Problem 624

Let $R$ and $R’$ be commutative rings and let $f:R\to R’$ be a ring homomorphism.
Let $I$ and $I’$ be ideals of $R$ and $R’$, respectively.

(a) Prove that $f(\sqrt{I}\,) \subset \sqrt{f(I)}$.

(b) Prove that $\sqrt{f^{-1}(I’)}=f^{-1}(\sqrt{I’})$

(c) Suppose that $f$ is surjective and $\ker(f)\subset I$. Then prove that $f(\sqrt{I}\,) =\sqrt{f(I)}$

Proof.

(a) Prove that $f(\sqrt{I}\,) \subset \sqrt{f(I)}$.

Let $x\in f(\sqrt{I}\,)$ be an arbitrary element. Then there is $a\in \sqrt{I}$ such that $f(a)=x$. As $a\in \sqrt{I}$, there exists a positive integer $n$ such that $a^n\in I$.

It follows that we have
\begin{align*}
x^n=f(a)^n=f(a^n)\in f(I).
\end{align*}

This implies that $x\in \sqrt{f(I)}$.
Hence we have $f(\sqrt{I}\,) \subset \sqrt{f(I)}$.

(b) Prove that $\sqrt{f^{-1}(I’)}=f^{-1}(\sqrt{I’})$

$(\subset)$ Let $x\in \sqrt{f^{-1}(I’)}$. Then there is a positive integer $n$ such that $x^n\in f^{-1}(I’)$ and thus $f(x^n)\in I’$.
As $f$ is a ring homomorphism, it follows that $f(x)^n=f(x^n)\in I’$.

Hence $f(x)\in \sqrt{I’}$, and then $x\in f^{-1}(\sqrt{I’})$.
This proves that $\sqrt{f^{-1}(I’)} \subset f^{-1}(\sqrt{I’})$.

$(\supset)$ Let $x\in f^{-1}(\sqrt{I’})$. Then $f(x)\in \sqrt{I’}$. It follows that there exists a positive integer $n$ such that $f(x^n)=f(x)^n\in I’$.
Hence $x^n\in f^{-1}(I’)$, and we deduce that $x\in \sqrt{f^{-1}(I’)}$.

This proves that $f^{-1}(\sqrt{I’}) \subset \sqrt{f^{-1}(I’)}$.
Combining this with the previous inclusion yields that $\sqrt{f^{-1}(I’)}=f^{-1}(\sqrt{I’})$.

(c) Suppose that $f$ is surjective and $\ker(f)\subset I$. Then prove that $f(\sqrt{I}\,) =\sqrt{f(I)}$

We now suppose that $f$ is surjective and $\ker(f)\subset I$. We proved $f(\sqrt{I}\,) \subset \sqrt{f(I)}$ in part (a). To show the reverse inclusion, let $x\in \sqrt{f(I)}\subset R’$.
Then there is a positive integer $n$ such that $x^n\in f(I)$.
So there exists $a\in I$ such that $f(a)=x^n$.

Since $f:R\to R’$ is surjective, there exists $y\in R$ such that $f(y)=x$.
Then we have
\begin{align*}
f(a)=x^n=f(y)^n=f(y^n),
\end{align*}
and hence $f(a-y^n)=0$.
Thus $a-y^n\in \ker(f) \subset I$ by assumption.
As $a\in I$, it follows that $y^n\in I$ as well.

We deduce that $y\in \sqrt{I}$ and
$x=f(y)\in f(\sqrt{I}),$ which completes the proof that $\sqrt{f(I)} \subset f(\sqrt{I})$.

Putting together this inclusion and the inclusion in (a) yields the required equality $f(\sqrt{I}\,) =\sqrt{f(I)}$.

Let $I=(x, 2)$ and $J=(x, 3)$ be ideal in the ring $\Z[x]$. (a) Prove that $IJ=(x, 6)$. (b) Prove that...