$R$ commutative ring with with unity. Then: “$(x)$ prime $iff R$ domain” and “$(x)$ maximal $iff R$ is...












0












$begingroup$


This is question 7.4.6 from Dummit and Foote. Just need someone to verify my answer.



$$R[x]/(x) stackrel{pi}leftrightsquigarrow R$$



$$a_0 + (a_1 + dots a_nx^{n-1})x = a_0 + (x) stackrel{pi}leftrightsquigarrow a_0$$



This is clearly an isomorphism with kernel $ker pi = (x) = bar{0} in R[x]/(x)$. This will answer both question since if $(x)$ is maximal, it is prime. Then $R[x]/(x)$ is a field and hence $R$. Similarly $R[x]/(x)$ would be an integral domain, and so is $R$.






abstract-algebra proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    That's correct.
    $endgroup$
    – Wojowu
    Dec 21 '18 at 9:56










  • $begingroup$
    The ring isomorphism is fine.
    $endgroup$
    – Wuestenfux
    Dec 21 '18 at 9:58










  • $begingroup$
    thank you to the both of you.
    $endgroup$
    – Hawk
    Dec 21 '18 at 10:05










  • $begingroup$
    @Wuestenfux actually wait my $pi$ is not well defined.
    $endgroup$
    – Hawk
    Dec 21 '18 at 11:12
















0












$begingroup$


This is question 7.4.6 from Dummit and Foote. Just need someone to verify my answer.



$$R[x]/(x) stackrel{pi}leftrightsquigarrow R$$



$$a_0 + (a_1 + dots a_nx^{n-1})x = a_0 + (x) stackrel{pi}leftrightsquigarrow a_0$$



This is clearly an isomorphism with kernel $ker pi = (x) = bar{0} in R[x]/(x)$. This will answer both question since if $(x)$ is maximal, it is prime. Then $R[x]/(x)$ is a field and hence $R$. Similarly $R[x]/(x)$ would be an integral domain, and so is $R$.






abstract-algebra proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    That's correct.
    $endgroup$
    – Wojowu
    Dec 21 '18 at 9:56










  • $begingroup$
    The ring isomorphism is fine.
    $endgroup$
    – Wuestenfux
    Dec 21 '18 at 9:58










  • $begingroup$
    thank you to the both of you.
    $endgroup$
    – Hawk
    Dec 21 '18 at 10:05










  • $begingroup$
    @Wuestenfux actually wait my $pi$ is not well defined.
    $endgroup$
    – Hawk
    Dec 21 '18 at 11:12














0












0








0





$begingroup$


This is question 7.4.6 from Dummit and Foote. Just need someone to verify my answer.



$$R[x]/(x) stackrel{pi}leftrightsquigarrow R$$



$$a_0 + (a_1 + dots a_nx^{n-1})x = a_0 + (x) stackrel{pi}leftrightsquigarrow a_0$$



This is clearly an isomorphism with kernel $ker pi = (x) = bar{0} in R[x]/(x)$. This will answer both question since if $(x)$ is maximal, it is prime. Then $R[x]/(x)$ is a field and hence $R$. Similarly $R[x]/(x)$ would be an integral domain, and so is $R$.






abstract-algebra proof-verification







share|cite|improve this question











$endgroup$




This is question 7.4.6 from Dummit and Foote. Just need someone to verify my answer.



$$R[x]/(x) stackrel{pi}leftrightsquigarrow R$$



$$a_0 + (a_1 + dots a_nx^{n-1})x = a_0 + (x) stackrel{pi}leftrightsquigarrow a_0$$



This is clearly an isomorphism with kernel $ker pi = (x) = bar{0} in R[x]/(x)$. This will answer both question since if $(x)$ is maximal, it is prime. Then $R[x]/(x)$ is a field and hence $R$. Similarly $R[x]/(x)$ would be an integral domain, and so is $R$.






abstract-algebra proof-verification




abstract-algebra proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 11:35









quid♦

37.2k95193




37.2k95193










asked Dec 21 '18 at 9:54









HawkHawk

5,5801140110




5,5801140110












  • $begingroup$
    That's correct.
    $endgroup$
    – Wojowu
    Dec 21 '18 at 9:56










  • $begingroup$
    The ring isomorphism is fine.
    $endgroup$
    – Wuestenfux
    Dec 21 '18 at 9:58










  • $begingroup$
    thank you to the both of you.
    $endgroup$
    – Hawk
    Dec 21 '18 at 10:05










  • $begingroup$
    @Wuestenfux actually wait my $pi$ is not well defined.
    $endgroup$
    – Hawk
    Dec 21 '18 at 11:12


















  • $begingroup$
    That's correct.
    $endgroup$
    – Wojowu
    Dec 21 '18 at 9:56










  • $begingroup$
    The ring isomorphism is fine.
    $endgroup$
    – Wuestenfux
    Dec 21 '18 at 9:58










  • $begingroup$
    thank you to the both of you.
    $endgroup$
    – Hawk
    Dec 21 '18 at 10:05










  • $begingroup$
    @Wuestenfux actually wait my $pi$ is not well defined.
    $endgroup$
    – Hawk
    Dec 21 '18 at 11:12
















$begingroup$
That's correct.
$endgroup$
– Wojowu
Dec 21 '18 at 9:56




$begingroup$
That's correct.
$endgroup$
– Wojowu
Dec 21 '18 at 9:56












$begingroup$
The ring isomorphism is fine.
$endgroup$
– Wuestenfux
Dec 21 '18 at 9:58




$begingroup$
The ring isomorphism is fine.
$endgroup$
– Wuestenfux
Dec 21 '18 at 9:58












$begingroup$
thank you to the both of you.
$endgroup$
– Hawk
Dec 21 '18 at 10:05




$begingroup$
thank you to the both of you.
$endgroup$
– Hawk
Dec 21 '18 at 10:05












$begingroup$
@Wuestenfux actually wait my $pi$ is not well defined.
$endgroup$
– Hawk
Dec 21 '18 at 11:12




$begingroup$
@Wuestenfux actually wait my $pi$ is not well defined.
$endgroup$
– Hawk
Dec 21 '18 at 11:12










1 Answer
1






active

oldest

votes


















1












$begingroup$

Well, I'd consider the substitution homomorphism $phi:R[x]rightarrow R$ defined by $phi:f(x)mapsto f(0)$. The kernel is the ideal $I=langle xrangle$ and so by the homomorphism theorem, $R[x]/ker(phi) = R[x]/langle xrangle$ is isomorphic to $phi(R[x])$ which is $R$, since $phi$ is surjective. The isomorphism is defined by $f(x)+langle xrangle mapsto f(0)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @Hawk If $a_0+(x)=a_1+(x)$, then $a_0,a_1$ differ by a polynomial divisible by $x$. Since $a_0,a_1$ have degree $0$, this polynomial can only be zero, so indeed $a_0=a_1$.
    $endgroup$
    – Wojowu
    Dec 21 '18 at 11:41










  • $begingroup$
    @Wojowu so since this polynomial $p(x) = 0$, this means $a_0 - a_1 in (0)$?
    $endgroup$
    – Hawk
    Dec 21 '18 at 13:49












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1 Answer
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1 Answer
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active

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active

oldest

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active

oldest

votes









1












$begingroup$

Well, I'd consider the substitution homomorphism $phi:R[x]rightarrow R$ defined by $phi:f(x)mapsto f(0)$. The kernel is the ideal $I=langle xrangle$ and so by the homomorphism theorem, $R[x]/ker(phi) = R[x]/langle xrangle$ is isomorphic to $phi(R[x])$ which is $R$, since $phi$ is surjective. The isomorphism is defined by $f(x)+langle xrangle mapsto f(0)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @Hawk If $a_0+(x)=a_1+(x)$, then $a_0,a_1$ differ by a polynomial divisible by $x$. Since $a_0,a_1$ have degree $0$, this polynomial can only be zero, so indeed $a_0=a_1$.
    $endgroup$
    – Wojowu
    Dec 21 '18 at 11:41










  • $begingroup$
    @Wojowu so since this polynomial $p(x) = 0$, this means $a_0 - a_1 in (0)$?
    $endgroup$
    – Hawk
    Dec 21 '18 at 13:49
















1












$begingroup$

Well, I'd consider the substitution homomorphism $phi:R[x]rightarrow R$ defined by $phi:f(x)mapsto f(0)$. The kernel is the ideal $I=langle xrangle$ and so by the homomorphism theorem, $R[x]/ker(phi) = R[x]/langle xrangle$ is isomorphic to $phi(R[x])$ which is $R$, since $phi$ is surjective. The isomorphism is defined by $f(x)+langle xrangle mapsto f(0)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @Hawk If $a_0+(x)=a_1+(x)$, then $a_0,a_1$ differ by a polynomial divisible by $x$. Since $a_0,a_1$ have degree $0$, this polynomial can only be zero, so indeed $a_0=a_1$.
    $endgroup$
    – Wojowu
    Dec 21 '18 at 11:41










  • $begingroup$
    @Wojowu so since this polynomial $p(x) = 0$, this means $a_0 - a_1 in (0)$?
    $endgroup$
    – Hawk
    Dec 21 '18 at 13:49














1












1








1





$begingroup$

Well, I'd consider the substitution homomorphism $phi:R[x]rightarrow R$ defined by $phi:f(x)mapsto f(0)$. The kernel is the ideal $I=langle xrangle$ and so by the homomorphism theorem, $R[x]/ker(phi) = R[x]/langle xrangle$ is isomorphic to $phi(R[x])$ which is $R$, since $phi$ is surjective. The isomorphism is defined by $f(x)+langle xrangle mapsto f(0)$.






share|cite|improve this answer









$endgroup$



Well, I'd consider the substitution homomorphism $phi:R[x]rightarrow R$ defined by $phi:f(x)mapsto f(0)$. The kernel is the ideal $I=langle xrangle$ and so by the homomorphism theorem, $R[x]/ker(phi) = R[x]/langle xrangle$ is isomorphic to $phi(R[x])$ which is $R$, since $phi$ is surjective. The isomorphism is defined by $f(x)+langle xrangle mapsto f(0)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 21 '18 at 11:30









WuestenfuxWuestenfux

5,5131513




5,5131513












  • $begingroup$
    @Hawk If $a_0+(x)=a_1+(x)$, then $a_0,a_1$ differ by a polynomial divisible by $x$. Since $a_0,a_1$ have degree $0$, this polynomial can only be zero, so indeed $a_0=a_1$.
    $endgroup$
    – Wojowu
    Dec 21 '18 at 11:41










  • $begingroup$
    @Wojowu so since this polynomial $p(x) = 0$, this means $a_0 - a_1 in (0)$?
    $endgroup$
    – Hawk
    Dec 21 '18 at 13:49


















  • $begingroup$
    @Hawk If $a_0+(x)=a_1+(x)$, then $a_0,a_1$ differ by a polynomial divisible by $x$. Since $a_0,a_1$ have degree $0$, this polynomial can only be zero, so indeed $a_0=a_1$.
    $endgroup$
    – Wojowu
    Dec 21 '18 at 11:41










  • $begingroup$
    @Wojowu so since this polynomial $p(x) = 0$, this means $a_0 - a_1 in (0)$?
    $endgroup$
    – Hawk
    Dec 21 '18 at 13:49
















$begingroup$
@Hawk If $a_0+(x)=a_1+(x)$, then $a_0,a_1$ differ by a polynomial divisible by $x$. Since $a_0,a_1$ have degree $0$, this polynomial can only be zero, so indeed $a_0=a_1$.
$endgroup$
– Wojowu
Dec 21 '18 at 11:41




$begingroup$
@Hawk If $a_0+(x)=a_1+(x)$, then $a_0,a_1$ differ by a polynomial divisible by $x$. Since $a_0,a_1$ have degree $0$, this polynomial can only be zero, so indeed $a_0=a_1$.
$endgroup$
– Wojowu
Dec 21 '18 at 11:41












$begingroup$
@Wojowu so since this polynomial $p(x) = 0$, this means $a_0 - a_1 in (0)$?
$endgroup$
– Hawk
Dec 21 '18 at 13:49




$begingroup$
@Wojowu so since this polynomial $p(x) = 0$, this means $a_0 - a_1 in (0)$?
$endgroup$
– Hawk
Dec 21 '18 at 13:49


















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