# If a finitely-generated ideal is prime, does it follow that the generators are prime?

If a finitely-generated ideal is prime, does it follow that the generators are prime? This is clearly true if it is generated by $1$ element, but is it true for any $n \ge 2$? If not, are there prime ideals generated by elements, none of which are prime?

• If $I=(x_1,\ldots,x_n)$ then also $I=(x_1,\ldots,x_n,x_1^2)$ and $x_1^2$ is not prime. Jun 5 '18 at 13:17

In $\mathbb{Z}[X]$, the ideal generated by $2X$ and $3X$ is $(X)$ because $X = 3X-2X$, so it is prime.

However, $2X$ is not prime because $2X$ divides $2\times X$, but it does not divide $2$ nor $X$ in $\mathbb{Z}[X]$. The same goes for $3X$.

• Thank you. Nice example
– user544680
Jun 5 '18 at 12:55
• You're welcome :) Jun 5 '18 at 12:58

It is not even true for principal ideals, as you say: $0$ is a generator of $\{0\}$ in any domain, but $0$ is not prime.

And in general, generating sets are flexible. You can add redundancy to them, and you generate the same ideal. So, suppose $x$ is in a generating set for a finitely generated ideal. Then adding $x^2$ to the same generating set still yields a generating set, and $x^2$ is not prime.

These are useful things to point out for the question as stated. If you meant to include some other hypotheses about the generating set being minimal or the ideal being nonzero, then I believe that the other solutions have covered this well.

Take the ring $\mathbb F_2^3$ (with componentwise addition and multiplication). Then the prime ideals are $\mathbb F_2\times \mathbb F_2\times \{0\}$, $\mathbb F_2 \times \{0\}\times \mathbb F_2$ and $\{0\}\times \mathbb F_2\times \mathbb F_2$.

For example $\mathbb F_2\times \mathbb F_2\times \{0\}$ is generated by $e_1 = (1,0,0)$ and $e_2 = (0,1,0)$, and this is the only generating set. But neither $e_1$ nor $e_2$ are prime: For example, $e_1$ divides $0 = e_2\cdot e_3$ (where $e_3 = (0,0,1)$), but $e_1$ divides neither $e_2$ nor $e_3$. Therefore $e_1$ is not prime.

Edit: As pointed out in the comment below, the above generating set is not unique, since also $e_1+e_2 = (1,1,0)$ is a generator, which is even prime! Hence, the above example (though answering the question) does not answer the question I wanted to answer.

Let me make another attempt: Consider the ring $\mathbb Z[\sqrt{-5}]$. It is known to be a Dedekind ring. In particular each non-zero prime ideal is a maximal ideal. The ideal $(2,1+\sqrt{-5})$ is not principal. If it would contain a non-zero prime element, say $x$, then $(x)$ would be a non-zero prime ideal. But, as noted above, $(x)$ then has to be a maximal ideal and hence $(x) = (2,1+\sqrt{-5})$: a contradiction.
In particular, $(2,1+\sqrt{-5})$ is a prime ideal which is not generated by prime elements.

• This example also shows that there are rings with prime ideals, such that every generating set consists of non-prime elements. Jun 5 '18 at 13:13
• Your example is a principal ideal ring: the ideal you've given is also generated by $(1,1,0)$, which is prime, so the remark in the comment above isn't really true... Jun 5 '18 at 13:43
• whoops… I hadn't noticed that. Thank you for pointing this out! Jun 5 '18 at 13:51

It is not clear what you mean by the generators. In $\mathbb Z$, $(4, 6)=(2)$ is prime, but neither $4$ nor $6$ are.