Given two polynomials $f$ and $g \in \mathbb{Q}[X]$, prove that $(f) + (g) = (h)$ and $(f)\bigcap(g) = (k)$

Question

Given two polynomials $f(X) = 3X^2 + 7X - 6$ and $g(X) = 2X^2 + 5X - 3 \in \mathbb{Q}[X]$, prove that there exist $(h)$ and $(k) \in \mathbb{Q}[X]$ such that $(f) + (g) = (h)$ and $(f)\bigcap(g) = (k)$. Also give such $h$ and $k$.

I don't know how to correctly solve this problem. I know that for $I$ and $J$ ideals, $I + J$ is also an ideal. So does that mean that because of that fact, there exists a $h$ such that $(f) + (g) = (h)$?

The same can be said about the intersection, I think.

If my reasoning is correct, how do I find such $h$ and $k$?

My guess would be: $(h) = 5X^2 + 12X - 9$ and $(k) = (1)$ because $\gcd(f,g) = 1$.

If anyone can help me better understand and solve this problem, I would really appreciate it.

Answer 1

To find $h(X)$: $-3$ is a root of both $f(X)$ and $g(X)$ so $X+3$ divides both $f(X)$ and $g(X)$. Also you can write $x+3$ as a linear combination of $f(X)$ and $g(X)$. So let $h(X)=X+3$ then $(f)+(g)=(h)$. Note that $h$ is the greatest common divisor of $f$ and $g$.

To find $k$, factor $f$ and $g$, then you get $f(X)=h(X)f_1(X)$ and $g(X)=h(X)g_1(X)$ (with $f_1$ and $g_1$ linear). So $(f)\cap (g)=(hf_1g_1)$. This is because $hf_1g_1$ is the lowest common multiple.

Answer 2

It is more general: every ideal in $K[X]$ ($K$ a field) is principal, since on a polynomial ring over a field, you have a Euclidean division.

Indeed, if $I\ne 0$ is an ideal, consider a polynomial $p$ in $I$ of smallest positive degree. For any polynomial $f\in I$, write the Euclidean division by $p$: there exist polynomials $q,r$ such that $$f=qp+r,\quad r=0\enspace\text{or}\enspace \deg r <\deg p.$$ Now $r=f-qp\in I$ since $f$ and $p\in I$. As $\deg p$ is minimal, this implies $r=0$, i.e. $p\mid r$.