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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?

user26857
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MathematicianByMistake
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3 Answers3

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

Community
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user26857
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  • Thank you for your answer! – MathematicianByMistake Mar 14 '16 at 17:49
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    @MathematicianByMistake In fact it's easy to see that a supposed generator is a non-zero constant (see the accepted answer in the linked thread) and this mean that $I=\mathbb R[x,y]$, so $1\in(x,y)$. Now write $1=xf+yg$ and for $x=y=0$ get $1=0$, a contradiction. (I hope everything is clear now.) – user26857 Mar 14 '16 at 17:56
  • Indeed it is, and again thank you for bringing clarity to what must appear as a rather mundane problem to those familiar with the concepts involved. Apparently I forgot that the principal ideal is generated by a single element. – MathematicianByMistake Mar 14 '16 at 18:02
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Hint: No, since $R[x,y]/I$ is isomorphic to $R$.

Tsemo Aristide
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  • Thank you. This is a very direct but clear approach! I think I can show this easily. – MathematicianByMistake Mar 14 '16 at 16:24
  • I don't understand this cryptical hint, but two users already did it. (I suppose that this is the reason for upvoting this answer.) Could some of them show me how to conclude from here that $I$ is not principal? (I also think $R$ should be $\mathbb R$.) – user26857 Mar 14 '16 at 17:51
  • @user26857 I don't know the original answerer's intent, but intuitively if $I$ is principal, then $\mathbb{R}[x,y]/I$ is the coordinate ring of a curve which should have dimension $1$. More formally, I think this follows from Krull's Hauptidealsatz. If $I$ were principal, then it would have height $1$. Since every maximal chain of primes of $\mathbb{R}[x,y]$ has length $2$, then $\mathbb{R}[x,y]/I$ would have dimension $1$. But $\mathbb{R}[x,y]/I \cong \mathbb{R}$ has dimension $0$, contradiction. – Viktor Vaughn Mar 15 '16 at 22:58
  • @SpamIAm I think I proved somewhere on M.SE that the maximal ideals of $K[x,y]$ can't be principal using even more elementary arguments. But look around and figure out where we are here: could such a hint help the OP? – user26857 Mar 15 '16 at 23:05
  • Here is the above mentioned proof: http://math.stackexchange.com/a/1625562/121097. – user26857 Apr 11 '16 at 00:22
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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.

Dietrich Burde
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