Let $f = (x^4 - 1)(x + 2)$ and let $g = (x^2 - 4)(x + 1)$. Find a single polynomial that generates the ideal $I = (f,g)$ of $\mathbb{R}[x]$

Question

The Problem:

Let $f = (x^4 - 1)(x + 2)$ and let $g = (x^2 - 4)(x + 1)$. Find a single polynomial that generates the ideal $I = (f,g)$ of $\mathbb{R}[x]$.

My Approach:

This seemed a bit too easy, which made me think I'm misunderstanding the problem...

Since $f = (x+1)(x-1)(x^2 +1)(x+2)$ and $g= (x+2)(x-2)(x+1)$, it follows that $f$ and $g$ are both in the ideal generated by $p = x+1$ (or, for that matter, the ideal generated by $q = x-1$). Thus $p$ generates $I$.

Is that all there is to it?

EDIT: I had misdefined $g$.

Answer 1

It is well-known that $\Bbb R[x]$ is a principal ideal domain; thus the ideal

$(f(x), g(x)) = (d(x)) \tag 1$

for some

$d(x) \in \Bbb R[x]; \tag 2$

it immediately follows that

$d(x) = r(x)f(x) + s(x)g(x) \tag 3$

for some

$r(x), s(x) \in \Bbb R[x]; \tag 4$

furthermore, it is clear from (1) that

$d(x) \mid f(x), \; d(x) \mid g(x); \tag 5$

it is equally clear via (3) that any

$p(x) \in \Bbb R[x] \tag 6$

such that

$p(x) \mid f(x), \; p(x) \mid g(x) \tag 7$

will also satisfy

$p(x) \mid d(x); \tag 8$

(5) and (8) together show that $d(x)$ by definition is in fact given by

$d(x) = \gcd(f(x), g(x)); \tag 9$

that is, $d(x)$ is the greatest common divisor of $f(x)$, $g(x)$ in $\Bbb R[x]$; this fact gives us a tool by which $d(x)$ may be had: calculating the irreducible, hence prime, factors of $f(x)$ and $g(x)$; we find that

$f(x) = (x^4 - 1)(x + 2) = (x^2 + 1)(x + 1)(x - 1)(x + 2), \tag{10}$

$g(x) = (x^2 - 4)(x + 1) = (x - 2)(x + 2)( x + 1); \tag{11}$

inspection of (10), (11) reveals that every factor on the right-hand sides of these equations is a prime (irreducible) in $\Bbb R[x]$, being either linear or the quadratic $x^2 + 1$ which has no roots in $\Bbb R$; the factors they have in common are $x + 1$ and $x + 2$; therefore, since $\Bbb R[x]$ is a unique factorization domain (as is any principal ideal domain), the unique prime factors of $d(x)$ must be these two; that is

$d(x) = (x + 1)(x + 2). \tag{12}$

$OE\Delta$.

Answer 2

No: you have shown that $(f,g)$ is contained in $(p)$ (or $(q)$), but you need to show $(f,g)$ is equal to the ideal generated by some single polynomial. In other words, you would need to show the reverse inclusion as well, that $(f,g)$ contains $(p)$. You have not proved that, and in fact it is not actually true for the $p$ you have chosen.