I am working through some algebra exercises and got stuck with the following problem:
I am working in the finite field $\mathbb{Z}_5$. Let $f \in \mathbb{Z}_5[x]$ and $f(x) = x^2 + 2$. By simple trial and error, I checked that this polynomial is irreducible, hence $\mathbb{Z}_5/(f) \cong GF(5^2)$, denoting the Galois field ($f$ is of degree $2$).
Let $L$ be a field extension of $\mathbb{Z}_5$ with $f(\alpha) = 0$ for $\alpha \in L$.
Is there a monomorphism $\varphi: \mathbb{Z}_5[x]/(f) \rightarrow L$ which maps $x + (f) \mapsto \alpha$?
My ideas so far: I checked that the multiplicative order of $x + (f)$ is $8$, which is a proper divisor of $24$, the order of the unit group of $GF(5^2)$. Injective homomorphisms do preserve orders, hence $\alpha$ must be of order $8$. This means that $\alpha^8 = 1$. $1$ is a solution as well, hence I perform a polynomial division by $(\alpha-1)$ and get $\alpha^7 + \cdots + \alpha + 1$ is the irreducible polynomial of this field extension. Is this of any help? How can I get to a conclusion?
When teaching this topic I emphasize the following route. I am not sure it is the most illuminating for all students, but the same approach comes in handy when explaining the emergence of automorphisms in Galois theory, so it is economical.
You may recall that evaluating a polynomial at a point gives a homomorphism of rings. More specifically, if $L$ is an extension field of $K$, and we fix an element $z\in L$, then the mapping $$ ev_z: K[x]\to L, ev_z(f(x))=f(z) $$ that calculates the value of a polynomial at $z$ is a homomorphism of rings. (This is nearly always explained first in the case where $z\in K$, but we easily see that it actually works with any commutative ring containing $K$ in its place.)
The other ingredient we need is the basic homomorphism theorem (don't know if it has an established name in English): If $f:R\to R'$ is a homomorphism of rings, and $I=\operatorname{ker}(f)$, then there is an isomorphism of rings $R/I$ and $\operatorname{im}(f)$ defined by $a+I\mapsto f(a)$. Equivalently, we get a well defined injective homomorphism of rings $F:R/I\to R'$ such that $F\circ p=f$, where $p:R\to R/I$ is the natural projection.
How to apply that here? Clearly $\Bbb{Z}_5$ should take the role of $K$, and $\alpha$ should take the role of $z$, so we have a ring homomorphism $$ev_\alpha:\Bbb{Z}_5[x]\to L, f(x)\mapsto f(\alpha).$$
So far so good, but the domain is not what we wanted. Let's think about the kernel of $ev_\alpha$, denote it by $I$. It is an ideal of $\Bbb{Z}_5[x]$ because the kernel of any ring homomorphism is. More or less by definition we also see that $f(x)=x^2+2\in I$. Therefore the ideal generated by $f(x)$, call it $J$, must be contained in $I$, so $J\subseteq I$.
At this point we already have a homomorphism of rings $\Bbb{Z}_5[x]/J\to \Bbb{Z}_5[x]/I\to L$ gotten by composing the projection $g(x)+J\mapsto g(x)+I$ with $ev_\alpha$. But you were asked to produce an injective homomorphism! The last hurdle we need to clear is to prove that we actually have equality $I=J$. How to see that? You already proved that the polynomial $f(x)=x^2+2$ is irreducible. We can then recall a theorem (a consequence of the division algorithm of polynomials) saying that an irreducible polynomial (over a field $K$) generates a maximal ideal of $K[x]$. This implies that the ring $\Bbb{Z}_5[x]$ has no ideals strictly between $J$ and $\Bbb{Z}_5[x]$. Because the constant polynomial $1$ is not in the kernel $I$, we can deduce that $I$ is not all of $\Bbb{Z}_5[x]$. We are left with $I=J$ as the only remaining alternative. In other words, we are done!