Preimage of maximal ideal is maximal (why is this answer wrong?)












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$begingroup$


Let $phi:Rto S$ Be a surjective ring homomorphism and $M subset S$ is maximal, then $phi^{-1}(M)$ is maximal in $R$.



Here is what I wrote (it is known the preimage is an ideal, so prove only maximality)




Assume $phi^{-1}(M) subset W subset R$. Then first $phi(phi^{-1}M) = S$ because $phi$ is surjective. Hence $phi(phi^{-1}M) subset phi(W) subset phi(R) implies S subset phi(W) subset S implies phi(W) = S$.



Now let $w in W$ and $phi(w) in phi(W) = S = phi(phi^{-1}M)$, so $w in phi^{-1}M$? Or this only shows $x in phi^{-1}(M)$ such that $phi(x) = phi(w)$?




Is there a way to complete the proof with what I wrote?



(I already have witness the solution, basically $phi^{-1}M subset R$ is an ideal. Then by isomorphism theorems $S/M approx R/ker(pi circ phi)$ where $pi$ is the quotient map and usually field argument implies $ker (pi circ phi) = phi^{-1}M$ because fields contain only two ideals)






abstract-algebra proof-verification







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    $phi(phi^{-1}(M)) subset M$ and $M$ must be a strict subset of $S$ to be maximal. Your claim that $phi(phi^{-1}M) = S$ is incorrect.
    $endgroup$
    – Aniruddh Agarwal
    Dec 22 '18 at 9:20












  • $begingroup$
    @AniruddhAgarwal $phi$ is surjective, so we get equality
    $endgroup$
    – Hawk
    Dec 22 '18 at 9:24










  • $begingroup$
    Yes, but in general for any function $f : X rightarrow Y$ and any $S subset Y$, we have that $f(f^{-1}(S)) subset S$. In your case, what is true is that $phi(phi^{-1}(M)) = M$.
    $endgroup$
    – Aniruddh Agarwal
    Dec 22 '18 at 9:26












  • $begingroup$
    Ah okay I see. Never mind then.
    $endgroup$
    – Hawk
    Dec 22 '18 at 9:31
















2












$begingroup$


Let $phi:Rto S$ Be a surjective ring homomorphism and $M subset S$ is maximal, then $phi^{-1}(M)$ is maximal in $R$.



Here is what I wrote (it is known the preimage is an ideal, so prove only maximality)




Assume $phi^{-1}(M) subset W subset R$. Then first $phi(phi^{-1}M) = S$ because $phi$ is surjective. Hence $phi(phi^{-1}M) subset phi(W) subset phi(R) implies S subset phi(W) subset S implies phi(W) = S$.



Now let $w in W$ and $phi(w) in phi(W) = S = phi(phi^{-1}M)$, so $w in phi^{-1}M$? Or this only shows $x in phi^{-1}(M)$ such that $phi(x) = phi(w)$?




Is there a way to complete the proof with what I wrote?



(I already have witness the solution, basically $phi^{-1}M subset R$ is an ideal. Then by isomorphism theorems $S/M approx R/ker(pi circ phi)$ where $pi$ is the quotient map and usually field argument implies $ker (pi circ phi) = phi^{-1}M$ because fields contain only two ideals)






abstract-algebra proof-verification







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    $phi(phi^{-1}(M)) subset M$ and $M$ must be a strict subset of $S$ to be maximal. Your claim that $phi(phi^{-1}M) = S$ is incorrect.
    $endgroup$
    – Aniruddh Agarwal
    Dec 22 '18 at 9:20












  • $begingroup$
    @AniruddhAgarwal $phi$ is surjective, so we get equality
    $endgroup$
    – Hawk
    Dec 22 '18 at 9:24










  • $begingroup$
    Yes, but in general for any function $f : X rightarrow Y$ and any $S subset Y$, we have that $f(f^{-1}(S)) subset S$. In your case, what is true is that $phi(phi^{-1}(M)) = M$.
    $endgroup$
    – Aniruddh Agarwal
    Dec 22 '18 at 9:26












  • $begingroup$
    Ah okay I see. Never mind then.
    $endgroup$
    – Hawk
    Dec 22 '18 at 9:31














2












2








2





$begingroup$


Let $phi:Rto S$ Be a surjective ring homomorphism and $M subset S$ is maximal, then $phi^{-1}(M)$ is maximal in $R$.



Here is what I wrote (it is known the preimage is an ideal, so prove only maximality)




Assume $phi^{-1}(M) subset W subset R$. Then first $phi(phi^{-1}M) = S$ because $phi$ is surjective. Hence $phi(phi^{-1}M) subset phi(W) subset phi(R) implies S subset phi(W) subset S implies phi(W) = S$.



Now let $w in W$ and $phi(w) in phi(W) = S = phi(phi^{-1}M)$, so $w in phi^{-1}M$? Or this only shows $x in phi^{-1}(M)$ such that $phi(x) = phi(w)$?




Is there a way to complete the proof with what I wrote?



(I already have witness the solution, basically $phi^{-1}M subset R$ is an ideal. Then by isomorphism theorems $S/M approx R/ker(pi circ phi)$ where $pi$ is the quotient map and usually field argument implies $ker (pi circ phi) = phi^{-1}M$ because fields contain only two ideals)






abstract-algebra proof-verification







share|cite|improve this question









$endgroup$




Let $phi:Rto S$ Be a surjective ring homomorphism and $M subset S$ is maximal, then $phi^{-1}(M)$ is maximal in $R$.



Here is what I wrote (it is known the preimage is an ideal, so prove only maximality)




Assume $phi^{-1}(M) subset W subset R$. Then first $phi(phi^{-1}M) = S$ because $phi$ is surjective. Hence $phi(phi^{-1}M) subset phi(W) subset phi(R) implies S subset phi(W) subset S implies phi(W) = S$.



Now let $w in W$ and $phi(w) in phi(W) = S = phi(phi^{-1}M)$, so $w in phi^{-1}M$? Or this only shows $x in phi^{-1}(M)$ such that $phi(x) = phi(w)$?




Is there a way to complete the proof with what I wrote?



(I already have witness the solution, basically $phi^{-1}M subset R$ is an ideal. Then by isomorphism theorems $S/M approx R/ker(pi circ phi)$ where $pi$ is the quotient map and usually field argument implies $ker (pi circ phi) = phi^{-1}M$ because fields contain only two ideals)






abstract-algebra proof-verification




abstract-algebra proof-verification






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asked Dec 22 '18 at 9:03









HawkHawk

5,5251139105




5,5251139105








  • 1




    $begingroup$
    $phi(phi^{-1}(M)) subset M$ and $M$ must be a strict subset of $S$ to be maximal. Your claim that $phi(phi^{-1}M) = S$ is incorrect.
    $endgroup$
    – Aniruddh Agarwal
    Dec 22 '18 at 9:20












  • $begingroup$
    @AniruddhAgarwal $phi$ is surjective, so we get equality
    $endgroup$
    – Hawk
    Dec 22 '18 at 9:24










  • $begingroup$
    Yes, but in general for any function $f : X rightarrow Y$ and any $S subset Y$, we have that $f(f^{-1}(S)) subset S$. In your case, what is true is that $phi(phi^{-1}(M)) = M$.
    $endgroup$
    – Aniruddh Agarwal
    Dec 22 '18 at 9:26












  • $begingroup$
    Ah okay I see. Never mind then.
    $endgroup$
    – Hawk
    Dec 22 '18 at 9:31














  • 1




    $begingroup$
    $phi(phi^{-1}(M)) subset M$ and $M$ must be a strict subset of $S$ to be maximal. Your claim that $phi(phi^{-1}M) = S$ is incorrect.
    $endgroup$
    – Aniruddh Agarwal
    Dec 22 '18 at 9:20












  • $begingroup$
    @AniruddhAgarwal $phi$ is surjective, so we get equality
    $endgroup$
    – Hawk
    Dec 22 '18 at 9:24










  • $begingroup$
    Yes, but in general for any function $f : X rightarrow Y$ and any $S subset Y$, we have that $f(f^{-1}(S)) subset S$. In your case, what is true is that $phi(phi^{-1}(M)) = M$.
    $endgroup$
    – Aniruddh Agarwal
    Dec 22 '18 at 9:26












  • $begingroup$
    Ah okay I see. Never mind then.
    $endgroup$
    – Hawk
    Dec 22 '18 at 9:31








1




1




$begingroup$
$phi(phi^{-1}(M)) subset M$ and $M$ must be a strict subset of $S$ to be maximal. Your claim that $phi(phi^{-1}M) = S$ is incorrect.
$endgroup$
– Aniruddh Agarwal
Dec 22 '18 at 9:20






$begingroup$
$phi(phi^{-1}(M)) subset M$ and $M$ must be a strict subset of $S$ to be maximal. Your claim that $phi(phi^{-1}M) = S$ is incorrect.
$endgroup$
– Aniruddh Agarwal
Dec 22 '18 at 9:20














$begingroup$
@AniruddhAgarwal $phi$ is surjective, so we get equality
$endgroup$
– Hawk
Dec 22 '18 at 9:24




$begingroup$
@AniruddhAgarwal $phi$ is surjective, so we get equality
$endgroup$
– Hawk
Dec 22 '18 at 9:24












$begingroup$
Yes, but in general for any function $f : X rightarrow Y$ and any $S subset Y$, we have that $f(f^{-1}(S)) subset S$. In your case, what is true is that $phi(phi^{-1}(M)) = M$.
$endgroup$
– Aniruddh Agarwal
Dec 22 '18 at 9:26






$begingroup$
Yes, but in general for any function $f : X rightarrow Y$ and any $S subset Y$, we have that $f(f^{-1}(S)) subset S$. In your case, what is true is that $phi(phi^{-1}(M)) = M$.
$endgroup$
– Aniruddh Agarwal
Dec 22 '18 at 9:26














$begingroup$
Ah okay I see. Never mind then.
$endgroup$
– Hawk
Dec 22 '18 at 9:31




$begingroup$
Ah okay I see. Never mind then.
$endgroup$
– Hawk
Dec 22 '18 at 9:31










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$begingroup$

The issue with your proof was discussed in the comments. We have $phi(phi^{-1}(M))=Mneq S$.



For the proof you mention, let $f:Sto S/M$ be the canonical surjective map onto the field $S/M$. Then $fcircphi$ has kernel exactly $phi^{-1}(M)$, hence $S/Mcong R/phi^{-1}(M)$. Since this is a field, it follows that $phi^{-1}(M)$ is a maximal ideal.






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    1 Answer
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    1 Answer
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    $begingroup$

    The issue with your proof was discussed in the comments. We have $phi(phi^{-1}(M))=Mneq S$.



    For the proof you mention, let $f:Sto S/M$ be the canonical surjective map onto the field $S/M$. Then $fcircphi$ has kernel exactly $phi^{-1}(M)$, hence $S/Mcong R/phi^{-1}(M)$. Since this is a field, it follows that $phi^{-1}(M)$ is a maximal ideal.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The issue with your proof was discussed in the comments. We have $phi(phi^{-1}(M))=Mneq S$.



      For the proof you mention, let $f:Sto S/M$ be the canonical surjective map onto the field $S/M$. Then $fcircphi$ has kernel exactly $phi^{-1}(M)$, hence $S/Mcong R/phi^{-1}(M)$. Since this is a field, it follows that $phi^{-1}(M)$ is a maximal ideal.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The issue with your proof was discussed in the comments. We have $phi(phi^{-1}(M))=Mneq S$.



        For the proof you mention, let $f:Sto S/M$ be the canonical surjective map onto the field $S/M$. Then $fcircphi$ has kernel exactly $phi^{-1}(M)$, hence $S/Mcong R/phi^{-1}(M)$. Since this is a field, it follows that $phi^{-1}(M)$ is a maximal ideal.






        share|cite|improve this answer









        $endgroup$



        The issue with your proof was discussed in the comments. We have $phi(phi^{-1}(M))=Mneq S$.



        For the proof you mention, let $f:Sto S/M$ be the canonical surjective map onto the field $S/M$. Then $fcircphi$ has kernel exactly $phi^{-1}(M)$, hence $S/Mcong R/phi^{-1}(M)$. Since this is a field, it follows that $phi^{-1}(M)$ is a maximal ideal.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 22 '18 at 11:41









        Matt SamuelMatt Samuel

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