$K'$ 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 $h\circ g_2 = g_3$.


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

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

  • $\begingroup$ I surmise that you suggest me to prove that $ker(peval(u'))$ is a maximal ideal. However, because you do not mention $peval(u')$ at all, I can not accept your answer. $\endgroup$ – beroal Aug 12 '11 at 3:54
  • $\begingroup$ @beroal have you even tried to conclude with what I wrote? The point is that you don't need to show $\ker(peval)$ is maximal, and I tell you exactly how to proceed! All you need to show is that the kernel is a prime ideal i.e. the quotient ring is a domain, which it is because it is isomorphic to a subring of a field... It's all written up in my answer. $\endgroup$ – Olivier Bégassat Aug 12 '11 at 9:01
  • $\begingroup$ Actually I do not see your musings as a proof at all. Sorry. @Olivier Bégassat: $\endgroup$ – beroal Aug 12 '11 at 12:57
  • $\begingroup$ @boreal: dear boreal, I have taken time to answer your question, I think you should have the decency to point out what it is in my answer you don't understand. I argue that in order to show that the ideal $(h)$ is maximal, you only need to show that it is a prime ideal, because the ring $F[X]$ is a PID. This is a general theorem valid in all PIDs, a non-zero ideal in a PID is maximal iff it is prime. This can be seen easily in the special case where the PID is a polynomial ring with coefficients in a field. Grant me this for the moment being. $\endgroup$ – Olivier Bégassat Aug 12 '11 at 22:54
  • $\begingroup$ Ok, with this in mind, let's show that $(h)$ is prime, i.e., that the quotient ring $\frac{F[X]}{(h)}$ is a domain. Well, consider the homomorphism of $F$-algebras $\varphi:K[X]\to K,P(X)\mapsto P(u)$ (you call it $peval$). Since $u$ is algebraic over $F$, its kernel is a non-zero ideal of the PID $K[X]$, say $\ker(\varphi)=(h)$. $h$ is by definition a non zero polynomial, and without loss of generality, we can assume $h$ to be unitary, i.e. have leading coefficient equal to $1$. This $h$ is also the minimal polynomial of $u$ over $F$. Indeed, let $\mu(X)$ be $u$'s minimal polynomial. $\endgroup$ – Olivier Bégassat Aug 12 '11 at 23:04

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.

  • $\begingroup$ You proof does not rely on maximal ideals and also gives an algorithm of inversion in $F[u']$. @Dylan Moreland: $\endgroup$ – beroal Aug 12 '11 at 12:27

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