5$ so it is $F_{5^2}$. The image of $X$ in $K$ is a root of $P$, since degree P=2, and $P$ has a root in $K$, $P$ splits in $K$.

Answer 2

Because fields are UFDs, we know that polynomials over the field is also a UFD, so since all elements of the group $F^\times$ satisfy $x^{|F^\times|}-1=0$ by Lagrange, and there can be at most $n$ solutions to any polynomial of degree $n$ because polynomials factor into linear factors over their algebraic closures, it must be that the finite group $F^\times$ (for a finite field, $F$) is cyclic. But then map a generator of the unit group of $\Bbb F_5[x]/(p(x))$ to a generator of the unit group of $\Bbb F_{25}$ and map $0\to 0$ and clearly this is a ring homomorphism and a bijection, hence an isomorphism.

To see there are two roots, simply note there is definitely one root, and so $p(x) = (x-\alpha)g(x)$ but then $\deg g(x) = 1$ because the degree of the product is the sum of the degrees, hence $g(x) = x-\beta$ and that shows the other root is also there.

Answer 3

When a polynomial of degree two has a root $a$ in a field extension, then it has both, because you can divide by $X-a$ remaining in the field.

The field $\mathbb{F}_{5}[X]/(P(X))$ has dimension $2$ over $\mathbb{F}_5$, so it has $25$ elements.

Now the problem is proving that any two fields with $25$ elements are isomorphic.

Consider a splitting field $K$ for $Q(X)=X^{25}-X$. Note that the roots of $Q(X)$ are distinct, because the (formal) derivative is $Q'(X)=-1$. The set of roots of $Q(X)$ is actually a subfield, because $$ (a+b)^{25}=((a+b)^5)^5=(a+b)^5=a+b $$ so if $a$ and $b$ are roots of $Q(X)$, then also $a+b$ is. Similarly for checking about $ab$, $-a$, $a^{-1}$, $0$ and $1$. Therefore $K$ is the set of roots of $X^{25}-X$.

Conversely, in a field with $25$ elements we have $a^{25}=a$ for all $a$, because the multiplicative group has order $24$ and therefore $a^{24}=1$ if $a\ne0$.

Thus a field with $25$ elements is the splitting field of $Q(X)$ and it's a general theorem that splitting fields of the same polynomial are isomorphic.

Of course this is a particular case of the general theorem that a field with $p^n$ elements is the splitting field over $\mathbb{F}_p$ of $X^{p^n}-X$, so there is a unique, up to isomorphisms, field with $p^n$ elements. The proof is the same with the suitable changes.