Being maximal ideal follows from being a kernel 
The ideal of all polynomials in $k[x_1,\ldots,x_n]$ with zero constant term is maximal (since it is the kernel of the homomorphism $k[x_1,\ldots,x_n]\to k$ which maps $f\mapsto f(0)$).

I understand why it is a maximal ideal. But I don't get why being the kernel of that homomorphism implies that it is maximal.
 A: Remember that in a commutative ring with unity $R$, you have the following condition for maximal ideals:
An ideal $I$ of $R$ is maximal $\iff R/I$ is a field
Then if $R = k[x_1, \dots, x_n]$, you're considering the map $ \phi: k[x_1, \dots, x_n] \to k $ given by $\phi(f(x_1, \dots, x_n)) = f(0, \dots, 0)$, that is, the evaluation at zero map. Since this map is clearly surjective, by the first isomorphism theorem you get
$$
k[x_1, \dots, x_n]/\ker{(\phi)} \cong k
$$
But then since $\ker{(\phi)} = \langle x_1, \dots, x_n \rangle$ is the ideal of all polynomials in $k[x_1, \dots, x_n]$ with zero constant term, and since $k$ is a field, by the above stated result, this shows that the ideal is maximal.
A: Note that


*

*"is an ideal" is the same as "is a kernel of a ring epimorphism to a ring"

*"is a prime ideal" is the same as "is a kernel of a ring epimorphism to an integral domain"

*"is a maximal ideal" is the same as "is a kernel of a ring epimorphism to a field"


If $R$ is a (commuative) ring with unity and $A$ an ideal, then $R/A$ is a ring and is an integral domain iff $A$ is prime and is a field iff $A$ is maximal. 
