Describe the smallest ideal (x, y) in k[x, y] containging both x and y. I am stumped on a problem from abstract algebra.
The problem is asking to describe what the elements look like in the smallest ideal (x,y) in k[x,y] where k is a field. Below is the complete question. 
Suppose k is a field.
Define k[x,y] = (k[x])[y]. Describe the smallest ideal (x,y) in k[x,y] containing
both x and y. (What do its elements look like?)
Thanks!
 A: Well, what is the smallest ideal containing $x$? This is $k[x, y] \cdot x$, which means all of the polynomials multiplied by $x$. Thus, the smallest ideal containing $x$ is the set of all polynomials that contain $x$ as a factor. Similarly, the smallest ideal containing $y$ is $k[x, y] \cdot y$, or the set of all polynomials that contain $y$ as a factor.
Now, we need to combine the ideals $k[x, y] \cdot x$ and $k[x, y] \cdot y$. It is well-known that to combine two ideals, we just add them, so our final answer is just $k[x, y] \cdot x+k[x, y] \cdot y$. Here are some examples of elements in this ideal:
$$x+y$$
$$x+y^2x+y+x^2$$
$$x+xy^3+y^3$$
Notice how to be in this ideal, all of the terms must have either an $x$ in them or a $y$ in them. This is because $k[x, y] \cdot x+k[x, y] \cdot y$ basically means a multiple of $x$ plus a multiple of $y$. None of the polynomials in this ideal have constant terms (except for $0$).
A: It contain $X$, so as it is an ideal, it contain $XX = X^2$, and by a quick reccurence, it contain every $X^n$, with $n>0$
It contain $Y$, so for the same reason, it contain every $Y^n$, $n>0$
It follow that it contain every $X^nY^k$, $n+k>0$
And as an ideal is a subgroup, it contain every linear combinaison of such elements.
It means that it contain every polynom without constant part. But this is an ideal, hence the result 
A: Show that $\phi: k[x,y] \to k$ given by $\phi(f) = f(0,0)$ is a ring-homomorphism.
Then show $\text{ker }\phi = (x,y)$. It's pretty easy to see that $(x,y) \subseteq \text{ker }\phi$, since $x,y \in \text{ker }\phi$, and $(x,y)$ is the minimal ideal containing $x$ and $y$. Use the definition of $\phi$ to show that if $\phi(f(x,y)) = 0$, the constant term is $0$, and thus $f(x,y) = x(a(x,y)) + y(b(x,y)) \in (x,y)$.
