1

I know that over any infinite field $F$ there are no nonzero polynomials in $F[x_1,\cdots,x_n]$ which vanish in all $F^n$ (Proof is by induction with basis step given by the fact that polynomials in $F[x]$ only vanish in a finite number of points).

But in quaternions, the polynomial $x^2+1$ vanishes in an uncountable set. Might there be one which vanishes in all $\mathbb{H}$?

LeviathanTheEsper
  • 3,163
  • 2
    How do you define a polynomial with quaternion coefficients? How do you define the related polynomial function (necessary to test whether the polynomial vanishes at a given point)? There are some subtleties. Anyway, if we use the definition that a polynomial $\Bbb{H}\to\Bbb{H}$ is a mapping of the form $$x\mapsto \sum_{i=0}^nq_ix^i,$$ where $n$ is a natural number, the coefficients $q_i$ are some fixed quaternions, $q_n\neq0$, then we can how that such a function does not vanish identically. – Jyrki Lahtonen Jun 21 '16 at 17:26
  • 2
    (cont'd) This is because the quaternionic norm is multiplicative, and satisfies the triangle inequality. So for large enough $||x||$ the highest degree term $q_nx^n$ dominates the sum of the others. The claim follows. This argument loses its footing, if you include maps like $x\mapsto q_1x^2+xq_2x+x^2q_3$ among polynomials. Because for the purposes of using the norm all three terms are of degree two, and may cancel each other under some circumstances. That's why I wanted you to be a bit more specific. – Jyrki Lahtonen Jun 21 '16 at 17:28
  • I would define the polynomial ring supposing that $x^k q=q x^k$, then only taking the sums $\sum q_i x^i$, sum as usual and product by extending $(q x^n) (r x^m)=(qr) x^{n+m}$ by bilinearity. – LeviathanTheEsper Jun 21 '16 at 22:06
  • I could also define it freely but it wpuld be very tricky in my opinion to manage it. – LeviathanTheEsper Jun 21 '16 at 22:17
  • 1
    @LeviathanTheEsper The OP wants to be able to evaluate the polynomial at elements of $\Bbb H$. In that case, $x^kq=qx^k$ is inconsistent with that goal: for example, if $q=j$, then you couldn't evaluate at $x=i$ properly. – rschwieb Jun 23 '16 at 13:01
  • 2
    Another way of looking at @rschwieb's comment is that while you do get a polynomial ring $\Bbb{H}[x]$ with $x$ an indeterminate commuting with the coefficients, the price you pay is that evaluating polynomials at a given point is no longer a homomorphism of rings. In other words: if $h(x)=f(x)g(x)$ as polynomials, then the rule $h(q)=f(q)g(q)$ does not hold for all quaternions $q$. In fact, this is what allows $x^2+1$ to have infinitely many zeros in $\Bbb{H}$. See Arturo Magidin's answer for more details. – Jyrki Lahtonen Jun 23 '16 at 15:32
  • Oh, that would be a big problem. Thanks. – LeviathanTheEsper Jun 24 '16 at 17:21

0 Answers0