$bigcap_{ninmathbb{N}} I^n = (0)$ if and only if no zero divisor of $R$ is of the form $1-z$ with $zin I$.
$begingroup$
Full problem, suppose that $R$ is a commutative Noetherian ring and $I$ is an ideal of $R$. We wish to prove that
$$bigcap_{n=1}^{infty} I^n=(0)$$ if and only if no zerodivisor of $R$ is of the form $1-z$ with $zin I$.
First I'll suppose that the intersection is $(0)$. Let $zin I$ and let $0neq rin R$ such that $r(1-z)=0$. Then $r=rz$ and so $rin I$. Is this useful here? I'm not sure how to use the Noetherian condition of $R$ since the chain of $I^n$ is descending, not ascending.
Any help would be much appreciated! I'm studying for a qual and need all the help I can get.
abstract-algebra ring-theory commutative-algebra ideals noetherian
edited Dec 7 '18 at 18:28
user10354138
7,3772925
asked Dec 7 '18 at 18:23
GengarGengar
1949
$endgroup$
add a comment |
$begingroup$
Full problem, suppose that $R$ is a commutative Noetherian ring and $I$ is an ideal of $R$. We wish to prove that
$$bigcap_{n=1}^{infty} I^n=(0)$$ if and only if no zerodivisor of $R$ is of the form $1-z$ with $zin I$.
First I'll suppose that the intersection is $(0)$. Let $zin I$ and let $0neq rin R$ such that $r(1-z)=0$. Then $r=rz$ and so $rin I$. Is this useful here? I'm not sure how to use the Noetherian condition of $R$ since the chain of $I^n$ is descending, not ascending.
Any help would be much appreciated! I'm studying for a qual and need all the help I can get.
abstract-algebra ring-theory commutative-algebra ideals noetherian
edited Dec 7 '18 at 18:28
user10354138
7,3772925
asked Dec 7 '18 at 18:23
GengarGengar
1949
$endgroup$
$begingroup$
Maybe take a look at Krulls intersection theorem?
$endgroup$
– red_trumpet
Dec 7 '18 at 18:38
$begingroup$
That assumes that R is local, though @red_trumpet
$endgroup$
– Gengar
Dec 7 '18 at 18:47
$begingroup$
Right, I missed that :(
$endgroup$
– red_trumpet
Dec 7 '18 at 19:37
add a comment |
$begingroup$
Full problem, suppose that $R$ is a commutative Noetherian ring and $I$ is an ideal of $R$. We wish to prove that
$$bigcap_{n=1}^{infty} I^n=(0)$$ if and only if no zerodivisor of $R$ is of the form $1-z$ with $zin I$.
First I'll suppose that the intersection is $(0)$. Let $zin I$ and let $0neq rin R$ such that $r(1-z)=0$. Then $r=rz$ and so $rin I$. Is this useful here? I'm not sure how to use the Noetherian condition of $R$ since the chain of $I^n$ is descending, not ascending.
Any help would be much appreciated! I'm studying for a qual and need all the help I can get.
abstract-algebra ring-theory commutative-algebra ideals noetherian
edited Dec 7 '18 at 18:28
user10354138
7,3772925
asked Dec 7 '18 at 18:23
GengarGengar
1949
$endgroup$
Full problem, suppose that $R$ is a commutative Noetherian ring and $I$ is an ideal of $R$. We wish to prove that
$$bigcap_{n=1}^{infty} I^n=(0)$$ if and only if no zerodivisor of $R$ is of the form $1-z$ with $zin I$.
First I'll suppose that the intersection is $(0)$. Let $zin I$ and let $0neq rin R$ such that $r(1-z)=0$. Then $r=rz$ and so $rin I$. Is this useful here? I'm not sure how to use the Noetherian condition of $R$ since the chain of $I^n$ is descending, not ascending.
Any help would be much appreciated! I'm studying for a qual and need all the help I can get.
abstract-algebra ring-theory commutative-algebra ideals noetherian
abstract-algebra ring-theory commutative-algebra ideals noetherian
edited Dec 7 '18 at 18:28
user10354138
7,3772925
asked Dec 7 '18 at 18:23
GengarGengar
1949
edited Dec 7 '18 at 18:28
user10354138
7,3772925
asked Dec 7 '18 at 18:23
GengarGengar
1949
edited Dec 7 '18 at 18:28
user10354138
7,3772925
edited Dec 7 '18 at 18:28
user10354138
7,3772925
edited Dec 7 '18 at 18:28
user10354138
7,3772925
7,3772925
asked Dec 7 '18 at 18:23
GengarGengar
1949
asked Dec 7 '18 at 18:23
GengarGengar
1949
asked Dec 7 '18 at 18:23
GengarGengar
1949
1949
$begingroup$
Maybe take a look at Krulls intersection theorem?
$endgroup$
– red_trumpet
Dec 7 '18 at 18:38
$begingroup$
That assumes that R is local, though @red_trumpet
$endgroup$
– Gengar
Dec 7 '18 at 18:47
$begingroup$
Right, I missed that :(
$endgroup$
– red_trumpet
Dec 7 '18 at 19:37
add a comment |
$begingroup$
Maybe take a look at Krulls intersection theorem?
$endgroup$
– red_trumpet
Dec 7 '18 at 18:38
$begingroup$
That assumes that R is local, though @red_trumpet
$endgroup$
– Gengar
Dec 7 '18 at 18:47
$begingroup$
Right, I missed that :(
$endgroup$
– red_trumpet
Dec 7 '18 at 19:37
$begingroup$
Maybe take a look at Krulls intersection theorem?
$endgroup$
– red_trumpet
Dec 7 '18 at 18:38
$begingroup$
Maybe take a look at Krulls intersection theorem?
$endgroup$
– red_trumpet
Dec 7 '18 at 18:38
$begingroup$
That assumes that R is local, though @red_trumpet
$endgroup$
– Gengar
Dec 7 '18 at 18:47
$begingroup$
That assumes that R is local, though @red_trumpet
$endgroup$
– Gengar
Dec 7 '18 at 18:47
$begingroup$
Right, I missed that :(
$endgroup$
– red_trumpet
Dec 7 '18 at 19:37
$begingroup$
Right, I missed that :(
$endgroup$
– red_trumpet
Dec 7 '18 at 19:37
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes, the observation that $r=rz$ $implies$ $rin I$ is very useful, since we can use the equation again to get that $rin I^2$, and thus $rin I^3$, and so on. Hence $rin bigcap_{n=1}^infty I^n$, so $r=0$, contradiction. No need to use Noetherianness here. (It may be necessary for the converse, but I haven't thought that far, since it's not clear from your question if you're also asking about that.)
Edit
Worked out my thoughts on the converse. I was being dumb. It's Nakayama's lemma (the general, not local ring version).
Let $$I^infty = bigcap_{n=1}^infty I^n.$$
Observe that clearly $I(I^infty) = I^infty$. Then, since $R$ is Noetherian, $I^infty$ is finitely generated, so Nakayama's lemma (Statement 1) applies.
Thus there exists $rin R$ with $r-1in I$ such that $rI^infty =0$. But then $r-1=i$ for some $iin I$, and $r=1+i$. Then $r$ is not a zero divisor by assumption, hence the fact that $rI^infty = 0$ implies that $I^infty=0$ as desired.
answered Dec 7 '18 at 18:35
jgonjgon
13.6k22041
$endgroup$
$begingroup$
Do you have any ideas on the other direction? Thank you so much!
$endgroup$
– Gengar
Dec 7 '18 at 18:41
$begingroup$
@Gengar, for the converse red_trumpet's suggestion to use Krull's intersection theorem looks viable. I'll flesh out the details in my answer.
$endgroup$
– jgon
Dec 7 '18 at 18:47
$begingroup$
I appreciate it so much @jgon
$endgroup$
– Gengar
Dec 7 '18 at 18:47
$begingroup$
@Gengar actually I have to go, and I can't complete my thoughts rn, I'll post what I have.
$endgroup$
– jgon
Dec 7 '18 at 19:09
$begingroup$
@Gengar, got it, I was being silly.
$endgroup$
– jgon
Dec 8 '18 at 0:30
|
show 3 more comments
$begingroup$
If there exists some y $in$ I such that (1-y) is a zero divisor, then there exists an x $not=$ 0 such that x(1-y) = 0. Then x = xy $in$ I. Thus x $in$ I $cap$ I$^2$. It follows that x $in$ $displaystylebigcap_n$ I$^n$.
answered Dec 7 '18 at 19:43
Joel PereiraJoel Pereira
74519
$endgroup$
$begingroup$
Yeah, jgon already said that
$endgroup$
– Gengar
Dec 7 '18 at 19:53
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3030209%2fbigcap-n-in-mathbbn-in-0-if-and-only-if-no-zero-divisor-of-r-is-of%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, the observation that $r=rz$ $implies$ $rin I$ is very useful, since we can use the equation again to get that $rin I^2$, and thus $rin I^3$, and so on. Hence $rin bigcap_{n=1}^infty I^n$, so $r=0$, contradiction. No need to use Noetherianness here. (It may be necessary for the converse, but I haven't thought that far, since it's not clear from your question if you're also asking about that.)
Edit
Worked out my thoughts on the converse. I was being dumb. It's Nakayama's lemma (the general, not local ring version).
Let $$I^infty = bigcap_{n=1}^infty I^n.$$
Observe that clearly $I(I^infty) = I^infty$. Then, since $R$ is Noetherian, $I^infty$ is finitely generated, so Nakayama's lemma (Statement 1) applies.
Thus there exists $rin R$ with $r-1in I$ such that $rI^infty =0$. But then $r-1=i$ for some $iin I$, and $r=1+i$. Then $r$ is not a zero divisor by assumption, hence the fact that $rI^infty = 0$ implies that $I^infty=0$ as desired.
answered Dec 7 '18 at 18:35
jgonjgon
13.6k22041
$endgroup$
$begingroup$
Do you have any ideas on the other direction? Thank you so much!
$endgroup$
– Gengar
Dec 7 '18 at 18:41
$begingroup$
@Gengar, for the converse red_trumpet's suggestion to use Krull's intersection theorem looks viable. I'll flesh out the details in my answer.
$endgroup$
– jgon
Dec 7 '18 at 18:47
$begingroup$
I appreciate it so much @jgon
$endgroup$
– Gengar
Dec 7 '18 at 18:47
$begingroup$
@Gengar actually I have to go, and I can't complete my thoughts rn, I'll post what I have.
$endgroup$
– jgon
Dec 7 '18 at 19:09
$begingroup$
@Gengar, got it, I was being silly.
$endgroup$
– jgon
Dec 8 '18 at 0:30
|
show 3 more comments
$begingroup$
Yes, the observation that $r=rz$ $implies$ $rin I$ is very useful, since we can use the equation again to get that $rin I^2$, and thus $rin I^3$, and so on. Hence $rin bigcap_{n=1}^infty I^n$, so $r=0$, contradiction. No need to use Noetherianness here. (It may be necessary for the converse, but I haven't thought that far, since it's not clear from your question if you're also asking about that.)
Edit
Worked out my thoughts on the converse. I was being dumb. It's Nakayama's lemma (the general, not local ring version).
Let $$I^infty = bigcap_{n=1}^infty I^n.$$
Observe that clearly $I(I^infty) = I^infty$. Then, since $R$ is Noetherian, $I^infty$ is finitely generated, so Nakayama's lemma (Statement 1) applies.
Thus there exists $rin R$ with $r-1in I$ such that $rI^infty =0$. But then $r-1=i$ for some $iin I$, and $r=1+i$. Then $r$ is not a zero divisor by assumption, hence the fact that $rI^infty = 0$ implies that $I^infty=0$ as desired.
answered Dec 7 '18 at 18:35
jgonjgon
13.6k22041
$endgroup$
$begingroup$
Do you have any ideas on the other direction? Thank you so much!
$endgroup$
– Gengar
Dec 7 '18 at 18:41
$begingroup$
@Gengar, for the converse red_trumpet's suggestion to use Krull's intersection theorem looks viable. I'll flesh out the details in my answer.
$endgroup$
– jgon
Dec 7 '18 at 18:47
$begingroup$
I appreciate it so much @jgon
$endgroup$
– Gengar
Dec 7 '18 at 18:47
$begingroup$
@Gengar actually I have to go, and I can't complete my thoughts rn, I'll post what I have.
$endgroup$
– jgon
Dec 7 '18 at 19:09
$begingroup$
@Gengar, got it, I was being silly.
$endgroup$
– jgon
Dec 8 '18 at 0:30
|
show 3 more comments
$begingroup$
Yes, the observation that $r=rz$ $implies$ $rin I$ is very useful, since we can use the equation again to get that $rin I^2$, and thus $rin I^3$, and so on. Hence $rin bigcap_{n=1}^infty I^n$, so $r=0$, contradiction. No need to use Noetherianness here. (It may be necessary for the converse, but I haven't thought that far, since it's not clear from your question if you're also asking about that.)
Edit
Worked out my thoughts on the converse. I was being dumb. It's Nakayama's lemma (the general, not local ring version).
Let $$I^infty = bigcap_{n=1}^infty I^n.$$
Observe that clearly $I(I^infty) = I^infty$. Then, since $R$ is Noetherian, $I^infty$ is finitely generated, so Nakayama's lemma (Statement 1) applies.
Thus there exists $rin R$ with $r-1in I$ such that $rI^infty =0$. But then $r-1=i$ for some $iin I$, and $r=1+i$. Then $r$ is not a zero divisor by assumption, hence the fact that $rI^infty = 0$ implies that $I^infty=0$ as desired.
answered Dec 7 '18 at 18:35
jgonjgon
13.6k22041
$endgroup$
Yes, the observation that $r=rz$ $implies$ $rin I$ is very useful, since we can use the equation again to get that $rin I^2$, and thus $rin I^3$, and so on. Hence $rin bigcap_{n=1}^infty I^n$, so $r=0$, contradiction. No need to use Noetherianness here. (It may be necessary for the converse, but I haven't thought that far, since it's not clear from your question if you're also asking about that.)
Edit
Worked out my thoughts on the converse. I was being dumb. It's Nakayama's lemma (the general, not local ring version).
Let $$I^infty = bigcap_{n=1}^infty I^n.$$
Observe that clearly $I(I^infty) = I^infty$. Then, since $R$ is Noetherian, $I^infty$ is finitely generated, so Nakayama's lemma (Statement 1) applies.
Thus there exists $rin R$ with $r-1in I$ such that $rI^infty =0$. But then $r-1=i$ for some $iin I$, and $r=1+i$. Then $r$ is not a zero divisor by assumption, hence the fact that $rI^infty = 0$ implies that $I^infty=0$ as desired.
answered Dec 7 '18 at 18:35
jgonjgon
13.6k22041
edited Dec 8 '18 at 0:30
answered Dec 7 '18 at 18:35
jgonjgon
13.6k22041
answered Dec 7 '18 at 18:35
jgonjgon
13.6k22041
answered Dec 7 '18 at 18:35
jgonjgon
13.6k22041
13.6k22041
$begingroup$
Do you have any ideas on the other direction? Thank you so much!
$endgroup$
– Gengar
Dec 7 '18 at 18:41
$begingroup$
@Gengar, for the converse red_trumpet's suggestion to use Krull's intersection theorem looks viable. I'll flesh out the details in my answer.
$endgroup$
– jgon
Dec 7 '18 at 18:47
$begingroup$
I appreciate it so much @jgon
$endgroup$
– Gengar
Dec 7 '18 at 18:47
$begingroup$
@Gengar actually I have to go, and I can't complete my thoughts rn, I'll post what I have.
$endgroup$
– jgon
Dec 7 '18 at 19:09
$begingroup$
@Gengar, got it, I was being silly.
$endgroup$
– jgon
Dec 8 '18 at 0:30
|
show 3 more comments
$begingroup$
Do you have any ideas on the other direction? Thank you so much!
$endgroup$
– Gengar
Dec 7 '18 at 18:41
$begingroup$
@Gengar, for the converse red_trumpet's suggestion to use Krull's intersection theorem looks viable. I'll flesh out the details in my answer.
$endgroup$
– jgon
Dec 7 '18 at 18:47
$begingroup$
I appreciate it so much @jgon
$endgroup$
– Gengar
Dec 7 '18 at 18:47
$begingroup$
@Gengar actually I have to go, and I can't complete my thoughts rn, I'll post what I have.
$endgroup$
– jgon
Dec 7 '18 at 19:09
$begingroup$
@Gengar, got it, I was being silly.
$endgroup$
– jgon
Dec 8 '18 at 0:30
$begingroup$
Do you have any ideas on the other direction? Thank you so much!
$endgroup$
– Gengar
Dec 7 '18 at 18:41
$begingroup$
Do you have any ideas on the other direction? Thank you so much!
$endgroup$
– Gengar
Dec 7 '18 at 18:41
$begingroup$
@Gengar, for the converse red_trumpet's suggestion to use Krull's intersection theorem looks viable. I'll flesh out the details in my answer.
$endgroup$
– jgon
Dec 7 '18 at 18:47
$begingroup$
@Gengar, for the converse red_trumpet's suggestion to use Krull's intersection theorem looks viable. I'll flesh out the details in my answer.
$endgroup$
– jgon
Dec 7 '18 at 18:47
$begingroup$
I appreciate it so much @jgon
$endgroup$
– Gengar
Dec 7 '18 at 18:47
$begingroup$
I appreciate it so much @jgon
$endgroup$
– Gengar
Dec 7 '18 at 18:47
$begingroup$
@Gengar actually I have to go, and I can't complete my thoughts rn, I'll post what I have.
$endgroup$
– jgon
Dec 7 '18 at 19:09
$begingroup$
@Gengar actually I have to go, and I can't complete my thoughts rn, I'll post what I have.
$endgroup$
– jgon
Dec 7 '18 at 19:09
$begingroup$
@Gengar, got it, I was being silly.
$endgroup$
– jgon
Dec 8 '18 at 0:30
$begingroup$
@Gengar, got it, I was being silly.
$endgroup$
– jgon
Dec 8 '18 at 0:30
|
show 3 more comments
$begingroup$
If there exists some y $in$ I such that (1-y) is a zero divisor, then there exists an x $not=$ 0 such that x(1-y) = 0. Then x = xy $in$ I. Thus x $in$ I $cap$ I$^2$. It follows that x $in$ $displaystylebigcap_n$ I$^n$.
answered Dec 7 '18 at 19:43
Joel PereiraJoel Pereira
74519
$endgroup$
$begingroup$
Yeah, jgon already said that
$endgroup$
– Gengar
Dec 7 '18 at 19:53
add a comment |
$begingroup$
If there exists some y $in$ I such that (1-y) is a zero divisor, then there exists an x $not=$ 0 such that x(1-y) = 0. Then x = xy $in$ I. Thus x $in$ I $cap$ I$^2$. It follows that x $in$ $displaystylebigcap_n$ I$^n$.
answered Dec 7 '18 at 19:43
Joel PereiraJoel Pereira
74519
$endgroup$
$begingroup$
Yeah, jgon already said that
$endgroup$
– Gengar
Dec 7 '18 at 19:53
add a comment |
$begingroup$
If there exists some y $in$ I such that (1-y) is a zero divisor, then there exists an x $not=$ 0 such that x(1-y) = 0. Then x = xy $in$ I. Thus x $in$ I $cap$ I$^2$. It follows that x $in$ $displaystylebigcap_n$ I$^n$.
answered Dec 7 '18 at 19:43
Joel PereiraJoel Pereira
74519
$endgroup$
If there exists some y $in$ I such that (1-y) is a zero divisor, then there exists an x $not=$ 0 such that x(1-y) = 0. Then x = xy $in$ I. Thus x $in$ I $cap$ I$^2$. It follows that x $in$ $displaystylebigcap_n$ I$^n$.
answered Dec 7 '18 at 19:43
Joel PereiraJoel Pereira
74519
answered Dec 7 '18 at 19:43
Joel PereiraJoel Pereira
74519
answered Dec 7 '18 at 19:43
Joel PereiraJoel Pereira
74519
answered Dec 7 '18 at 19:43
Joel PereiraJoel Pereira
74519
74519
$begingroup$
Yeah, jgon already said that
$endgroup$
– Gengar
Dec 7 '18 at 19:53
add a comment |
$begingroup$
Yeah, jgon already said that
$endgroup$
– Gengar
Dec 7 '18 at 19:53
$begingroup$
Yeah, jgon already said that
$endgroup$
– Gengar
Dec 7 '18 at 19:53
$begingroup$
Yeah, jgon already said that
$endgroup$
– Gengar
Dec 7 '18 at 19:53
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3030209%2fbigcap-n-in-mathbbn-in-0-if-and-only-if-no-zero-divisor-of-r-is-of%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Maybe take a look at Krulls intersection theorem?
$endgroup$
– red_trumpet
Dec 7 '18 at 18:38
$begingroup$
That assumes that R is local, though @red_trumpet
$endgroup$
– Gengar
Dec 7 '18 at 18:47
$begingroup$
Right, I missed that :(
$endgroup$
– red_trumpet
Dec 7 '18 at 19:37