Dear Uncle Colin,

If you know that $a$ is an algebraic number, how do you prove that $ai$ is also algebraic?

Am Lost! Get Equations But Real Analysis Is Confusing

Hi, ALGEBRAIC, and thanks for your message!

I’m in the same boat – my real analysis lectures were at the other end of town from my French classes, so I always pitched up ten minutes late, out of puff and missing the context of whatever Professor Falconer was talking about.


So, with that in mind, I typically have to work stuff like this up from the ground using examples, and try to generalise from them.

In this case, I went for the simplest thing I could think of (although I’ve since come up with a simpler example - of which more later): I know that 2 and 3 are algebraic because $(x-2)(x-3) = x^2 - 5x + 6$ – and an algebraic number is one that’s a zero of any polynomial with real, rational coefficients.

How can I generate a satisfactory polynomial that has $2i$ and $3i$ as zeros? $(x-2i)(x-3i)$ doesn’t work, because $x^2 - 5ix - 6$ doesn’t have real coefficients.

However, there’s a lemma I can use: if $p + iq$ is a root of a real polynomial, so is $p - qi$ – so I need to include $-2i$ and $-3i$ as roots as well. That give $(x-2i)(x-3i)(x+2i)(x+3i) = x^4 + 13x^2 + 36$, which is the kind of polynomial we want.

And what I’ve done there is take a polynomial $p(x)$ with roots $A$ and $B$, then work out $p(ix)p(-ix)$ – which has roots $\pm Ai$ and $\pm Bi$.

So much for an example. Does it always work?

A sketchy proof

Recap: we’re trying to prove that if $a$ is algebraic, then so is $ai$.

We know (by definition) that there’s a polynomial with real, rational coefficients, $p(x)$, such that $p(a)=0$.

We want to show that $q(x) = p(ix)p(-ix)$ has two properties:

  • It has $ai$ as a root
  • It has rational, real coefficients.

The first of those is trivial: $q(ai) = p(-a)p(a)$, and $p(a)=0$.

The second is a little trickier, but not by much. What do the coefficients of $p(ix)$ look like? The coefficients of even powers of $x$ remain rational, but the odd-power coefficients become imaginary (but still rational). We can write $p(ix)$ as $s(x) + it(x)$, where $s$ and $t$ both have real, rational coefficients.

The same is true of $p(-ix)$, only with a change of sign in the odd terms: $p(-ix) = s(x) - it(x)$.

Their product is $\left[s(x)\right]^2 + \left[t(x)\right]^2$, which is a polynomial with real, rational coefficients – and with $ai$ as a root.

I hope that helps!

- Uncle Colin