Let $\mathbb{F}$ be a field, $a \in \mathbb{F}$ and set $p(x)=(x-a)^2$. Prove that $\mathbb{F}[x]/\langle p(x)\rangle$ is not a field.
I'm not sure where to start on this one. We were told to pick a nonzero class which is not invertible.
Let $\mathbb{F}$ be a field, $a \in \mathbb{F}$ and set $p(x)=(x-a)^2$. Prove that $\mathbb{F}[x]/\langle p(x)\rangle$ is not a field
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abstract-algebra
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2$[x-a]$ is certainly not zero. But when you square it... – 2017-02-01
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0It is reducible polynomial over $\mathbb{F},$ as $(x-a)^2=(x-a)(x-a).$ $\mathbb{F}[x]/\langle p(x) \rangle $ is a field if and only if $p(x)$ is irreducible over $\mathbb{F}$ – 2017-02-01
2 Answers
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Recall that $F[x]/(p)$ is a field iff $(p)$ is a maximal ideal of $F[x]$. Can you find $q\in F[x]$ such that $(q) \subsetneq (p) \subsetneq F[x]$?
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0I highly doubt the OP would have to ask for help if they knew that more advanced result. – 2017-02-01
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0@BillDubuque they may just tell me and I'll elaborate or use some less advanced result. – 2017-02-01
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The thought process should be here that: $$k[x]/\langle f(x)\rangle\text{ is a field}\iff \langle f(x)\rangle\text{ is maximal}\iff f(x)\text{ is irreducible}$$ So, as $f(x)$ isn't irreducible (as $(x-a)^2 = (x-a)(x-a)$), we have that $k[x]/\langle f(x)\rangle$ isn't a field.