Proposition

Suppose that $k$ is a field and $\phi: k[x_1, \cdots, x_n] \rightarrow k[x_1]$ is a ring homomorphism that is the identity on $k$ and maps $x_1$ to $x_1$. Given an ideal $I \subset k[x_1, \cdots, x_n]$, prove that $\phi(I) \subset k[x_1]$ is an ideal.

Solution

Since $\phi$ is a ring homomorphism, $\phi(I)$ is an additive subgroup of $k[x_1]$. Let $f \in k[x_1]$. Then $f$ is a polynomial in $x_1$. $f$ can be seen as an element of $k[x_1, \cdots, x_n]$, and moreover, $\phi(f) = f$. Since $I$ is an ideal, $fI \subset I$. Then $f\phi(I) = \phi(f)\phi(I) = \phi(fI) \subset \phi(I)$, so $\phi(I)$ is indeed an ideal.