# Is the sum of two prime ideals a prime ideal?

Is the sum of two prime ideals (of a ring) again a prime ideal?

Please can someone give me a hint? I know that the intersection of two prime ideals is again a prime ideal iff one is contained in the other, but I'm not sure about the sum.

• Prime ideals are neither closed under sums, nor under intersection. Look at prime ideals in $\mathbb{Z}$. – Martin Brandenburg Sep 19 '13 at 13:14
• Are you sure about the intersection? Take the two prime ideals $2 \mathbb{Z}$ and $3 \mathbb{Z}$ in $\mathbb{Z}$. – Andreas Caranti Sep 19 '13 at 13:14
• $2\mathbb{Z}\cap 3\mathbb{Z}=6\mathbb{Z}$ which is not a prime ideal. – Edoardo Lanari Sep 19 '13 at 13:34

Th answer is no, consider the ring $\mathbb{Z}$ and the prime ideals $(2)$ and $(5)$. Since two and five are relative primes then every number can be written as a linear combination over $\mathbb{Z}$ of 2 and 5. To see this. note that if if $k$ is any interger, since we know that $$5-(2)2=1$$ by multiplying by $k$ we obtain the expression $$k5-(2k)2=k$$ which is a $\mathbb{Z}$-linear combination of $k$ in therms of $2$ and $5$. Note that $k5\in(5)$ and $-2k(2)=-4k\in(2)$. This means that $(2)+(5)=\mathbb{Z}$, which is not a prime ideal.

Note that the sum of ideals is an ideal when one is contained in the other. This is not possible in $\mathbb{Z}$ with the exemption when of the ideals is the prime ideal $(0)$, but other rings can give more interesting examples. Have a look at polynomials rings for example.

• how is that if two and 5 are relatively prime yhen every ... of 2 and 5. – abc Sep 19 '13 at 13:27
• two integers are relative prime iff you can express the number 1 as a linear combination of the two numbers, it follows from euclid's algorithm. – Mauricio G Tec Sep 19 '13 at 13:30

I think (correct me if I'm wrong) that if you are not so restrictive in your definition of prime ideals (allowing 1 to belong to a prime ideal) the answer is positive.

For instance, let $p_{1}$ and $p_{2}$ be two prime ideals of $R$, and take the natural projection map $\pi: R \longrightarrow R/p_{1}$, then $\pi(p_{2})$ is a prime ideal because $\pi$ is a surjective map and $p_{2}$ is a prime ideal. But then, $\pi^{-1}(\pi(p_{2}))$ is a prime ideal in R, and $\pi^{-1}(\pi(p_{2}))=p_{1}+p_{2}$.

• 1. The ideal $(1)$ is not prime. 2. In $\mathbb{C}[x,y]$ the ideals $(y)$ and $(x^2+y^2-1)$ are prime but their sum is not. Images don't preserve primeness. That is, $\pi(p_2)$ is not necessarily prime. – Zach Teitler Aug 22 '17 at 1:51
• I started my answer by saying that "if one is not so restrictive in the definition of the prime ideals (allowing 1 to belong to a prime ideal)", I'm not saying every prime ideal must have a 1, but rather that I'm allowing (1) to also be a prime ideal – Santiago Estupiñán Aug 22 '17 at 2:08
• Images do preserve primeness if the map is surjective – Santiago Estupiñán Aug 22 '17 at 2:14
• You may want to read the definition of a prime ideal from Wolfram mathworld.wolfram.com/PrimeIdeal.html where he says that whether it is proper or not depends on the author – Santiago Estupiñán Aug 22 '17 at 16:11
• Only for prime ideas that contain the kernel. – Zach Teitler Aug 22 '17 at 17:45