I am writing a dissertation currently. This is not a bad thing in itself, I actually rather enjoy writing, however there is a small problem with the subject, and general system for writing. I am a mathematician, and thus my dissertation contains a not insubstantial quantity of mathematics, furthermore, I am a pure mathematician, which means the maths is obscure and abstract, and is therefore represented in obscure and abstract ways.

As such, I have to make use of a typing system called LaTeX, which is extremely good in that it formats documents in a very standardised way, and is able to handle mathematical symbols very well, spacing them accurately and generally doing a good job of making things look correct. However the nature of the program is that instead of type a/b for a fraction, you have to type something like: $frac{a}{b}$, for it to show up correctly. Then you compile the stuff and it makes a nice pdf file which looks as it should.

This sounds like a good idea, and indeed it is. However the result of this approach to typing is that what I write looks like pure evil. Here is a short excerpt from what I have written so far in it’s raw pre-compilation format.

begin{ttt}

Let $i:K to K’$ be an isomorphism of fields, and let $Sigma$ be a splitting field for some $f$ over $K$, and $Sigma’$ be a splitting field for $i(f)$ over $K’$. Then there exists an isomorphism $j:Sigma to Sigma’$ such that $j|_k=i$. That is to say, the extensions $Sigma:K$ and $Sigma’:K’$ are isomorphic.

end{ttt}

begin{pf}

We have the following situation:$$xymatrix{

Kar[r]ar[d]_i & Sigmaar[d]^j \

K’ar[r] & Sigma’

}$$

and we must show first that $j$ exists, and then that it is in fact an isomorphism.

To begin with we will show by induction on the degree of $f$ that there exists a monomorphism $j:Sigma to Sigma’$ such that $j|_K = i$. $Sigma$ is a splitting field for $f$, so $f$ can be written as a polynomial over $Sigma$,

[ f(x) = (x – alpha_1) hdots (x – alpha_n), ]

with $alpha_1, hdots , alpha_n in Sigma$ being the zeros of $f$ in $Sigma$. There will then be a minimum polynomial, $m_{alpha_{1}}$ of $alpha_1$, and this $m_{alpha_{1}}$ will be and irreducible factor of $f$. This $i(m_{alpha_{1}})$ will also divide $i(f)$, and hence will split over $Sigma’$, so it can be written;

[ i(m_{alpha_{1}}) = (x – beta_{alpha_{1}1}) hdots (x – beta_{alpha_{1}r}), ]

with $beta_{alpha_{1}1}, hdots , beta_{alpha_{1}r} in Sigma’$. We also know that $i(m_{alpha_{1}})$ is irreducible over $K’$, so it is a minimum polynomial of $beta_{alpha_{1}1}$ over $K’$. Hence by ref{th3.9}, we have an isomorphism

[ j_1 : K(alpha_1) to K'(beta_{alpha_{1}1}) ]

such that $j_1|_K = i$ and $j_1(alpha_1) = beta_{alpha_{1}1}$. It is clear that $Sigma$ is also a splitting field of $g = f/(x-alpha_1)$ over $K(alpha_1)$, and so by induction we can form an isomorphism

[ j:K(alpha_1, hdots, alpha_n) to K'(beta_{alpha_{1}1}, hdots , beta_{alpha_{n}1}). ]

As we saw in Definition 1.3, $K(alpha_1, hdots, alpha_n)$ is precisely the splitting field for $f$ over $K$, $Sigma$. And since each $beta_{alpha_{1}1}, hdots, beta_{alpha_{n}1}$ is a zero of $i(f)$, $K'(beta_{alpha_{1}1}, hdots, beta_{alpha_{n}1}) subseteq Sigma’$. Hence $j:Sigma to Sigma’$ is a monomorphism, and by construction $j|_{K(alpha_{1})} = j_1$. So we now have that $j|_K = i$.

All that remains to be shown is that this $j$ is in fact an isomorphism. This follows easily, since $j(Sigma)$ is a splitting field for $i(f)$ over $K’$, and is contained in $Sigma’$, and $Sigma’$ is also a splitting field for $i(f)$ over $K’$. So we have that $j(Sigma) = Sigma’$, so $j$ is surjective, and hence is an isomorphism.

end{pf}