2

Problem: If $u\in \mathbb C$ is algebraic over $\mathbb Q$, then the span of $\{1, u, u^2, ..., u^{n-1}\}$ is a field.

Attempt: Since every element in the span is in $\mathbb C$ it suffices to show that we have a subfield of $\mathbb C$. Closure under addition and multiplication is clear to me. It's also clear that every element in the span has an additive inverse; namely, it's negative. But showing that every nonzero linear combination has a multiplicative inverse is throwing me off.

Question: Suppose $f(u)$ belongs to the span. Then $f(u) = a_0 + a_1u + \cdots + a_{n-1}u^{n-1}$ where the coefficients are rational numbers. What would the inverse look like or how would I construct it?

dxiv
  • 76,497
Ungar Linski
  • 757
  • 2
    It depends what tools you can use. The standard way is to define the homomorphism $\varphi:\mathbb{Q}[x]\to\mathbb{C}$ by $\varphi(f)=f(u)$. The image is exactly your span, which is usually denoted by $\mathbb{Q}[u]$. The kernel is generated by the minimal polynomial of $u$, call it $m(x)$. (you should have noted in your question that $n=\deg(m)$) Then $\mathbb{Q}[x]/(m)\cong\mathbb{Q}[u]$. Since the minimal polynomial is irreducible, this quotient ring is a field. – Mark Apr 06 '23 at 16:10
  • 1
    It is difficult to construct the inverse directly. A non-constructive argument begins as follows: let $\psi_{f(u)}$ be the map $K \to K, x \mapsto xf(u)$, where $K$ is the span over $\mathbb{Q}$. Can you see how this might lead to solution? – legionwhale Apr 06 '23 at 16:15
  • An alternative is to use the fact that a finite dimensional integral domain is a field. – Jyrki Lahtonen Apr 06 '23 at 16:15
  • 1
    BTW, this is a duplicate (in all the different ways). I want input from others before I vote to close as a duplicate, because my vote would be immediately binding (dupehammer), and the choice of the target is not easy (though we can use all of them). – Jyrki Lahtonen Apr 06 '23 at 16:18
  • @Mark "Since the minimal polynomial is irreducible, this quotient ring is a field" - This result I don't have. This is a linear algebra course, believe it or not. – Ungar Linski Apr 06 '23 at 16:19
  • @legionwhale No, I don't quite see it. Show that the map is a bijection maybe? Or injection is enough since it's a map to itself. – Ungar Linski Apr 06 '23 at 16:21
  • 1
    @UngarLinski Yes, show it is an injection is sufficient, but only because it's a finite-dimensional vector space over $\mathbb{Q}$. dxiv has probably given the nicest answer actually. – legionwhale Apr 06 '23 at 16:56
  • @legionwhale It is not hard to explicitly construct an inverse see here: https://math.stackexchange.com/a/1363867/789929 – Vivaan Daga Apr 06 '23 at 17:03
  • @Shinrin-Yoku True. I forgot that the construction there works for arbitrary elements. – legionwhale Apr 06 '23 at 17:04
  • See the linked dupes (and their links) for all the common arguments, Please delete this question since we already have too many copies of these arguments, which makes it difficult to locate the best answers by search. You will need to unaccept the answer before deleting (click the checkmark). – Bill Dubuque Apr 06 '23 at 19:37
  • 2
    @BillDubuque This question is significantly narrower in scope than the ones you linked as duplicates. Just because it's a (very) particular case of the general result doesn't make it a dupe IMHO. – dxiv Apr 06 '23 at 19:43
  • @dxiv It si not really "narrower". But even if you think so, see abstract duplicates. Your answer (and likely any other) is dupe of hundreds of others and only serves to clutter searxch results on this huge FAQ, making it difficult (if not impossible) for users to locate the "best" answers by search. – Bill Dubuque Apr 06 '23 at 19:53
  • @BillDubuque I laid no claim it was original research ;-) But, in the context of the narrow question here, it is an elementary and constructive answer, which btw I do not see as such in the dupes you linked, on a cursory look. – dxiv Apr 06 '23 at 20:00
  • 1
    @dxiv You ddin't read carefully them, since links to constructive methods for inversion such as using the extended Euclidean algorithm to compute the Bezout equation are already in this answer in one of the dupes, along with linked woirked examples. We have many huindreds of posts on that as you should surely know. – Bill Dubuque Apr 06 '23 at 20:06

1 Answers1

1

For a constructive proof, let $p(x)$ be the $n^{th}$ degree rational polynomial having $u$ as a root. It can be assumed WLOG that $p(x)$ is irreducible, so either $f(x)$ is a rational multiple of $p(x)$ or otherwise $\gcd(p(x),f(x))=1$. The former case means $f(u)=0$ which is not invertible, in the latter case by Bezout's lemma (1, 2, 3, 4) there exist rational polynomials $a(x),b(x)$ such that $a(x)f(x)+b(x)p(x)$ $=\gcd(p(x),f(x))=1$. Setting $x=u$ gives $\require{cancel}a(u)f(u)+\cancel{b(u)p(u)}=1$ which both proves the existence of and explicitly calculates the inverse $\big(f(u)\big)^{-1}=a(u)$.

dxiv
  • 76,497
  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. This already occurs here many tens (if not hundreds) of times already. – Bill Dubuque Apr 06 '23 at 19:34
  • Bezout is not needed since we get an inverse immediately by rearranging the equation $p(u)=0$ - see here. @OP see the linked dupes for more conceptual answers that will help you better understand the essence of the matter. – Bill Dubuque Apr 06 '23 at 20:01
  • 1
    @BillDubuque With the notation from the linked post, finding $,\rm f(x),$ is not trivial. This is why I started my answer here with "for a constructive proof". It proves the inverse exists and it calculates it, using nothing more than euclidean division. – dxiv Apr 06 '23 at 20:12
  • @BillDubuque It's a different $f$ there. Your $f(x)$ would be the rational polynomial that OP's $f(u)$ satisfies. – dxiv Apr 06 '23 at 20:47
  • My first comment needs only the existence of $p(x)$ with $p(\alpha) = 0,\ \alpha = f(u),$ (immediate by fin. dim). In any case, both the common existential and constructive methods are already in the dupes and their links (and in hundreds of other answers). Their is no need to repeat them ad infintum (doing so clutters search results so greatly that people give up searching for answers - meaning all of our work over the past $13$ years is mostly for nought), – Bill Dubuque Apr 06 '23 at 21:09
  • @BillDubuque FWIW my first comment here was in reply to the "Bezout is not needed" part. As I wrote, the reason Bezout was used here is because my answer set out to provide a constructive proof. I am well aware that proving the existence does not require constructing the inverse. But there is nothing wrong with a constructive proof, either. I know similar proofs must have been posted before. But for someone less versed, it can be easier to understand a 5-line self-contained proof, rather than follow links 2-3 levels deep. We can agree to disagree, of course, and I'll leave it at that. – dxiv Apr 06 '23 at 21:20
  • 1
    Imo it is much better (as in this linked dupe) to give a few onsite links (including links to worked examples) than to give a single offsite (Wikipeda) link as you do. The linked Wikipedia section is very poor pedagogically and will ony serve to obfuscate the essence of the matter. Otoh, the linked dupes are much more helpful pedagogically. We should aim to maximize pedagogical quality, not minimize #clicks. – Bill Dubuque Apr 06 '23 at 21:35