Let $$B=\langle 3x+y-m, 4x+y\rangle⊆ ℝ[x,y].$$
Find a parameter $m$ so that $$B=ℝ[x,y].$$
My attempt so far:
If $1_{ℝ[x,y]}\in B$ then we get the desired equivalence, since any ideal containing the unit is equal to the whole ring. But what is the unit element here? Is it true that $1_{ℝ[x,y]}=1?$
If so, then it must hold that for $p(x,y),q(x,y)\in \mathbb R[x,y]$ we have: $$(3x+y-m)p(x,y)+(4x+y)q(x,y)=1$$ and from this it follows that $m≠0$ so that if we denote $p_0$ the constant term of $p(x,y)$ we must have $p_0m=1$.
Is this approach wrong? I am rather unfamiliar with multivariate rings and I think I am missing something trivial here.
$$\mathbb R[x,y]/(3x+y-m,y+4x)\simeq\frac{\mathbb R[x,y]/(y+4x)}{(3x+y-m, 4x+y)/(y+4x)}\simeq\mathbb R[x]/(x+m)\simeq\mathbb R,$$ so there is no such $m\in\mathbb R$.
You can also see this from $(3x+y-m)p(x,y)+(4x+y)q(x,y)=1$ by sending $y$ to $-4x$ and getting $(x+m)p(x,-4x)=-1$.
Edit. If $m\in\mathbb R[x,y]$ then you can choose $m=-x+a$ with $a\in\mathbb R$, $a\ne0$.