8

Let $K$ be a field, and let $u$ be algebraic over $K$. Show that $K[u]$ is a field.

Context/progress so far: This is a generalization of a problem in the first few pages of Peterson's "Linear Algebra." If possible, I'd like to avoid using the machinery from abstract algebra and show directly that elements of $K[u]$ are invertible. I can show that $\{1, u, ..., u^{n-1}\}$ is a basis for $K[u]$ over $K$, where $n$ is the degree of the minimal polynomial of $u$. Using the same polynomial, I can show that $u^{-1}$ exists. But I'm having trouble inverting general elements of $K[u]$. Any hints?

Mikasa
  • 67,374
Josh Keneda
  • 2,907
  • Hint: take the minimal polynomial--which must have nonzero constant term. Use this to express $u^{-1}$ as a polynomial in $u$. – Alex Youcis Sep 03 '13 at 05:28
  • @AlexYoucis Thanks for the hint. But I must be being dense: I can see how this would be used to invert any of the basis elements, but I don't see how this helps us to invert linear combinations of them. How could we use it to invert something like $u+1$, for example? – Josh Keneda Sep 03 '13 at 05:33
  • 2
    @JoshKeneda I don't understand the hint either. For your specific case you can invert $u+1$ in the same way that you would something like $1+\sqrt{2}$: multiply by the conjugates. If $u_2,...,u_n$ are the other zeros of the minimal polynomial, then $(u+1)(u_2 + 1)...(u_n+1)$ lies in $K$. You can see this directly by multiplying it out, noticing that the sums and products that appear are the coefficients of said minimal polynomial. General elements can be inverted in the same way but you may need some machinery to prove that – Cocopuffs Sep 03 '13 at 05:58
  • That's a really cool strategy, @Cocopuffs. Does that work for any polynomial $p \in K[u]$? I mean, is $p(u)p(u_2)\cdots p(u_n)\in K$? Or does the strategy get more complicated as you generalize? – Josh Keneda Sep 03 '13 at 06:11
  • @JoshKeneda I guess I took for granted you knew that $u$ algebraic implied that $K[u]$ was algebraic. I also wrote $u$ above, when I meant it to mean a generic element $\alpha$ of $K[u]$, not the actual $u$ itself :) Sorry for the confusion! – Alex Youcis Sep 03 '13 at 06:21
  • @AlexYoucis No worries! Thanks for all of the help. Is there an obvious way to see that "$u$ algebraic implies $K[u]$ algebraic," or do you just have to know that algebraic numbers are closed under addition and multiplication? – Josh Keneda Sep 03 '13 at 06:26
  • 1
    @JoshKeneda Note that since, for any $\alpha\in K[u]$, you have that $K[\alpha]\subseteq K[u]$, you see that for all $\alpha$ in $K[u]$ one has that $[K[\alpha]:K]<\infty$. Then, you merely note the map $K[T]\to K[\alpha]$ has to have non-trivial kernel. – Alex Youcis Sep 03 '13 at 06:31
  • @AlexYoucis Thank you. That's a great approach. – Josh Keneda Sep 03 '13 at 06:35
  • 1
    @JoshKeneda sorry for the delay. Yes $p(u)p(u_2)...p(u_n)$ will always lie in $K$. It's a symmetric polynomial in the roots, and there is an algorithm for writing it as a polynomial expression in elementary symmetric polynomials (or: the coefficients of the mipo), which lie in $K$. This is not obvious and it's the machinery I was referring to – Cocopuffs Sep 03 '13 at 22:23
  • @Cocopuffs This is late, but how do we actually get the inverse from the fact that $(u+1)(u_2+1)\dots(u_n+1) \in K$? – A.B Feb 03 '21 at 16:43
  • @A.B It's been a long time since I've thought about this, but I think the point was the following. I was asking about how to invert something like $u+1$, and if $$(u+1)\Pi_{i=2, ..., n}(u_i+1) = k \in K,$$ then multiplying both sides by $k^{-1}$ gives $$(u+1)\Pi_{i>1} (u_i+1)k^{-1} = 1,$$ so $(u+1)^{-1} = \Pi_{i>1} (u_i+1)k^{-1}$. The point is just that once you've multiplied by something and landed in $K$, you're only a slight adjustment away from the unit. – Josh Keneda Feb 07 '21 at 00:34

4 Answers4

6

I am not sure if this proof is what you want, but here goes :

Let $p(x)\in K[x]$ be the minimal polynomial of $u$, then $$ K[u] = K[x]/(p(x)) $$ Consider a polynomial $f(X) \in K[X]$ such that $\overline{f}\neq \overline{0}$ in $K[u]$. Replacing $f$ by its remainder on division with $p$, we may assume without loss of generality that $deg(f) < deg(p)$.

Since $p$ is prime, $$ gcd(f,p) = 1 $$ Thus, there exists $g(x), r(x) \in K[x]$ such that $$ fg + rp = 1 $$ Thus $$ \overline{f}\overline{g} = \overline{1} $$ And hence $\overline{f}$ is invertible.

Prahlad Vaidyanathan
  • 31,554
  • Thank you! I think this is a bit different from the suggested approach (which involved computing $u^-1$ directly), but I really like this proof. – Josh Keneda Sep 03 '13 at 06:17
3

Suppose we have any element $x$ algebraic over $K,$ with minimal polynomial $x^n + a_1 x^{n-1} + \ldots + a_n =0.$ Then $x$ has inverse $(-a_n)^{-1} (x^{n-1}+ a_1 x^{n-2} + \cdots + a_{n-1}).$ Now the key is that sums or products of algebraic elements is algebraic.

Ragib Zaman
  • 35,127
3

This result is only true if $u$ is supposed to live in an integral domain containing $K$ (for instance in a field extension, a context suggested by the term "algebraic"). I'll suppose this hypothesis.

That $u$ is algebraic means that $K[u]$ is a finite dimensional vector space over$~K$. Multiplication by any nonzero element $a\in K[u]$ is an injective (because $K[u]$ is an integral domain) $K$-linear map $K[u]\to K[u]$. By finite dimensionality, such $K$-linear maps are surjective as well; in particular the element$~1\in K[u]$ is in the image of multiplication by$~a$, which means that $a$ is invertible.

If you need to compute the inverse of $a$ explicitly, you can proceed as follows. Let $\mu\in K[X]$ be the minimal polynomial of$~u$ over$~K$, which is an irreducible monic polynomial of degree$~n$ (if it were reducible $\mu=PQ$, then $P[u]Q[u]=\mu[u]=0$ would contradict that $K[u]$ is an integral domain), and write $a$ as a polynomial (that can be taken to be of degree${}<n$) in$~u$, say $a=A[u]$. Then since $a\neq0$ and $\mu$ irreducible one has $\gcd(A,\mu)=1$, so if $S,T\in K[X]$ are Bezout coefficients $1=SA+T\mu$, then $S[u]$ is an inverse of$~a$ in$~K[u]$.

Marc van Leeuwen
  • 115,048
1

Will this work?:

Let $\phi(x) \in K[x]$ be the minimal polynomial of $u$. Consider $f(u) \in K[u]$ where $f(x) \in K[x]$. I claim we can take $\deg f < \deg \phi$ without loss of generality. Why? If $\deg f \ge \deg \phi$, we can use the division algorithm in $K[x]$ to write $f(x) = \phi(x) q(x) + r_f(x)$, where $\deg r_f < \deg \phi$. Then $f(u) = \phi(u) q(u) + r_f(u)$, whence, since $\phi(u) = 0$, $f(u) = r_f(u)$. I claim that $\phi(x)$ must be irreducible in $K[x]$. Why? If $\phi(x) = \alpha(x) \beta(x)$ in $K[x]$, and neither $\alpha(x)$ nor $\beta(x)$ is constant, i.e. of degree $0$, then we would have $\alpha(u) \beta(u) = \phi(u) = 0$, whence either $\alpha(u) = 0$ or $\beta(u) = 0$. But $0 < \deg \alpha < \deg \phi$, $0 < \deg \beta < \deg \phi$, so this contradicts the minimality of $\phi(x)$. Since $\phi(x)$ is irreducible, it is prime in $K[x]$, a principal ideal domain. Since $\phi(x)$ is prime, we must have $\gcd(\phi(x), r_f(x)) = 1$, since we cannot have $\phi(x) \vert r_f(x)$ by virtue of the fact that $\deg r_f < \deg \phi$, whence there exist $g(x), h(x) \in K[x]$ such that $r_f(x)g(x) + \phi(x)h(x) = 1$. But then $r_f(u)g(u) = 1$, since $\phi(u) = 0$. QED.

P.S. After the above breakdown, I must point out that the evaluation homomorphism $\theta:K[x] \to K[u]$ has kernel precisely $(\phi(x))$ in $K[x]$, whence $K[u]$ is isomorphic to $K[x]/(\phi(x))$, a field since $\phi(x)$ is prime etc. etc. etc. Less typing! ;)

Robert Lewis
  • 71,180