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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:

  1. 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?
  2. 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.
  3. 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

  1. S. MacLane, G. Birkhoff. Algebra.

2 Answers 2

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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 $F(u)=F[u]$.

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    There is a typo$I$can't correct anymore two comments above,$I$meant to write "Take $\bar{P}=P(X)+I\in\frac{F[X]}{(h)}\setminus\lbrace 0 \rbrace$. Also, in the post where$I$start showing that non zero prime ideals are maximal in $F[X]$, it should read "... and any *non-zero* ideal $0\neq I=(h)\subset F[X]$,..."2011-08-12
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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.

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    You proof does not rely on maximal ideals and also gives an algorithm of inversion in $F[u']$. @Dylan Moreland:2011-08-12