K' is a field extension of $F$, $h\in F[x]$, $h$ is minimal for u'\in K', F(u') is a field generated by F\cup \{u'\}, K'=F(u'). In [1. XIII. Galois theory. 2. Algebraic and transcendental elements. Theorem 1.] F[x]/(h)\cong F(u') is proved in the following way (the underlined text is mine):
…substitution of u' for $x$ in the polynomial ring gives a homomorphism \underline{peval(u'):} F[x]\to F(u') with kernel $(h)$ and hence (by universality properties of the quotient ring) an isomorphism F(u)=F[x]/(h)\cong F(u').
IMHO there are gaps in the proof:
- I suppose that the proof relies on the fact that initial objects are isomorphic. Then the universal property of the quotient ring given in [1. III. Rings. 3. Quotient rings. Theorem 8. Main theorem on quotient rings.]. Instead, my formulation below is applicable. Is my formulation presented somewhere else?
- We still must prove that ker(peval(u'))=(h). Because $h$ is irreducible and h(u')=0, f(u')=0 \to h\mid f. Maybe this is considered trivial.
- We still must prove that peval(u') is surjective. I can not find any trace of a proof. This is not trivial, because F(u') is a generated field, but elements of $F[x]$ are polynomials, and polynomials consist of ring operations. We must somehow convert every field term into a ring term.
Am I correct?
The universal property of the quotient ring
Let $R_0, R_1$ be rings. Every surjective homomorphism $f:R_0\to R_1$ is an initial object in the following category:
- object: $(R_2, g)$ such that $g:R_0\to R_2$ and $ker(f)\subseteq ker(g)$;
- morphism: $h:(R_2, g_2)\to (R_3, g_3)$ is a function $h:R_2\to R_3$ such that $h\circ g_2 = g_3$.
References
- S. MacLane, G. Birkhoff. Algebra.