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I am trying to prove that $K[x,y]$ is not principal, $K$ a field.

I have taken in consider the polynomial $I=\langle x,y \rangle$. If $I$ is principal, then $I=\langle d\rangle$, for $d \in K[x,y]$. But in this case $d$ is a unit, it contains $1$ and hence $\langle d\rangle=K[x,y]$.

Now I have to show that $\langle d\rangle$ is distinct from $\langle x,y \rangle$. I think to prove this by showing that the polynomial $1$ does not lie in $\langle x,y \rangle$. If $1$ lies in $\langle x,y \rangle$, then $1=px+qy$, for some $p, q \in K[x,y]$, but I cannot arrive at a contradiction.

Would you help me, please? Thank you in advance.

user404634
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2 Answers2

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The constant term of $px+qy$ is $0$ not $1$, because every term has at least one $x$ or $y$ in it.

Lukas Heger
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  • Thank you! So, the degree of (px+qy) cannot be zero, since every term contains at least one x or y in it contrary to the degree of 1 which is zero. – user404634 Jun 18 '17 at 20:00
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Hint $ $ If $\,(x,y) = (f)\,$ then $\,f = \underbrace{ \gcd(x,y) = 1}_{\large {\rm by\ prime}\ x\,\nmid\, y}$ so $1 = x g + y h\,$ $\Rightarrow1 = 0$ by eval at $x,y = 0$

Bill Dubuque
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  • $x$ is prime by $K[x,y]/(x) \cong K[y]$ a domain, and $x\nmid y $ else $xf = y,\Rightarrow, 0 = 1,$ by eval at $,x,y=0,1$ – Bill Dubuque Jun 18 '17 at 20:14