isomorphism between specific generated field and specific quotient ring — gap in a proof $K'$ 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.

 A: I think $3)$ needs some explanation. What has been proven, is that $\frac{F[X]}{(h)}\approx F[u]$, where $F[u]$ stands for the $F$ subalgebra of $K$ generated by $F$ and $u$. What is missing is the fact that $F[u]=F(u)$ when $u$ is algebraic over $F$. This follows form the following fact that is surely found somewhere in your book.

Let $\Lambda$ be a principal ideal domain (i.e. commutative, unital, non zero, domain, and such that every ideal is principal, that is, of the form $(\lambda)$ for some $\lambda\in\Lambda$). You will be interested in $\Lambda =F[X]$. Take $I=(\lambda)$ a non zero ideal, then the following assertions are equivalent: 
  
  
*
  
*$\lambda$ is irreducible,
  
*$I$ is a prime ideal (i.e. $\Lambda / I$ is a domain)
  
*$I$ is a maximal ideal (i.e. $\Lambda / I$ is a field)
  

This allows you to show that $F[u]=F(u)$ when $u$ is algebraic over $F$, since $F[u]$ is a subalgebra of $K$, thus it's a domain, and it follows via the theorem that $F[u]$ is actually a field. Finally, $F[u]\subset F(u)$, and $F(u)$ is the smallest subfield of $K$ containing both $F$ and $u$, and therefore $F(u)=F[u]$.
A: The first two gaps are good to think about, but I think that at this point in an algebra book it's okay to omit them.
There are a few ways to see that the map $F[u'] = F(u')$ (How did you define the minimal polynomial of an algebraic element?). The following might make sense: If $f \in F[x]$ and $f(u') \neq 0$, then the irreducible polynomial $h$ does not divide $f$, and since $F[x]$ is a principal domain there exist $a(x), b(x) \in F[x]$ such that $a(x)h(x) + b(x)f(x) = 1$, and hence $b(u')f(u') = 1$, so $F[u']$ is already a field.
Put briefly, a non-zero prime ideal of a principal domain is maximal, so the quotient is a field.
