$\mathbb{R}[X]$ is an integral extension of $\mathbb{R}[X^2-1]$

Question

I am trying to prove that every polynomial of $\mathbb{R}[X]$ satisfies a monic polynomial equation with coeffients in $\mathbb{R}[X^2-1]$ that is every polynomial $b(x)= x^m+b_{m-1}x^m-1+...+b_{0}$ satisfies that there exits an $ n\in N$ such that $b(x)^n+a_{n-1}(x^2-1)b(x)^{n-1}+...+b_{0}$=0 is satisfied being $a_{i}(x^2-1)$ a polynomial in $\mathbb{R}[X^2-1]$ .

I have tried to distinguish the polynomials $b(X)$ depending of its degree so that I can choose $a_{i}(x^2-1)$ that satisfies the equation but I don't know if I am in the correct path.

Thank you.

@VeganMaths Why do you want to find explicit polynomials? To check that it is an integral extension, you just need to show that $X \in R[X]$ is integral over $R[X^2 - 1]$. Remember that sums an products of integral elements are integral. — Badam Baplan, May 24 '20 at 22:47

Bernard · Answer 1 · 2020-05-24T16:01:04.127

1

Hint:

$\mathbf R[X^2-1]$ contains $X^2$, hence $\mathbf R[X^2]$.

edited May 24 '20 at 16:01

answered May 24 '20 at 13:59

Bernard

175,478

How can I show the inclusion? – Vegan Maths May 24 '20 at 14:05
The inclusion of $X^2$ or of $R[X^2-1]$? – Bernard May 24 '20 at 14:07
I asked for which inclusion you wanted details. – Bernard May 24 '20 at 14:13
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Can I affirm that $\mathbb{R}[X^2]=\mathbb{R}[X^2-1]$ – Vegan Maths May 24 '20 at 14:18
Ues, the reverse inclusion is obvious since $X^2-1=\dots$, well $(X^2)-1$. – Bernard May 24 '20 at 14:21
And how can I find a monic polynomia such that it is integral any element of $\mathbb{R}[X]$ – Vegan Maths May 24 '20 at 15:57
That is another problem. Maybe writing any polynomial as $P(X^2)+Q(X^2)X first? – Bernard May 24 '20 at 16:04
But , the main aim is to find such polynomial in order to affirm that $\mathbb{R}[x] $ is an integral extension right? – Vegan Maths May 24 '20 at 16:12
You have other non explicit ways, using general properties of integral extensions, e.g. $\mathbf R[X]$ is an $\mathbf R[X^2]$-module of finite type. – Bernard May 24 '20 at 16:15

score 1 · Answer 2 · answered May 25 '20 at 14:15

Note that for a ring extension $A \subseteq B$, the elements of $B$ that are integral over $A$ form a ring (this is a standard theorem and many proofs are available online).

Clearly every element of $\mathbb{R}$ is integral over $\mathbb{R}[X^2 - 1]$.

Also, $X$ is a root of the monic polynomial $f(t) = t^2 - (X^2 - 1) - 1 \in \mathbb{R}[X^2 -1][t]$, so $X$ is integral over $\mathbb{R}[X^2 -1]$.

Since $\mathbb{R}[X]$ is generated as a ring by $\mathbb{R}$ and $X$, the result follows from the fact in the first paragraph.

$\mathbb{R}[X]$ is an integral extension of $\mathbb{R}[X^2-1]$

2 Answers2