Why shouldn't this prove the Prime Number Theorem?












3












$begingroup$


Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_{n=1}^{infty} frac{mu(n)}{n^k}$ can be interpreted as the probability that a randomly chosen integer is $k$-free.



Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form



$$sum_{n=1}^{infty} frac{mu(n)}{n}=0,$$



since the probability that an integer is ``$1$-free'' is zero ?






nt.number-theory prime-numbers







share|cite|improve this question







New contributor




Fourton. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 11




    $begingroup$
    It is true that the PNT is equivalent to $sum_{n leq x} frac{mu(n)}{n} = o(1)$. It is also relatively easy to prove that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = 0$. The hard part is proving that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = lim_{x to infty} sum_{n leq x} frac{mu(n)}{n}$. This is highly nontrivial!
    $endgroup$
    – Peter Humphries
    5 hours ago






  • 11




    $begingroup$
    In general, limit of sums of series $neq$ sum of limits of series. In this particular case, the equality does hold, but it requires intricate arguments to prove, which you see in any proof of PNT.
    $endgroup$
    – Wojowu
    5 hours ago






  • 6




    $begingroup$
    I think this question should be reopened, and the comments made by Peter Humphries and Wojowu posted as an answer. The question might be borderline too elementary for MO but it is natural and I'm sure I'm not the only one to have been confused by this at some (embarrassingly recent) point, it's a bit silly to close when, in effect, the answer is there.
    $endgroup$
    – Gro-Tsen
    4 hours ago






  • 1




    $begingroup$
    I agree with Fourton and have voted accordingly
    $endgroup$
    – Yemon Choi
    4 hours ago
















3












$begingroup$


Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_{n=1}^{infty} frac{mu(n)}{n^k}$ can be interpreted as the probability that a randomly chosen integer is $k$-free.



Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form



$$sum_{n=1}^{infty} frac{mu(n)}{n}=0,$$



since the probability that an integer is ``$1$-free'' is zero ?






nt.number-theory prime-numbers







share|cite|improve this question







New contributor




Fourton. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 11




    $begingroup$
    It is true that the PNT is equivalent to $sum_{n leq x} frac{mu(n)}{n} = o(1)$. It is also relatively easy to prove that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = 0$. The hard part is proving that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = lim_{x to infty} sum_{n leq x} frac{mu(n)}{n}$. This is highly nontrivial!
    $endgroup$
    – Peter Humphries
    5 hours ago






  • 11




    $begingroup$
    In general, limit of sums of series $neq$ sum of limits of series. In this particular case, the equality does hold, but it requires intricate arguments to prove, which you see in any proof of PNT.
    $endgroup$
    – Wojowu
    5 hours ago






  • 6




    $begingroup$
    I think this question should be reopened, and the comments made by Peter Humphries and Wojowu posted as an answer. The question might be borderline too elementary for MO but it is natural and I'm sure I'm not the only one to have been confused by this at some (embarrassingly recent) point, it's a bit silly to close when, in effect, the answer is there.
    $endgroup$
    – Gro-Tsen
    4 hours ago






  • 1




    $begingroup$
    I agree with Fourton and have voted accordingly
    $endgroup$
    – Yemon Choi
    4 hours ago














3












3








3





$begingroup$


Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_{n=1}^{infty} frac{mu(n)}{n^k}$ can be interpreted as the probability that a randomly chosen integer is $k$-free.



Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form



$$sum_{n=1}^{infty} frac{mu(n)}{n}=0,$$



since the probability that an integer is ``$1$-free'' is zero ?






nt.number-theory prime-numbers







share|cite|improve this question







New contributor




Fourton. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_{n=1}^{infty} frac{mu(n)}{n^k}$ can be interpreted as the probability that a randomly chosen integer is $k$-free.



Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form



$$sum_{n=1}^{infty} frac{mu(n)}{n}=0,$$



since the probability that an integer is ``$1$-free'' is zero ?






nt.number-theory prime-numbers




nt.number-theory prime-numbers






share|cite|improve this question







New contributor




Fourton. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




Fourton. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




Fourton. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 5 hours ago









Fourton.Fourton.

422




422




New contributor




Fourton. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Fourton. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Fourton. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 11




    $begingroup$
    It is true that the PNT is equivalent to $sum_{n leq x} frac{mu(n)}{n} = o(1)$. It is also relatively easy to prove that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = 0$. The hard part is proving that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = lim_{x to infty} sum_{n leq x} frac{mu(n)}{n}$. This is highly nontrivial!
    $endgroup$
    – Peter Humphries
    5 hours ago






  • 11




    $begingroup$
    In general, limit of sums of series $neq$ sum of limits of series. In this particular case, the equality does hold, but it requires intricate arguments to prove, which you see in any proof of PNT.
    $endgroup$
    – Wojowu
    5 hours ago






  • 6




    $begingroup$
    I think this question should be reopened, and the comments made by Peter Humphries and Wojowu posted as an answer. The question might be borderline too elementary for MO but it is natural and I'm sure I'm not the only one to have been confused by this at some (embarrassingly recent) point, it's a bit silly to close when, in effect, the answer is there.
    $endgroup$
    – Gro-Tsen
    4 hours ago






  • 1




    $begingroup$
    I agree with Fourton and have voted accordingly
    $endgroup$
    – Yemon Choi
    4 hours ago














  • 11




    $begingroup$
    It is true that the PNT is equivalent to $sum_{n leq x} frac{mu(n)}{n} = o(1)$. It is also relatively easy to prove that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = 0$. The hard part is proving that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = lim_{x to infty} sum_{n leq x} frac{mu(n)}{n}$. This is highly nontrivial!
    $endgroup$
    – Peter Humphries
    5 hours ago






  • 11




    $begingroup$
    In general, limit of sums of series $neq$ sum of limits of series. In this particular case, the equality does hold, but it requires intricate arguments to prove, which you see in any proof of PNT.
    $endgroup$
    – Wojowu
    5 hours ago






  • 6




    $begingroup$
    I think this question should be reopened, and the comments made by Peter Humphries and Wojowu posted as an answer. The question might be borderline too elementary for MO but it is natural and I'm sure I'm not the only one to have been confused by this at some (embarrassingly recent) point, it's a bit silly to close when, in effect, the answer is there.
    $endgroup$
    – Gro-Tsen
    4 hours ago






  • 1




    $begingroup$
    I agree with Fourton and have voted accordingly
    $endgroup$
    – Yemon Choi
    4 hours ago








11




11




$begingroup$
It is true that the PNT is equivalent to $sum_{n leq x} frac{mu(n)}{n} = o(1)$. It is also relatively easy to prove that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = 0$. The hard part is proving that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = lim_{x to infty} sum_{n leq x} frac{mu(n)}{n}$. This is highly nontrivial!
$endgroup$
– Peter Humphries
5 hours ago




$begingroup$
It is true that the PNT is equivalent to $sum_{n leq x} frac{mu(n)}{n} = o(1)$. It is also relatively easy to prove that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = 0$. The hard part is proving that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = lim_{x to infty} sum_{n leq x} frac{mu(n)}{n}$. This is highly nontrivial!
$endgroup$
– Peter Humphries
5 hours ago




11




11




$begingroup$
In general, limit of sums of series $neq$ sum of limits of series. In this particular case, the equality does hold, but it requires intricate arguments to prove, which you see in any proof of PNT.
$endgroup$
– Wojowu
5 hours ago




$begingroup$
In general, limit of sums of series $neq$ sum of limits of series. In this particular case, the equality does hold, but it requires intricate arguments to prove, which you see in any proof of PNT.
$endgroup$
– Wojowu
5 hours ago




6




6




$begingroup$
I think this question should be reopened, and the comments made by Peter Humphries and Wojowu posted as an answer. The question might be borderline too elementary for MO but it is natural and I'm sure I'm not the only one to have been confused by this at some (embarrassingly recent) point, it's a bit silly to close when, in effect, the answer is there.
$endgroup$
– Gro-Tsen
4 hours ago




$begingroup$
I think this question should be reopened, and the comments made by Peter Humphries and Wojowu posted as an answer. The question might be borderline too elementary for MO but it is natural and I'm sure I'm not the only one to have been confused by this at some (embarrassingly recent) point, it's a bit silly to close when, in effect, the answer is there.
$endgroup$
– Gro-Tsen
4 hours ago




1




1




$begingroup$
I agree with Fourton and have voted accordingly
$endgroup$
– Yemon Choi
4 hours ago




$begingroup$
I agree with Fourton and have voted accordingly
$endgroup$
– Yemon Choi
4 hours ago










1 Answer
1






active

oldest

votes


















2












$begingroup$

You ask:




Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_{n=1}^{infty} frac{mu(n)}{n^k}$ can be interpreted as the probability that a randomly chosen integer is $k$-free.



Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form



$$sum_{n=1}^{infty} frac{mu(n)}{n}=0,$$



since the probability that an integer is ``$1$-free'' is zero ?




As pointed out by the users @wojowu and @PeterHumphries,
it is true that the PNT is equivalent to



$$sum_{n=1}^{infty} frac{mu(n)}{n}=o(1),$$
and it is relatively easy to prove that



$$lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s}=0.$$
The real difficulty lies in proving that



$$lim_{xrightarrow infty} sum_{nleq x} frac{mu(n)}{n}=
lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s},$$
which is highly nontrivial and requires intricate arguments.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Analyst's life story: you have two limiting operations (limit, infinite sum, integral, derivative, etc), and if only you could interchange them, you'd have your result; but in order to justify doing so, you need some hard estimates...
    $endgroup$
    – Nate Eldredge
    9 mins ago














Your Answer








StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "504"
};
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
});


}
});






Fourton. is a new contributor. Be nice, and check out our Code of Conduct.










draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f328552%2fwhy-shouldnt-this-prove-the-prime-number-theorem%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

You ask:




Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_{n=1}^{infty} frac{mu(n)}{n^k}$ can be interpreted as the probability that a randomly chosen integer is $k$-free.



Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form



$$sum_{n=1}^{infty} frac{mu(n)}{n}=0,$$



since the probability that an integer is ``$1$-free'' is zero ?




As pointed out by the users @wojowu and @PeterHumphries,
it is true that the PNT is equivalent to



$$sum_{n=1}^{infty} frac{mu(n)}{n}=o(1),$$
and it is relatively easy to prove that



$$lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s}=0.$$
The real difficulty lies in proving that



$$lim_{xrightarrow infty} sum_{nleq x} frac{mu(n)}{n}=
lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s},$$
which is highly nontrivial and requires intricate arguments.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Analyst's life story: you have two limiting operations (limit, infinite sum, integral, derivative, etc), and if only you could interchange them, you'd have your result; but in order to justify doing so, you need some hard estimates...
    $endgroup$
    – Nate Eldredge
    9 mins ago


















2












$begingroup$

You ask:




Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_{n=1}^{infty} frac{mu(n)}{n^k}$ can be interpreted as the probability that a randomly chosen integer is $k$-free.



Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form



$$sum_{n=1}^{infty} frac{mu(n)}{n}=0,$$



since the probability that an integer is ``$1$-free'' is zero ?




As pointed out by the users @wojowu and @PeterHumphries,
it is true that the PNT is equivalent to



$$sum_{n=1}^{infty} frac{mu(n)}{n}=o(1),$$
and it is relatively easy to prove that



$$lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s}=0.$$
The real difficulty lies in proving that



$$lim_{xrightarrow infty} sum_{nleq x} frac{mu(n)}{n}=
lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s},$$
which is highly nontrivial and requires intricate arguments.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Analyst's life story: you have two limiting operations (limit, infinite sum, integral, derivative, etc), and if only you could interchange them, you'd have your result; but in order to justify doing so, you need some hard estimates...
    $endgroup$
    – Nate Eldredge
    9 mins ago
















2












2








2





$begingroup$

You ask:




Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_{n=1}^{infty} frac{mu(n)}{n^k}$ can be interpreted as the probability that a randomly chosen integer is $k$-free.



Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form



$$sum_{n=1}^{infty} frac{mu(n)}{n}=0,$$



since the probability that an integer is ``$1$-free'' is zero ?




As pointed out by the users @wojowu and @PeterHumphries,
it is true that the PNT is equivalent to



$$sum_{n=1}^{infty} frac{mu(n)}{n}=o(1),$$
and it is relatively easy to prove that



$$lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s}=0.$$
The real difficulty lies in proving that



$$lim_{xrightarrow infty} sum_{nleq x} frac{mu(n)}{n}=
lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s},$$
which is highly nontrivial and requires intricate arguments.






share|cite|improve this answer











$endgroup$



You ask:




Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_{n=1}^{infty} frac{mu(n)}{n^k}$ can be interpreted as the probability that a randomly chosen integer is $k$-free.



Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form



$$sum_{n=1}^{infty} frac{mu(n)}{n}=0,$$



since the probability that an integer is ``$1$-free'' is zero ?




As pointed out by the users @wojowu and @PeterHumphries,
it is true that the PNT is equivalent to



$$sum_{n=1}^{infty} frac{mu(n)}{n}=o(1),$$
and it is relatively easy to prove that



$$lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s}=0.$$
The real difficulty lies in proving that



$$lim_{xrightarrow infty} sum_{nleq x} frac{mu(n)}{n}=
lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s},$$
which is highly nontrivial and requires intricate arguments.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








answered 25 mins ago


























community wiki





kodlu









  • 1




    $begingroup$
    Analyst's life story: you have two limiting operations (limit, infinite sum, integral, derivative, etc), and if only you could interchange them, you'd have your result; but in order to justify doing so, you need some hard estimates...
    $endgroup$
    – Nate Eldredge
    9 mins ago
















  • 1




    $begingroup$
    Analyst's life story: you have two limiting operations (limit, infinite sum, integral, derivative, etc), and if only you could interchange them, you'd have your result; but in order to justify doing so, you need some hard estimates...
    $endgroup$
    – Nate Eldredge
    9 mins ago










1




1




$begingroup$
Analyst's life story: you have two limiting operations (limit, infinite sum, integral, derivative, etc), and if only you could interchange them, you'd have your result; but in order to justify doing so, you need some hard estimates...
$endgroup$
– Nate Eldredge
9 mins ago






$begingroup$
Analyst's life story: you have two limiting operations (limit, infinite sum, integral, derivative, etc), and if only you could interchange them, you'd have your result; but in order to justify doing so, you need some hard estimates...
$endgroup$
– Nate Eldredge
9 mins ago












Fourton. is a new contributor. Be nice, and check out our Code of Conduct.










draft saved

draft discarded


















Fourton. is a new contributor. Be nice, and check out our Code of Conduct.













Fourton. is a new contributor. Be nice, and check out our Code of Conduct.












Fourton. is a new contributor. Be nice, and check out our Code of Conduct.
















Thanks for contributing an answer to MathOverflow!


  • 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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f328552%2fwhy-shouldnt-this-prove-the-prime-number-theorem%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Ponta tanko

Image
Usona ponta tanko M60A1 Ponta tanko estas, simile kiel kirasita savveturilo, inter specialaj kirasitaj transportiloj. Kiel pontaj tankoj eblas konsideri ĉiujn memmovigaj rimedoj ebligantaj superi obstaklojn ĝis larĝeco 24 metroj rekte en spaco de bataloperacoj. Ties ĉefa tasko estas ebligi al la propraj taĉmentoj daŭrigi en rapida antaŭenirado sen senbezona reteniĝo. El tio pli-malpli rezultas taktikaj-teknikaj postuloj, kiujn ili devas plenumi. Precipe temas pri kapablo operacii komune kun unue-sinsekvaj taĉmentoj kaj transpontigi obstaklojn dum kelke da minutoj. This page is only for reference, If you need detailed information, please check here
Read more

Tantalo (mitologio)

Image
Ĉi tiu artikolo temas pri mitologio. Por aliaj signifoj vidu la paĝon Tantalo (apartigilo) . Tantalo , laŭ Gioacchino Assereto. Laŭ la helena mitologio, Tantalo (el la greka Τάνταλος , ŝancelanta aŭ tre malfeliĉa ) estis filo de Zeŭso kaj la oceanidino Pluto, laŭ la plej divastigata versio. Li estis reĝo de Frigio [1] aŭ de la monto Sipilo, en Lidio (Malgrandazio) [2] kaj post sia morto li suferis eternan torturadon ĉe la Tartaro, la plej profunda parto de Hadeso, rezervita por la plej perversaj homoj. Tamen, kelkaj rakontoj asertas ke li estis reĝo de Argo aŭ de Korinto [3] , aŭ ke li migris norden el la monto Sipilo por akiri la tronon de Paflagonio, el kie Ilo forpelis lin pro tio, ke Tantalo forkaptis kaj seksperfortis lian fraton Ganimedon [4] , kaj tiel ofendis la diojn. Verŝajne li ankaŭ regis urbon en okcidenta Anatolio, kiun oni nomis Tantalis [5] pro li, aŭ Sipilo , pro la monto sur kies flanko ankoraŭ staras ĝiaj ruinaĵoj [6] . Laŭ la geografiisto Strabo,...
Read more

Erzsébet Schaár

Image
Erzsébet Schaár Memportreto de Erzsébet Schaár en Pécs Personaj informoj Naskiĝo 27-an de julio 1908 ( 1908-07-27 ) en Budafok Morto 29-an de aŭgusto 1975 ( 1975-08-29 ) (67-jara) en Budapeŝto Ŝtataneco Hungario Profesio skulptisto Familianoj Edz(in)o Tibor Vilt v   •   d   •   r Erzsébet Schaár [erĵEbet ŝAr], laŭ hungarlingve kutima sinsekvo Schaár Erzsébet estis hungara skulptisto. Ŝia edzo estis Tibor Vilt. Déryné Széppataki Róza en Miskolc fare de Erzsébet Schaár Erzsébet Schaár [1] naskiĝis la 27-an de julio 1908 en Budafok, ŝi mortis la 29-an de aŭgusto 1975 en Budapeŝto. Enhavo 1 Biografio 2 Elektitaj solaj ekspozicioj 3 Elektitaj kolektivaj ekspozicioj 4 Elektitaj skulptaĵoj 5 Elektitaj disĉiploj 6 Fontoj 7 Referencoj Biografio | Erzsébet Schaár studis en Industriarta Lernejo, poste en Reĝa Belarta Altlernejo (1924-1926). Ŝia ĉefinstruisto estis Zsig...
Read more