Question
Show that (x, 5) is not a principal ideal in Z[x].
Answer :
Word Count : 177
Assume for contradiction that the ideal $(x,5)\subset \mathbb{Z}[x]$ is principal, say $(x,5)=(f)$ for some $f\in\mathbb{Z}[x]$. Then $x\in(f)$ and $5\in(f)$, so $f\mid x$ and $f\mid 5$ in $\mathbb{Z}[x]$. If $f$ is a nonconstant polynomial, then from $f\mid 5$ we would have a factorization $5=f\cdot g$ ______ __________ __________ ____ __________ _________ _____ ____.
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Assume for contradiction that the ideal $(x,5)\subset \mathbb{Z}[x]$ is principal, say $(x,5)=(f)$ for some $f\in\mathbb{Z}[x]$. Then $x\in(f)$ and $5\in(f)$, so $f\mid x$ and $f\mid 5$ in $\mathbb{Z}[x]$. If $f$ is a nonconstant polynomial, then from $f\mid 5$ we would have a factorization $5=f\cdot g$ ______ __________ __________ ____ __________ _________ _____ ____.
______ _____ ______ ______ _____ ___ ___ _______ ________ _________ ________ _____.
_________ _______ ______ ___ ___.
____ ____ _______ ___ ______ ________ _____ ______ ____ ___ ______ _________.
_______ ____ ____ ________ _________ _________ _________ __________ ___ ______ __________.
____ _________ __________ ______ _________ ____ _____ ________ ______ _____.
___ _______ _________ ___ ____ ___ ______ _____ ________ ____ _____.
______ _______ ________ _____ _______ _______ ______.
__________ _________ ___ __________ __________ __________ ____ ________ _____ ______ _______ ______.
______ __________ ________ ______ __________ __________.
___ _________ ____ ____ __________ ___ ______ ________ __________ ____ ______.
___ _________ ________ ________ __________ ________ _______.
_________ ____ ________ ________ ____ ______ ________.
_______ ______ ____ _____ _________ __________ ____ _______.
________ _____ _______ ___ _______ _________.
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