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Let $F$ be a field and let $p(x)\in F[x]$ be irreducible over $F$.If $a$ is a zero of $p(x)$ in some extension $E$ of $F$,then F[x]/Ker($\phi$)$\simeq F(a)$ .Furthermore,if $deg(p(x))=n$,then every member of $F(a)$ can be uniquely expressed in the form $$C_{n-1}a^{n-1}+C_{n-2}a^{n-2}+C_{n-3}a^{n-3}...+C_{1}a^{1}+C_0$$ Where,$C_0,C_1,C_2,...,C_{n-1}\in F$

Consider $\phi:F[x]\rightarrow F(a)$ given by $\phi (f(x))=f(a)$.(How this map is well defined?)

$\phi$ is a ring homomorphism.

Let $f(x),g(x)\in F[x]$.

$\phi(f(x)+g(x))=\phi((f+g)(x))=(f+g)(a)=f(a)+g(a)=\phi(f(x))+\phi (g(x))$.

&

$\phi(f(x)*g(x))=\phi((f*g)(x))=(f*g)(a)=f(a)*g(a)=\phi(f(x))*\phi (g(x))$.

Hence,$\phi $ is a ring homomorphism.

Claim:$Ker(\phi)=<p(x)>$

Since,$\phi(p(x))=p(a)=0\implies p(x)\in Ker(\phi)\implies <p(x)>\subset Ker(\phi)$.

Since,$p(x)\in F[x]$ is irreducible.So, is a maximal ideal.Also Ker($\phi$) $\neq$ $F[x]$ as $1$ belong to $F[x]$ but does not belongs to Ker($\phi$).

Hence,$Ker(\phi)=<p(x)>$

Since, $\phi:F[x]\rightarrow F(a)$ is a ring homomorphism with Ker($\phi$).Then by First Isomorphism theorem on Rings,we have

F[x]/Ker($\phi$)$\simeq F(a)$ .

Queries

-Is this proof correct upto here?

  • How to prove the if $deg(p(x))=n$,then every member of $F(a)$ can be uniquely expressed in the form $$C_{n-1}a^{n-1}+C_{n-2}a^{n-2}+C_{n-3}a^{n-3}...+C_{1}a^{1}+C_0$$ Where,$C_0,C_1,C_2,...,C_{n-1}\in F$?

  • If $\phi:F[x]\rightarrow F(a)$ is a ring homomorphism,then is it always true that $F[x] \simeq \phi (F[x])$?

  • How $\phi (F[x])$ contains both $F$ and $a$?

  • How {$1,a,a^2,...,a^{n-1}$} is a basis for $F(a)$ over $F$?
Styles
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    This is a theorem from Gallian's algebra.Chapter-Field extension. – Styles May 31 '17 at 12:25
  • For full credit: explain why the rule $(fg)(a)=f(a)g(a)$ fails if we drop the assumption that $E$ is commutative (some texts, notably French ones, allow non-commutative fields) – Jyrki Lahtonen May 31 '17 at 14:15
  • @JyrkiLahtonen:this seems to be an interesting problem as we encountered this thing in real analysis very often but frankly speaking i never thougt about this problem so,firstly thanks a lot for making me learn such a beautiful result.I 'll definitely try to go through this.Will you provide some reference where i can see some examples where this result fails? – Styles May 31 '17 at 14:42
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    Pk Styles, study this thread. Arturo Magidin's answer in particular. The short explanation is that in the polynomial ring the indeterminate $x$ necessarily commutes with coefficients from the field (otherwise we would need to distinguish between $cx, xc, cxc,\ldots$). But when we plug-in something in place of $x$ that may not be the case unless we have commutativity. – Jyrki Lahtonen May 31 '17 at 18:02

1 Answers1

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Short ideas/answers:

=== What do you think can happen for the map $\;\phi(f(x)):=f(a)\;$ not to be well defined? This seems to be straightforward...and thus

=== Yes, the proof is correct up to there.

=== If $\;p(x)=b_0+b_1x+\ldots+b_nx^n\;,\;\;b_n\neq0$ , then

$$p(a)=\sum_{k=0}^n b_ka^k=0\implies a^n=-\frac1{b_n}\left(b_0+b_1a+\ldots+b_{n-1}a^{n-1}\right)\in\text{Span}\,\{1,a,...,a^{n-1}\}$$

and inductively $\;a^m\in \text{Span}\,\{1,a,...,a^{n-1}\}\;,\;\;\forall\,m>n-1$

Since we also know that $\;F(a)=F[a]\;$ (why?), then every polynomial in $\;a\;$ can be written, by the above, as an element in $\;\text{Span}\,\{1,a,...,a^{n-1}\}\;$

=== If $\;\phi\;$ is an injective ring homomorphism $\;R\to S\;$ , it is always trie that $\;R\cong \phi(R)\;$ , and you can even get this with the first isomorphism theorem, again.

=== Finally: since canonically $\;F\le F[x]\;$ , then $\;F=\phi(F)\subset \phi(F[x])\;$ . That the image contains $\;a\;$ is trivial.

DonAntonio
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  • :If $\phi$ is not well defined then it can map $f(x)$ to atleast two elements of $F(a)$ but this not happening,that's why $\phi $ is well defined? – Styles May 31 '17 at 13:45
  • :I think in third point in place of $f(a)$ there should be $p(a)$? – Styles May 31 '17 at 13:55
  • @PKStyles Typo corrected, thanks. The map $;\phi;$ is well defined because a polynomial, when taken as a function, is well defined and thus $;p(a);$ is one, single number.... – DonAntonio May 31 '17 at 14:07
  • :$F(a)=F[a]$ can be directly obtained just by putting $x=a$ in RHS of $F(a) \simeq F[x]/<p(x)>$?but this seems to me quite absurd as the result you mentioned contains equality while the result i got by substituting $x=a$ contains isomorphism symbol? – Styles May 31 '17 at 14:19
  • @I don't really understand your doubt, but it is a reasonably easy exercise to prove that, with your notation, $;F(a)=F[a]\iff a;$ is algebraic over $;F;$ ... – DonAntonio May 31 '17 at 14:25
  • :I haven't studied the concept of $ "algebraic"$ yet.I think its in the next chapter.In my last comment i was trying to to deduce $F[a]=F(a) $ from $F(a) \simeq F[x]/<p(x)>$ .I will definitely try to prove it today only. – Styles May 31 '17 at 14:29
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    @PKStyles Well, that can be a problem without something else. It is clear that $;F[x]/\langle p(x)\rangle\cong F(a);$ is a field , but $;F[\beta];$ does not necessarily is a field... unless $;\beta;$ is algebraic ( or, without using this name, iff $;\beta;$ is a root of a non-zero polynomial in $;F[x];$) . – DonAntonio May 31 '17 at 16:52