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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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    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.2011-08-12
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    @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.2011-08-12
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    Actually I do not see your musings as a proof at all. Sorry. @Olivier Bégassat:2011-08-12
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    @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.2011-08-12
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    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.2011-08-12
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    Since $\varphi(\mu)=\mu(u)=0$, we have $\mu\in \ker (\varphi)=(h)$ i.e. $h|\mu$. On the other hand, if $h(X)=Q(X)\mu(X)+R(X)$ is the euclidean division of $h$ by $\mu$, with $Q(X),R(X)\in F[X]$ and $\mathrm{deg} (R(X))<\mathrm{deg} (\mu(X))$ by definition of euclidean division, upon applying $\varphi$ to this equation you get $0=\varphi(h)=\varphi(\mu)\times\varphi(Q)+\varphi(R)=0\times\varphi(Q)+\varphi(R)=\varphi(R)$. Thus, $R$ annihiliates $u$ but has degree strictly smaller than $\mu$, the minimal polynomial of $u$, and thus $R=0$. This concludes that $h=Q\mu$ i.e. $\mu|h$.2011-08-12
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    Thus, we have $h$ divides $\mu$, and $\mu$ divides $h$, they are both unitary polynomials with coefficients in a domain (actually the field $F$), so they are equal. This shows that $\ker(\varphi)=(h)=(\mu)$. Ok, by universal porperty, we get that there is an induced *injective* homomorphism of *rings* $\hat{\varphi}:\frac{F[X]}{(h)}=\frac{F[X]}{\ker(\varphi)}\to K$ whose image is the same as the image of $\varphi$, that is all of $F[u]$ by its definition and definition of $\varphi$. This yields, by restriction an injective, surjective ring homomorphism (I'll still call $\hat{\varphi}$)2011-08-12
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    of rings $$\hat{\varphi}:\frac{F[X]}{(h)}\to F[u]$$ : I repeat, $\hat{\varphi}$ is an **isomorphism** of rings. But from the first point I made and asked you to grant me for the moment, we can deduce that those are actually *fields* on both sides. Why? because the ideal $(h)$ is non zero, and the quotient ring $\frac{F[X]}{(h)}$ is isomorphic as rings to the subring $F[u]\subset K$ of the field $K$ which therefore is a *domain*. Hence $\frac{F[X]}{(h)}$ is a domain, and by the granted fact, it is a field.2011-08-12
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    We now know that the ring $F[u]$ is actually a field, and we are now in position to show that $F[u]=F(u)$. Indeed, by definition we already had $F\cup\lbrace u\rbrace\subset F[u]\subset F(u)$, and $F(u)$ is the smallest subfield of $K$ that contains $F\cup\lbrace u\rbrace$. But we know that $F[u]$ is a *subfield* of $K$, and it contains $F\cup\lbrace u\rbrace$, thus, by minimality $F(u)\subset F[u]$. This concludes $F(u)=F[u]\approx\frac{F[X]}{(h)}$. Is this clear now?2011-08-12
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    So let's show that for any field $F$, and any *non-zero* ideal $I=(h=\subset F[X]$, $I$ is maximal iff it is prime. regardless of the ground ring, maximal ideals are prime. We only need to show that if $I=(h)$ is prime, it is maximal. So suppose $I$ is prime. First of all, the polynomial $h$ has to be irreducible : it can't be a constant (it can't be $0$ since $I$ is non zero, and it can't be a non zero constant, otherwise we'd have $I=F[X]$ and this is not a prime ideal by definition. It can't be composite either, for else it wouldn't be a prime ideal : if $h=A\cdot B$ with $A,B$ non constant2011-08-12
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    then we'd have $0<\mathrm{deg}(A),\mathrm{deg}(B)<\mathrm{deg}(h)$, thus $h\nmid A,B$ i.e. $A,B\notin (h)$ yet $A\cdot B=h\in I$ which would contradict $I$ being prime. This shows that $h$ is irreducible. So we want to show that the ring $\frac{F[X]}{(h)}$ is actually a field. We know that $h$ is irreducible. Take $p(X)+I\in\frac{F[X]}{(h)}\setminus \lbrace 0\rbrace$, this means $P(X)\notin I$ if we take the euclidean division of $P$ by $h$, we get $P=Qh+p$ with $0\neq p\notin I$ and $\mathrm{deg}(p)<\mathrm{deg}(h)$. We also have $P+I=p+I$. Because $h$ is irreducible, and $p$ is non zero2011-08-12
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    and $\mathrm{deg}(p)<\mathrm{deg}(h)$, we have $\gcd(p,h)=1$ and by Bezout there are two polynomials $U,V\in F[X]$ with $p(X)U(X)+h(X)V(X)=1$. This relation gives tells us us $p(X)U(X)\in 1+I$, i.e.after projecting to $\frac{F[X]}{(h)}$, $\bar{p}\times\bar{U}=\bar{1}$ where I have abreviated $\bar{a}=a+I$. This shows that $P(X)+I \in \frac{F[X]}{(h)}\setminus\lbrace 0\rbrace$ has an inverse in $\frac{F[X]}{(h)}\setminus\lbrace 0\rbrace$ and finishes the proof that $\frac{F[X]}{(h)}$ is indeed a field.2011-08-12
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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