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An ideal $P$ of a ring $R$ is called prime (In ideal sense) if for ideals $I,J $ of $R$, $P$ contains the product ideal $IJ$ implies $P$ contains $I$ or $P$ contains $J$.

At this moment we call prime ( in ideal sense) as (ideally) prime.

I can't found a ring and one maximal ideal which is not (ideally) prime but also cannot disprove the exsitance of such an animal.

What I've got is If such an animal exists, it must lies in a non commutative ring, or a non unital ring (actually stronger, $R^2$ is strictly contained in $R$)

Thank you.

By the way, I have an example that an ideal is maximal but not prime in the usual sense ( for any two elements $a,b $ in $R$, $ab$ in $P$ implies $ a$ in $P $ or $b $ in $P$ ), unfortunately that non prime ideal is Ideally prime. It is the zero ideal in the matrix ring with entry in an division ring.

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  • $\begingroup$ I'm not familliar with the ideally prime definition however a maximal ideal is always prime in regards to your last paragraph $\endgroup$ Commented May 20, 2018 at 0:35
  • $\begingroup$ @SheelStueber no, zero is maximal but not prime in that ring $\endgroup$ Commented May 20, 2018 at 0:43
  • $\begingroup$ @SheelStueber Not quite - there's something subtle going on here. When the ring R is commutative, what you said is true, but in this case the two definitions are equivalent. (When the ring R is non-commutative, well, most people will still tell you that all maximal ideals are prime - but they are not working with the same definition of "prime" as the one given in this post!) $\endgroup$
    – Billy
    Commented May 20, 2018 at 0:45
  • $\begingroup$ @Billy even commutative is not enough, also need unital or R^2=R $\endgroup$ Commented May 20, 2018 at 0:46
  • $\begingroup$ ah i supposed all rings were unital+ commutative my fault $\endgroup$ Commented May 20, 2018 at 2:19

1 Answer 1

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When $R$ is unital, the answer is no.

Let $P$ be a maximal ideal, and choose any two ideals $I$ and $J$ such that $I\not\subseteq P$ and $J\not\subseteq P$. Then $P+I$ and $P+J$ are ideals of $R$ that contain $P$ and some stuff not in $P$ - and so they're bigger than $P$, so (as $P$ was maximal) they must be equal to $R$.

In particular, we can write $1\in R = P+I$ as $1 = p+i$ for some $p\in P$ and $i\in I$, and $1\in R = P+J$ as $1 = p'+j$ for some $p'\in P$ and $j\in J$. Multiplying these two expressions together, we get $$1 = (p+i)(p'+j) = pp' + ip' + pj + ij.$$ But it's obvious that $pp'$, $ip'$ and $pj$ are in $P$. Hence $ij\in P$ if and only if $1\in P$. But we know already that $1\not\in P$, because $P$ is a maximal ideal (and hence by definition $P\neq R$). So $ij\not\in P$, and hence $IJ\not\subseteq P$.

When $R$ is non-unital, the answer is yes. For example, the non-unital ring $R = 2\mathbb{Z}$ contains the maximal ideal $P = 4\mathbb{Z}$, and the ideals $I = J = 2\mathbb{Z}$. Then $IJ\subseteq P$, but $I, J\not\subseteq P$.

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  • $\begingroup$ Notice that R need not to be unital $\endgroup$ Commented May 20, 2018 at 0:44
  • $\begingroup$ Ah. Updated my answer to reflect this. $\endgroup$
    – Billy
    Commented May 20, 2018 at 0:49
  • $\begingroup$ (In fact, didn't you stumble on this answer yourself? I didn't understand what you meant by $R^2 \subset R$ until I came up with this example, but I'm presuming this is what you meant.) $\endgroup$
    – Billy
    Commented May 20, 2018 at 0:51
  • $\begingroup$ Wow, I have to say it's brilliant construction. I should have come up with this. Thank you a lot, I appreciate it. $\endgroup$ Commented May 20, 2018 at 0:51
  • $\begingroup$ haha actually I before I want to find this I haven't raise any example such that R^2 is strictly contained in R .. So I said I should have... $\endgroup$ Commented May 20, 2018 at 2:30

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