If $r(\mathfrak a)$ is maximal, then $ \mathfrak a $ is primary.

Question

I want to prove

If $r(\mathfrak a)$ (radical) is maximal, then $ \mathfrak a $ is primary.

Let $xy \in \mathfrak a$ thus $xy \in r(\mathfrak a)$. By primality $x \in r( \mathfrak a)$. Thus $x^n \in \mathfrak a$ and thus $ \mathfrak a$ is primary.
Whats wrong with above proof ? (Clearly something is wrong as we only used primality and not maximality!).
Above, I have followed notation of Atiyah Macdonald. Let me know if something is not clear.
Also I am quoting this result from Atiyah Macdonald.

Answer 1

The proof is incorrect for the following reason: the definition of primary is not symmetric, and you used the symmetry without realizing in your proof. Lets go over it line by line:

"Let $xy \in \mathfrak a$ thus $xy \in r(\mathfrak a)$. " So far it looks good.

"By primality $x \in r(\mathfrak a)$." Here is the issue. You mean By primality $x \in r(\mathfrak a)$ or $y \in r(\mathfrak a)$ .

"Thus $x^n \in \mathfrak a$". Here the line above becomes, "Thus, there exists some $n$ such that $x^n \in \mathfrak a$ OR there exists some $m$ such that $y^m \in \mathfrak a$".

But you need $n=1$ to make $\mathfrak a$ primary.

A counterexample, acording to wiki, is the following:

$$R=k[x,y,z]/(xy-z^2) ; \mathfrak a= (\bar{x}, \bar{z})^2$$

Then $rad(\mathfrak{a})=( \bar{x}, \bar{y})$ is prime, but $\mathfrak a$ is not primary, as $\bar{x} \bar{y} =\bar{z}^2 \in \mathfrak a$ but you have neither $\bar{x} \in \mathfrak {a}$ nor $\bar{y}^n \in \mathfrak a$.

With this example, you can see exactly where your proof goes wrong: $\bar{x} \bar{y} \in rad(\mathfrak{a})$ gives as above either $\bar{x}^n \in \mathfrak{a}$ or $\bar{y}^m \in \mathfrak{a}$. The second condition does not happen, and the first only happens for $n \geq 2$, but you need $n=1$.

P.S. The subtle mistake is related to the following: the definition of the primary is basically the following: $\mathfrak a$ is primary exactly when $xy \in \mathfrak{a}$ implies that one of the following three relations happen.

• $x \in \mathfrak{a}$
• $y \in \mathfrak{a}$
• $x \in rad(\mathfrak{a})$ AND $y \in rad(\mathfrak{a})$.

If you read your proof carefully, and not skip the second part where $xy \in rad(\mathfrak{a})$ implies $x \in rad(\mathfrak{a})$ OR $y \in rad(\mathfrak{a})$, then you will see that you proved that

$rad(\mathfrak{a})$ prime and $xy \in \mathfrak{a}$ implies that

• $x \in rad(\mathfrak{a})$ OR $y \in rad(\mathfrak{a})$.

The fact that you skipped this part, missled you into confusing the OR shish should had been there with an AND at the end of the proof.

Answer 2

Let us try to rewrite your "proof" in a detailed fashion.

Let $xy \in \mathfrak a$ and suppose $y\not\in\mathfrak a$. So we need to show that $x^n \in \mathfrak a$ for some $n\in\mathbb N$. Clearly $xy \in r(\mathfrak a)$. By primality, either $x \in r(\mathfrak a)$ or $y\in r(\mathfrak a)$. So either $x^n \in \mathfrak a$ or $y^m\in\mathfrak a$.

Suppose $y^m\in\mathfrak a$. But as you can see, there is no problem with this case!! So we cannot conclude anything.