In algebraic number theory, the existence of a Frobenius element at any prime $p$ in a Galois extension $K/{\mathbf Q}$ is crucial. That is, for any prime ideal $\mathfrak p$ lying over $p$ in $K$ there is some $\sigma \in {\rm Gal}(L/K)$ which looks like the $p$-th power map mod $\mathfrak p$: $$ \sigma(\alpha) \equiv \alpha^p \bmod \mathfrak p $$ for all $\alpha$ in the integers of $K$. (This can be jazzed up to the relative case, but I'll keep the base field as $\mathbf Q$ for simplicity here.)
In any modern reference I have seen which shows the existence of $\sigma$, first the decomposition field is introduced in order to make a reduction to the case where the base field is the decomposition field. But if you look at the original proof by Frobenius (1896) it is different, using multivariable polynomials in an interesting way and there is no decomposition field. The argument fits in one page; see http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/frobeniuspf.pdf, where I consider a fairly general setup using the method of Frobenius. This nice proof by Frobenius has been completely forgotten, even though it handles the general case. (Frobenius himself worked with base field ${\mathbf Q}$.)
What is the mathematical insight here? That you can prove this theorem with just your bare hands, so without having to speak. It mention decomposition fields (which also means that makes it easier for students new to the subject to follow the proof). I found this essential when you're teaching a course on algebraic number theory you do since it meant I did not have to introduce decomposition fields in the lectures at all; they could safely be left to homework assignments, if I so earlychose. The proof is also a nice illustration of the usefulness of multivariable polynomials, especially considering that a lot of basic algebraic number theory only requires polynomials in one variable.
