I have a PID, in particular $\mathbb{Q}[x]$ and I must find the generator for the ideal generated by $x^4-1$ and $x^3-x$. Now, the ideal generated by these two elements looks like $p(x)(x^4-1) + q(x)(x^3-x)$ for $p,q \in \mathbb{Q}[x]$, correct? Now, in order to find a generator, I think that we can use division to find a common generator, correct? as in,
$x^4-1 = x(x^3-x) + (x^2-1)$, and so $x^2-1 = (x^4-1) – x(x^3-x)$, and so $(x^4-1), -x(x^3-x) \in I$ so so is their sum, and so $x^2-1$ generates the whole ideal? Is this correct?
There is, however, a second part of the question which asks if, in $\mathbb{Q}[x,y]$, there exists a generator $p(x,y)$ that generates the ideal generated by the polynomials $<x>$ and $<y>$. I can show this ideal is maximal, but I don't know how to show its not expressible as a principal ideal.
Best Answer
The general problem of finding a generator for the ideal generated by two elements in a P.I.D. is not so easy, except if the P.I.D. is a Euclidean domain – which is the case of a polynomial ring (in one indeterminate) over a field: then we have the Euclidean algorithm.
However, iIn the present case, you can easily factor the given polynomials into irreducible polynmials: $$ x^4-1=(x-1)(x+1)(x^2+1),\qquad x^3-x=x(x-1)(x+1),$$ so the g.c.d. decomposes as the product of the common irreducible factors and you obtain indeed $$(x-1)(x+1)=x^2+1.$$
As to your second question, the ideal $(x,y)$ is not a principal ideal. If it were generated by $p(x,y)$, we would have polynomials $q(x,y)$ and $r(x,y)$ such that $$x=p(x,y)q(x,y),\qquad y=p(x,y)r(x,y).$$ Since the partial degrees are multiplicative, these equalities imply that $\;\deg_yp(x,y)=0\,$ and $\;\deg_xp(x,y)=0\,$ respectively. In pother words, $p(x,y)$ should be a constant, and we'd have polynomials $u(x,y)$ and $v(x,y)$ such that $$ x\,u(x,y)+y\,v(x,y)=1. $$ Setting $y=0$, we would get $\;x\,u(x,0)=1$, which is impossible for degrees reason.