Number of generators of a given ideal.

Question

Let $I=\langle 3x+y, 4x+y \rangle \subset \Bbb{R}[x,y]$. Can $I$ be generated by a single polynomial?

My approach: If $I$ can be generated by a single polynomial, then the two "apparent" generators, $(3x+y),(4x+y)$ are dependent in this fashion:
There exist two polynomials $p(x),q(x) \in \Bbb{R}[x,y]$ such that: $(3x+y)p(x)=(4x+y)q(x)$.
For $y=-3x$ in the above equivalence we have: $0=-xq(x)$ so $q(x)=0$.
Similarly, for $y=-4x$ we have $p(x)=0$.
So $(3x+y),(4x+y)$ are not dependent and $I$ cannot be generated by a single polynomial. Morover, since $(3x+y),(4x+y)$ are linear combinations of the variables $x,y$ the number of generators is exactly 2 and they are $x,y$.
Thus, $I=\langle x, y \rangle$.

I am not familiar with multivariate polynomials and I might be misinterpreting something here (basically "translating" linear independence in a wrong manner). Also, is the last statement about $(3x+y),(4x+y)$ being linear combinations sufficient for the proof, ommiting the preceding steps?

Answer 1

$I=\langle 3x+y, 4x+y \rangle=\langle 3x+y, x \rangle=\langle y, x \rangle$. Now refer to The ideal $I= \langle x,y \rangle\subset k[x,y]$ is not principal.

Answer 2

Hint: No, since $R[x,y]/I$ is isomorphic to $R$.

Answer 3

It is not enough to consider only polynomials $p(x)$ and $q(x)$ in one variable. Assume that $I$ is principal, and let $f$ be a generator in $I$. We need to prove or disprove that $f(x,y)=(3x+y)p(x,y)+(4x+y)q(x,y)$ for some polynomials $p,q \in \mathbb{R}[x,y]$. Indeed, we obtain a contradiction.