The fourth paragraph of the wikipedia article algebraic extension states that $K[a]$ is a field, which means every polynomial has an inverse. The inverse has to be a polynomial over $K$ as well. It seems it requires $K$ to be a non-integral domain. How do we resolve the confusion? How do we prove this paragraph?
Why Is Inverse of a Polynomial Still a Polynomial?
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$\begingroup$
polynomials
ring-theory
field-theory
extension-field
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0With $\varphi(X) = a$ you have a field isomorphism $K[X]/(P(X)) \to K(a)$ where $P$ is the minimal polynomial of $a$. From this it is obvious that $K(a) = \{\sum_{j=0}^{deg(P)-1} c_j a^j, c_j \in K\} = K[a]$ – 2017-01-01
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0@user1952009: What is $\varphi$? I still do not understand how to find the inverse which we need to show $Q[a]$ is a field. – 2017-01-01
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0$F = K[X]/(P(X))$ is a field by the Euclidean/Bezout algorithm. Take $Q \in K[X]$ if $gcd(Q,P) = 1$ then there is $U,V \in K[X]$ such that $Q(X)U(X)+P(X)V(X) = 1$ i.e. $QU=1$ in $F$ – 2017-01-01
2 Answers
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Polynomials over a field $K$, that is, elements of $K[x]$, don't generally (unless they have degree zero) have an inverse.
$K[a]$ is a different object, it is isomorphic to the quotient ring $K[x]/(m)$ where $m$ is the minimal polynomial of $a$ (you can see this via the first isomorphism theorem using the morphism $\varphi:K[x]\to K[a]$ which sends $\sum k_i x^i\mapsto\sum k_i a^i$), assuming $a$ lives inside a field extension of $K$ its minimal polynomial will be irreducible, so the ideal $(m)$ it generates will be maximal and the quotient $K[x]/(m)$ will be a field.
Inverses in this field can be computed via the usual euclidean algorithm, take a polynomial $f\in K[x]$ with $\gcd(f,m)=1$, then by Bezout's theorem there are $h$ and $g$ such that $fh+gm=1$, but in the quotient ring $gm=0$ so $h$ is the inverse of $f$
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First, $\;f(a)\;$ is not a polynomial but the evaluation of a polynomial at $\;a\;$ , meaning: a number.
The complete claim is: If $\;K\;$ is a field and $\;F\;$ is some extension of $\;K\;$ , and $\;a\in F\;$ is algebraic over $\;K\;$ , then
$$K[a]:=\left\{\,f(a)\;/\;f(x)\in K[x]\,\right\}\;\;\text{is a field, which means}\;\;K[a]=K(a)$$
The proof: since $\;a\;$ is algebraic over $\;K\;$ there exists $\;0\not\equiv m_a(x)\in K[x]\;$ of minimal degree s.t. $\;m_a(a)=0\;$ , but then:
$$m_a(x)=\sum_{k=0}^{n-1}b_kx^k+x^n\implies b_0=-a^n-\sum_{k=1}^{n-1}b_ka^k\;\;\color{red}{(*)}$$
(...Observe the claim is trivial if $\;a\in K\;$ , and we can assume $\;b_0\neq0\;$ (why?) ...)
Yet $\;0\neq a\in F\;$ and $\;F\;$ is a field, so $\;a^{-1}\in F\;$ , and then multiplying $\;\color{red}{(*)}\;$ by $\;a^{-1}\;$:
$$a^{-1}=-\frac1{b_0}\left(a^{n-1}+\sum_{k=1}^{n-1}b_ka^{k-1}\right)\in K[a]$$
Deduce now that $\;K(a)\subset K[a]\;$ , and since the other contention is trivial if we remember $\;K(a)\;$ is the fraction field of the integral domain $\;K[a]\;$ , we're done.
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2And you need to show that $K(a)= K[a,a^{-1}]$ (at first $K(a)$ is the smallest field containing $K$ and $a$) – 2017-01-01
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0@user1952009 Thanks, but I don't think that's necessary: by definition $\;K(a)\;$ is the fraction **field** of the ring $\;K[a]\;$ ...so if it contains $\;a\neq0\;$ it **must contain** $\;a^{-1}\;$ as well. – 2017-01-01
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0You are assuming that $K(a) = \{\sum_{n=-N}^N c_n a^n, c_n \in K,N \in \mathbb{Z}\}$ but I don't think this is obvious to the OP – 2017-01-01
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1We are to prove $F$ is a field, why do assume it in the middle of your proof? – 2017-01-01
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1@Hans No, we are *not* to prove such a thing. $\;F/K\;$ is a field extension: both $\;F,K\;$ *are fields* \ – 2017-01-01
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2It is not at all obvious that $K[a,a^{-1}]$ is a field: $a$ has an inverse, but there are plenty of other elements that might not have inverses. – 2017-01-01
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1@E I think some people here are confusing things...one of them could be me, but right now I doubt it: First, $\;K(a)\;$ is **the fraction field** of the integral domain $\;K[a]\;$ . Is this clear? The OP is doing extension fields (and perhaps later Galois theory), I think he must know some basic ring theory. Second, if we prove $\;a^{-1}\in K[a]\;$ , then *from this* one can prove $\;K[a]\;$ is a field. That what's implied in "Deduce now that..." . Some work is left to the OP to complete. Nothing here is really obvious *but* perhaps when seen within its proper context. – 2017-01-01
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0@DonAntonio : $K(X)$ is the set of rational functions in $X$, but $K(a)$ is the set of finite linear combination of $a^n, n \in \mathbb{Z}$ : two different things because $a$ is algebraic over $K$ while $X$ isn't. – 2017-01-01
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1@user1952009 For me, and I think for some other authors (I could check...) , $\;K(a)\;$ is the **field** of rational functions of the form $\;\frac{f(a)}{g(a)}\;,\;\;g(a)\neq0\;,\;\;f,g\in K[x]\;$ . What you describe is what I call $\;K[a]\;$ . – 2017-01-01
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1@DonAntonio yes but you didn't mention it in your answer... You used the linear combination of $a^{\pm n}$ definition, that's all we said ! And $K[a]$ is the set of finite linear combination of $a^n,n \ge 0$ – 2017-01-01
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0@user1952009 Yes, I did mention it in the answer: it is *exactly there* where I mention $\;K(a)\;$ is the **fraction field** of $\;K[a]\;$ ...which, **by definition**, is what I wrote in my past comment. I really don't understand: what I wrote (and hinted the OP to continue) is, as far as I remember, a rather easy, standard proof . I shall look for it in several books. – 2017-01-01
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0After that you said "we are done" whereas it is the case only after we showed $K(a) = K[a,a^{-1}]$ – 2017-01-01
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0And showing $K(a) = K[a,a^{-1}]=K[a]$ really needs the gcd algorithm in $K[X]$ (or the proof that integral domain / maximal ideal = field) – 2017-01-01
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0@EricWofsey What part of the hinted part you didn't understand? I am expecting the OP to complete that. If he has some doubt he can ask again. I didn't say *at no moment* the above is a complete, full answer...I, and many other members here, don't *usually* do answer like that, though sometimes it happens. Not in this case. – 2017-01-01
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0@user1952009 Exactly so, and that is what I was expecting the OP to complete. Am I really being this unclear in this? The OP MUST COMPLETE THE PROOF...and I think that there are left perhaps 3-4 lines more for that: something using Euclides algorithm, coprime polynomials and etc. – 2017-01-01
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0Well, the issue is that if that is the way you intend for the proof to be completed, then all the work you did showing that $a^{-1}\in K[a]$ is totally irrelevant. Knowing $a^{-1}\in K[a]$ doesn't make the Euclidean division argument any easier. (Alternatively, though, you could observe that any element of $K[a]$ is algebraic over $K$, and so the same argument applied to that element produces an inverse. Not sure if this is what you intended though.) – 2017-01-01
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0Is there a typo in the first equation (definition) where $K[a]$ is defined --- should it be $K(a)$? $K(a)$ is not defined there ---- I understand it is mentioned in your comment that it is the fraction field. Maybe the $/$ should be $|$ in the first definition? It is a bit confusing. – 2017-01-01
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0@EricWofsey Your opinion, and I won't mess with what I expected the OP to deduce (obviously the last part in your last comment...). And I am stopping all this: either the OP can take some helpful ideas for him or not. We've already ground a lot of water here, I see no point in conitnuing any more. Thanks all. – 2017-01-01
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0@Hans Too bad it is so confusing for you. If that is true then downvoting the answer is always an option. Anyway, you *already* have another answer. – 2017-01-01
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0@Hans $K(a)$ and $K[a]$ are the smallest field and ring containing $K$ and $a$. – 2017-01-01
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0Please do not misunderstand me. I DO really appreciate your answer. I am studying and like your answer as it is detailed. That is why asking the questions as I try to understand all the steps and symbols. I am suspecting there could be some typos. But I may well be wrong as I am a novice. Thank you, DonAntonio. – 2017-01-01
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0@user1952009: OK. Thank you. – 2017-01-01