Joe Gurr

"Fields are not as well-behaved categorically as most other common algebraic structures" - nCatLab

Recall that a morphism \(f: X \to Y\) in a category is a monomorphism if for any \(g, h: Z \to X\) then \( fg = fh \implies g = h\).

We want to understand what are the monomorphism in the category of fields.

Let Field denote the category of fields. Recall that Ob(Field) are fields and Mor(Field) are field homomorphisms.

Let \(f:X \to Y\) be a morphism. As \(f\) is a field homomorphism we know that the kernel of \(f\) is an ideal.

The only ideals in \(X\) are \(\{0\}\) and \(X\). If ker(\(f\)) was the whole field, then \(f\) would be the zero morphism, which is not a field homomorphism.

So ker(\(f\)) \( =\{0\}\), which tells us that \(f\) is injective.

Now, let \(h, k\) be a pair of parallel morphisms from \(Z\to X\) such that composition with \(f\) is sensible and \(fh = fk\).

Towards a contradiction, suppose that \(h \neq k\), then there exists an \(z \in Z\) such that \(h(z) \neq k(z)\).

As \(f\) is injective would then have \(f(h(z)) \neq f(k(z))\) violating the assumption that \(fh=fk\).

Therefore we must have \(h=k\) meaning that \(f\) is a monomorphism. Since our choice of \(f\) was arbitrary we have that all morphisms in Field are monomorphisms. \(\blacksquare\)