Ring Epimorphism and Nil Radical

[Sudharaka]

Hi everyone, :)

Here's a question that I failed to do correctly in an exam. I want to find the answer to this and understand it fully. Any comments, hints would be greatly appreciated.

Question:

If $\theta:\, R\rightarrow S$ is a ring epimorphism, prove that \(\theta(\mbox{Nil }( R))\subseteq\mbox{Nil }(S)\) where $\mbox{Nil}( R)$ is the nil radical (sum of all nil two sided ideals of $R$).

 

[Sudharaka]

Hi everyone, :)

Here's a question that I failed to do correctly in an exam. I want to find the answer to this and understand it fully. Any comments, hints would be greatly appreciated.

Question:

If $\theta:\, R\rightarrow S$ is a ring epimorphism, prove that \(\theta(\mbox{Nil }( R))\subseteq\mbox{Nil }(S)\) where $\mbox{Nil}( R)$ is the nil radical (sum of all nil two sided ideals of $R$).

I think I got the answer to this one. Take any element, $\theta(x)\in \theta(\mbox{Nil }( R))$. Then $x\in\mbox{Nil }( R)$ and therefore $x^n=0$ for some $n$. Since $\theta$ is a ring homomorphism, $[\theta(x)]^n=\theta(x^n)=\theta(0)=0$ and therefore $\theta(x)\in\mbox{Nil }( S)\Rightarrow \theta(\mbox{Nil }( R))\subseteq\mbox{Nil }(S)$. The thing I don't understand is where I have to use the fact that $\theta$ is an epimorphism ?

Edit: I think I know what is happening here. In the book this problem was taken there is another part that asks to show that $\theta(\mbox{Rad }( R)) \subseteq\mbox{Rad }(S)$ where $\mbox{Rad }(R )$ denotes the Jacobson radical of $R$. Perhaps this is where the fact that $\theta$ is an epimorphism is used. :)

 

[Deveno]

Suppose $I$ is a nil ideal of $R$. Then $\theta(I)$ is a nil ideal of $S$ (here it is vital that $\theta$ be surjective, or else the multiplicative property of an ideal may fail).

It follows that any sum of nil ideals in $R$ is mapped by $\theta$ to a nil ideal in $S$.

(Note: strictly speaking, ring epimorphisms need not be surjective maps, although some texts use the term in this sense. For example, the inclusion map: $\Bbb Z \to \Bbb Q$ is an epimorphism in the strict sense, but is not surjective, since any map: $\Bbb Q \to R$ is uniquely determined by its values on the integers).

 

[Sudharaka]

Suppose $I$ is a nil ideal of $R$. Then $\theta(I)$ is a nil ideal of $S$ (here it is vital that $\theta$ be surjective, or else the multiplicative property of an ideal may fail).

Ohhhh….. (Nod) So I should show that $\theta(\mbox{Nil }(R ))$ is an ideal before showing that it's included in $\mbox{Nil }(S)$. And for this I need the surjective (epimorphism) property. Is this what you meant?

It follows that any sum of nil ideals in $R$ is mapped by $\theta$ to a nil ideal in $S$.

(Note: strictly speaking, ring epimorphisms need not be surjective maps, although some texts use the term in this sense. For example, the inclusion map: $\Bbb Z \to \Bbb Q$ is an epimorphism in the strict sense, but is not surjective, since any map: $\Bbb Q \to R$ is uniquely determined by its values on the integers).

I know that there's this difference between epimorphisms and surjective maps, but in our case I am pretty sure that the term epimorphism is used to mean a surjective ring homomorphism. :)

 

[Deveno]

Yes! The nilradical of $S$ may be somewhat smaller than the set of all nilpotent elements (since this set may not even be closed under addition). So showing that $\theta(r)$ is nilpotent when $r$ is, is necessary, but not sufficient.

If, however, you show $\theta(r)$ lies in some nil ideal, it necessarily must lie in the nilradical of $S$.

Here is how the surjectivity comes into play:

Clearly, for any ideal $I,\ \theta(I)$ is an additive subgroup of $S$ (since $\theta$ is an abelian group homomorphism, as all ring homomorphisms are).

Now if we pick any $s \in S, y \in \theta(I)$, since $\theta$ is surjective:

$s = \theta(r)$ for some $r \in R$.

Thus:

$sy = \theta(r)\theta(x) = \theta(rx)$ for some $x \in I$.

Since $I$ is an ideal, $rx \in I$, thus $sy \in \theta(I)$.

 

[Sudharaka]

Yes! The nilradical of $S$ may be somewhat smaller than the set of all nilpotent elements (since this set may not even be closed under addition). So showing that $\theta(r)$ is nilpotent when $r$ is, is necessary, but not sufficient.

If, however, you show $\theta(r)$ lies in some nil ideal, it necessarily must lie in the nilradical of $S$.

Here is how the surjectivity comes into play:

Clearly, for any ideal $I,\ \theta(I)$ is an additive subgroup of $S$ (since $\theta$ is an abelian group homomorphism, as all ring homomorphisms are).

Now if we pick any $s \in S, y \in \theta(I)$, since $\theta$ is surjective:

$s = \theta(r)$ for some $r \in R$.

Thus:

$sy = \theta(r)\theta(x) = \theta(rx)$ for some $x \in I$.

Since $I$ is an ideal, $rx \in I$, thus $sy \in \theta(I)$.

Thank you so much, you are a lifesaver. Now that I have understood the gist of the problem, I will read the additional details you gave me. Thanks again. :)