A question about the statement of the Prime Avoidance Lemma.

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The Prime Avoidance Lemma states the following:




Suppose $I_1,I_2,dots,I_n,J$ are ideals of a ring $R$, such that $Jsubset cup U_i$. If $R$ contains an infinite field, or if at most $2$ of the ideals $I_1,dots,I_n$ are prime, then $Jsubset I_k$ for some $k$.




Consider $(4)cup (6)$ as ideals of the ring $BbbZ$. Although it does not contain in infinite field, the ideals $(6)$ and $(4)$ satisfy the criterion of at most two ideals being not prime. Hence, the condition of the prime avoidance lemma is satisfied. However, $(14)subset (4)cup (6)$, and $(14)$ is not contained within either of $(4)$ or $(6)$. Is this not a counter-example? Where am I going wrong?




commutative-algebra




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  • @Randall- Yeah sorry I was confusing $(4)cup (6)$ with $(4)+(6)$
    – fierydemon
    Jul 30 at 17:33














up vote
0
down vote

favorite












The Prime Avoidance Lemma states the following:




Suppose $I_1,I_2,dots,I_n,J$ are ideals of a ring $R$, such that $Jsubset cup U_i$. If $R$ contains an infinite field, or if at most $2$ of the ideals $I_1,dots,I_n$ are prime, then $Jsubset I_k$ for some $k$.




Consider $(4)cup (6)$ as ideals of the ring $BbbZ$. Although it does not contain in infinite field, the ideals $(6)$ and $(4)$ satisfy the criterion of at most two ideals being not prime. Hence, the condition of the prime avoidance lemma is satisfied. However, $(14)subset (4)cup (6)$, and $(14)$ is not contained within either of $(4)$ or $(6)$. Is this not a counter-example? Where am I going wrong?




commutative-algebra




share|cite|improve this question



















  • @Randall- Yeah sorry I was confusing $(4)cup (6)$ with $(4)+(6)$
    – fierydemon
    Jul 30 at 17:33












up vote
0
down vote

favorite









up vote
0
down vote

favorite











The Prime Avoidance Lemma states the following:




Suppose $I_1,I_2,dots,I_n,J$ are ideals of a ring $R$, such that $Jsubset cup U_i$. If $R$ contains an infinite field, or if at most $2$ of the ideals $I_1,dots,I_n$ are prime, then $Jsubset I_k$ for some $k$.




Consider $(4)cup (6)$ as ideals of the ring $BbbZ$. Although it does not contain in infinite field, the ideals $(6)$ and $(4)$ satisfy the criterion of at most two ideals being not prime. Hence, the condition of the prime avoidance lemma is satisfied. However, $(14)subset (4)cup (6)$, and $(14)$ is not contained within either of $(4)$ or $(6)$. Is this not a counter-example? Where am I going wrong?




commutative-algebra




share|cite|improve this question











The Prime Avoidance Lemma states the following:




Suppose $I_1,I_2,dots,I_n,J$ are ideals of a ring $R$, such that $Jsubset cup U_i$. If $R$ contains an infinite field, or if at most $2$ of the ideals $I_1,dots,I_n$ are prime, then $Jsubset I_k$ for some $k$.




Consider $(4)cup (6)$ as ideals of the ring $BbbZ$. Although it does not contain in infinite field, the ideals $(6)$ and $(4)$ satisfy the criterion of at most two ideals being not prime. Hence, the condition of the prime avoidance lemma is satisfied. However, $(14)subset (4)cup (6)$, and $(14)$ is not contained within either of $(4)$ or $(6)$. Is this not a counter-example? Where am I going wrong?





commutative-algebra





share|cite|improve this question










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asked Jul 30 at 17:24









fierydemon

4,32312151




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  • @Randall- Yeah sorry I was confusing $(4)cup (6)$ with $(4)+(6)$
    – fierydemon
    Jul 30 at 17:33
















  • @Randall- Yeah sorry I was confusing $(4)cup (6)$ with $(4)+(6)$
    – fierydemon
    Jul 30 at 17:33















@Randall- Yeah sorry I was confusing $(4)cup (6)$ with $(4)+(6)$
– fierydemon
Jul 30 at 17:33




@Randall- Yeah sorry I was confusing $(4)cup (6)$ with $(4)+(6)$
– fierydemon
Jul 30 at 17:33










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$(14)$ isn't contained in the union $(4) cup (6)$, which only contains numbers divisible by either $4$ or $6$, but only in the larger set given by the ideal sum $(4)mathbf+(6)=(2)$.






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  • Ah yes. I was confusing $(4)cup (6)$ with $(4)+(6)$. Thanks!
    – fierydemon
    Jul 30 at 17:29










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up vote
2
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$(14)$ isn't contained in the union $(4) cup (6)$, which only contains numbers divisible by either $4$ or $6$, but only in the larger set given by the ideal sum $(4)mathbf+(6)=(2)$.






share|cite|improve this answer





















  • Ah yes. I was confusing $(4)cup (6)$ with $(4)+(6)$. Thanks!
    – fierydemon
    Jul 30 at 17:29














up vote
2
down vote













$(14)$ isn't contained in the union $(4) cup (6)$, which only contains numbers divisible by either $4$ or $6$, but only in the larger set given by the ideal sum $(4)mathbf+(6)=(2)$.






share|cite|improve this answer





















  • Ah yes. I was confusing $(4)cup (6)$ with $(4)+(6)$. Thanks!
    – fierydemon
    Jul 30 at 17:29












up vote
2
down vote










up vote
2
down vote









$(14)$ isn't contained in the union $(4) cup (6)$, which only contains numbers divisible by either $4$ or $6$, but only in the larger set given by the ideal sum $(4)mathbf+(6)=(2)$.






share|cite|improve this answer













$(14)$ isn't contained in the union $(4) cup (6)$, which only contains numbers divisible by either $4$ or $6$, but only in the larger set given by the ideal sum $(4)mathbf+(6)=(2)$.







share|cite|improve this answer













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answered Jul 30 at 17:27









Chessanator

1,592210




1,592210











  • Ah yes. I was confusing $(4)cup (6)$ with $(4)+(6)$. Thanks!
    – fierydemon
    Jul 30 at 17:29
















  • Ah yes. I was confusing $(4)cup (6)$ with $(4)+(6)$. Thanks!
    – fierydemon
    Jul 30 at 17:29















Ah yes. I was confusing $(4)cup (6)$ with $(4)+(6)$. Thanks!
– fierydemon
Jul 30 at 17:29




Ah yes. I was confusing $(4)cup (6)$ with $(4)+(6)$. Thanks!
– fierydemon
Jul 30 at 17:29












 

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