The ideal $I= \langle x,y \rangle\subset k[x,y]$ is not principal 
The ideal $I= \langle x,y \rangle\subset k[x,y]$ is not a principal ideal.

I don't know how to consider it. Any suggestions?
 A: Suppose it would be principal. Then there would exist $p(x,y)$ such that for all polynomials $g(x,y)$ and $h(x,y)$ in $k[x,y]$, we would have some $i(x,y)$ such that 
$$
xg + yh = ip.
$$
In particular, $x = i_1 p$ and $y = i_2 p$, so that the degree of $ p$ is at most $1$. It can't be zero, because that would mean p is a unit or zero and we would lose the fact that $\langle p \rangle = \langle x,y \rangle$. So $\deg p = 1$, hence $p=ax+by+c$ for $a,b,c \in k$. But then $x = i_1(ax+by+c)$ means $b= 0$ and $ y =i_2 (ax+c)$ means $a=0$, hence $p$ has not degree $1$, contradiction.
Hope that helps,
A: Suppose it's generated by $P_0$. Then $x=PP_0$ and $y=QP_0$ for some $P,Q\in k[x,y]$. For each $y$, $P(x,y)P_0(x,y)$ is a polynomial of degree $1$ with respect to $x$ so the maximal degree with respect to $x$ of $P_0(x,y)$ is at most $1$, so $P_0(x,y)=a_0(y)+a_1(y)x$ where $a_0,a_1\in k[y]$. 
Doing the same for $y$ we get that the maximal degree with respect to $y$ of $P_0(x,y)$ is at most $1$ for each $x$ so the maximal degree of $a_0$ and $a_1$ is $1$. We have $P_0(x,y)=a_{0,0}+a_{0,1}y+a_{1,0}x+a_{1,1}xy$ where $a_{0,0},a_{1,0}, a_{0,1},a_{1,1}\in k$. Identifying the term of highest degree in the equality $x=P(x,y)(a_{0,0}+a_{0,1}y+a_{1,0}x+a_{1,1}xy)$ we get that $a_{1,1}=0$ and since $a_{0,1}y+a_{1,0}x\in\langle x,y\rangle$, $a_{0,0}=0$. We finally get that $a_{1,0}=a_{0,1}=0$, which is a contradiction.
A: Hint $ $ Among principal ideals: contains = divides; thus having no proper containing ideal (maximal) is equivalent to having no proper divisor (irreducible). Thus the only principal ideal containing $(x)$ is $(1)$. But $(x,y)\ne (1)$ by evaluating at $x=0=y.\:$ QED. Below is more detail
Observe the ideals $\: (x) \subsetneq (x,y) \subsetneq (1)\:$ are distinct $\: $  primes $\rm\: [\:or\ (1)\:]\:$ since
their  residue rings $\ k[y]\ \supsetneq\ k\:\ \supsetneq\:\ (0)\:$ are distinct domains $\rm[\:or\ (0)\:].$
So $(x,y)$ isn't principal since principal primes $\ne (0)$ are maximal among principal ideals.
This maximality is familiar from PIDs, e.g. in $\mathbb Z$ or $k[x]$.
It simply says that $\:\gcd(x,p) = p\ \ {\rm or}\:\ 1\:$ if $\:p\:$ is prime,
or, in ideal language, $(x,p) = (p)\ {\rm or}\ (1)\:$ so $(p)$ is maximal.
More generally, in any domain where ideals satisfy contains $\equiv$ divides (e.g. Dedekind domains), prime ideals $\ne 0$ are maximal. This characterizes PIDs, i.e. PIDs are precisely those UFDs where every prime ideal $\ne 0$ is maximal (i.e. Krull dimension $\le 1$). For a handful of other useful characterizations of when a UFD is a PID see this post.
A: $k[x,y]$ modulo a proper principal ideal is infinite dimensional, yet your ideal is of codimension $1$.
A: A generator $g$ of $\langle x,y \rangle$ would divide $x$ and $y$, which are irreducible and non-associated. Thus $g$ would be a unit, and  $\langle x,y \rangle$ would be the unit ideal, contradiction.
A: Here's a complete simple  proof. Put $\:\rm c \!=\! y,\ D = k[y]\:$ below, using $\rm\: y\:\!g(y) \ne 1\:$ by eval at $\rm\: y\!=\!0$.
Theorem $\ $ Let $\rm\: 0\ne c\in D\:\!$ a $\rm\color{#90f}{domain}$. In $\rm\: E = D[x]:\ (c,x) = (f)\ \Rightarrow\  c\ |\ 1\: $ in $\rm D.\ \, $ Proof:
$\rm\ \ f\ \in\ (c,x)\ \Rightarrow\ f = c\, g_1 + x\,h_1.\, $ Eval at $\rm\, x = 0\ \Rightarrow\ \color{#0a0}{f(0)} = cg_1(0) = \color{#0a0}{cd},\,\ d\in  C$
$\rm\  \ c\ \in\ (f)\ \Rightarrow\ c\ =\ f\, g, \ \,\color{#90f}{hence}\,\ \ \deg f\:\! =\:\! 0\:\Rightarrow\:  \color{#c00}f = \color{#0a0}{f(0) = cd}$
$\rm\ \ x\ \in\ (f)\ \Rightarrow\, x\ =\ \color{#c00}f\, h\ =\ \color{#0a0}{cd}\:\!h.\, $ Eval at $\rm\ x\! =\! 1\ \Rightarrow\ 1 = c\:\!d\,h(1)\ \Rightarrow\ c\mid 1\,\Rightarrow\, c\,$ unit in $\rm\,C$

Beware $ $ Our use of $\rm C$ is a $\rm\color{#90f}{domain}$ to infer $\,\deg c=0,\ f\mid c\Rightarrow \deg f = 0\,$ is crucial to the proof. This generally fails for non-domains, and so does the Lemma, e.g. in $\,\Bbb Z/6[x]\!:\ (2,x) = (2+3x)\,$ by $\,(2,x)=(2+3x,x)=(2+3x),\,$ by $\,2+3x\mid x,\,$ by $\,(2+3x)(3+2x)=x.\,$ Notice  here we have positive degree $ f\mid c,\,$  i.e. $\,2+3x\mid 2\,$ [by $\,-2(2+3x) = 2$].
