3
$\begingroup$

This question remained somehow incompletely solved. The OP also asked for an explicit form of the minimal polynomial of the sum (respectively, product) of two algebraic elements (in a field extension). Although such form is, in general, impossible to find, he proposed the polynomial $\Pi_j\Pi_i (x-\alpha_i-\beta_j)$ (respectively, $\Pi_j\Pi_i (x-\alpha_i\beta_j)$) as possible candidates.

The question is the following:

Have these two polynomials the coefficients in the base field?

I find it also useful for this related question.

$\endgroup$
2
  • 1
    $\begingroup$ Certainly the obvious sufficient condition would be that $\mathbb{Q}(\alpha)$ and $\mathbb{Q}(\beta)$ are linearly disjoint over $\mathbb{Q}$, and Galois. :) $\endgroup$ Dec 1 '13 at 10:01
  • 1
    $\begingroup$ AFAICT the answer to the greyed out question is affirmative. If the $\alpha_i\neq\alpha_{i'}$ whenever $i\neq i'$, and similarly the $\beta_j$:s are all distinct, then Galois theory gives you the answer. In the inseparable case there's a bit more work to do. $\endgroup$ Dec 1 '13 at 10:06
3
$\begingroup$

Let $p(x)=\prod_i (x-\alpha_i)$. Then every symmetric function in the $\alpha_i$ is a polynomial in the elementary symmetric functions, which are $\pm$ the coefficients of $p$. Hence, it must lie in the base field.

If $q(x)=\prod_j (x-\beta_j)$, then $\prod_{i,j} (x-\alpha_i - \beta_j)$ can be considered as a polynomial in the $x,\alpha_i,\beta_j$. It is symmetric in the $\alpha_i$, so that it is a polynomial in $x,\beta_j$ with coefficients in the base field. But it is also symmetric in the $\beta_j$, hence is a polynomial in $x$ with coefficients in the base field.

The same argument works for $\prod_{i,j} (x-\alpha_i \beta_j)$.

When $p,q$ are irreducible, it is not true in general that $\prod_{i,j} (x-\alpha_i-\beta_j)$ and $\prod_{i,j} (x-\alpha_i \beta_j)$ are irreducible. This can already been seen from their degrees, which are too large, for example if $\alpha_1=-\beta_1$ or $\alpha_1=\beta_1^{-1}$.

Actually we don't need fields here. The same argument can be used to show that the sum and the product of two integral elements is again integral, for arbitrary ring extensions.

PS: The proof above is really constructive. Let me illustrate this for two quadratic polynomials $x^2-px+q=(x-\alpha_1)(x-\alpha_2)$ and $x^2-rx+s=(x-\beta_1)(x-\beta_2)$, so that $p=\alpha_1+\alpha_2$, $q=\alpha_1 \alpha_2$ and $r=\beta_1+\beta_2$, $s=\beta_1 \beta_2$. Then we compute:

$(x - a_1 - b_1) (x - a_1 - b_2) (x - a_2 - b_1) (x - a_2 - b_2)$

$=x^4 - 2 b_1 x^3 - 2 b_2 x^3 - 2 x^3 (\mathbf{a_1 + a_2}) + b_1^2 x^2 + 4 b_1 b_2 x^2 + b_1 x^2 (\mathbf{3 a_1 + 3 a_2}) + b_2^2 x^2 + b_2 x^2 (\mathbf{3 a_1 + 3 a_2}) + x^2 (\mathbf{a_1^2 + 4 a_1 a_2 + a_2^2}) - 2 b_1^2 b_2 x - b_1^2 x (\mathbf{a_1 + a_2}) - 2 b_1 b_2^2 x - 4 b_1 b_2 x (\mathbf{a_1 + a_2}) - b_1 x (\mathbf{a_1^2 + 4 a_1 a_2 + a_2^2}) - b_2^2 x (\mathbf{a_1 + a_2}) - b_2 x (\mathbf{a_1^2 + 4 a_1 a_2 + a_2^2}) - 2 \mathbf{a_1 a_2} x (\mathbf{a_1 + a_2}) + b_1^2 b_2^2 + b_1^2 b_2 (\mathbf{a_1 + a_2}) + \mathbf{a_1 a_2} b_1^2 + b_1 b_2^2 (\mathbf{a_1 + a_2}) + b_1 b_2 (\mathbf{a_1 + a_2})^2 + \mathbf{a_1 a_2} b_1 (\mathbf{a_1 + a_2}) + \mathbf{a_1 a_2} b_2^2 + \mathbf{a_1 a_2} b_2 (\mathbf{a_1 + a_2}) + \mathbf{a_1^2 a_2^2}$

$=x^4 - 2 b_1 x^3 - 2 b_2 x^3 - 2 x^3 p + b_1^2 x^2 + 4 b_1 b_2 x^2 + b_1 x^2 (3 p) + b_2^2 x^2 + b_2 x^2 (3 p) + x^2 (p^2 + 2q) - 2 b_1^2 b_2 x - b_1^2 x p - 2 b_1 b_2^2 x - 4 b_1 b_2 x p - b_1 x (p^2 + 2q) - b_2^2 x p - b_2 x (p^2 + 2q) - 2 q x p + b_1^2 b_2^2 + b_1^2 b_2 p + q b_1^2 + b_1 b_2^2 p + b_1 b_2 p^2 + q b_1 p + q b_2^2 + q b_2 p + q^2$

$=x^4 - 2 p x^3 - 2 x^3 (\mathbf{b_1 + b_2}) + p^2 x^2 + p x^2 (\mathbf{3 b_1 + 3 b_2}) + 2 q x^2 + x^2 (\mathbf{b_1^2 + 4 b_1 b_2 + b_2^2}) - p^2 x (\mathbf{b_1 + b_2}) - 2 p q x - p x (\mathbf{b_1^2 + 4 b_1 b_2 + b_2^2}) - 2 q x (\mathbf{b_1 + b_2}) - 2 \mathbf{b_1 b_2} x (\mathbf{b_1 + b_2}) + \mathbf{b_1 b_2} p^2 + p q (\mathbf{b_1 + b_2}) + \mathbf{b_1 b_2} p (\mathbf{b_1 + b_2}) + q^2 + q (\mathbf{b_1^2 + b_2^2}) + \mathbf{b_1^2 b_2^2}$

$=x^4 - 2 p x^3 - 2 x^3 (r) + p^2 x^2 + p x^2 (3 r) + 2 q x^2 + x^2 (r^2+2s) - p^2 x (r) - 2 p q x - p x (r^2+2s) - 2 q x (r) - 2 b_1 b_2 x (r) + s p^2 + p q (r) + s p (r) + q^2 + q (r^2-2s) + s^2$

This polynomial has coefficients in $\mathbb{Z}[p,q,r,s]$.

$\endgroup$
5
  • $\begingroup$ What is the meaning of "symmetric function in the $ a_i $"? $\endgroup$
    – Riccardo
    Dec 1 '13 at 23:25
  • 1
    $\begingroup$ en.wikipedia.org/wiki/Symmetric_polynomial $\endgroup$ Dec 2 '13 at 0:27
  • $\begingroup$ Dear @MartinBrandenburg, how does this prove the sum and product of integral elements is integral for an arbitrary commutative ring? The monic witnesses need not be split. Do you pass to the universal splitting algebra? $\endgroup$
    – Arrow
    Sep 21 '21 at 10:13
  • $\begingroup$ @Arrow Yeah why not. $\endgroup$ Sep 21 '21 at 21:10
  • $\begingroup$ @MartinBrandenburg do you see a way to deduce the sum/product of separable elements is separable? $\endgroup$
    – Arrow
    Sep 21 '21 at 21:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy