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Show that $\langle x\rangle$ is a maximal ideal of $R[x]$, where $R$ is a field.

This is one of my assignment questions. It is supposed to be hard but what I did below is so easy but I couldn't find anything wrong. Can anybody help me check it out?

Since $R$ is a field and $f(x) = x$ is irreducible in $R[x]$, so $R[x]/\langle x\rangle$ is a field, and $\langle x\rangle$ is a ideal of $R[x]$, so $\langle x\rangle$ is maximal.

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You are correct that if you can prove that $R[x]/\langle x\rangle$ is a field, then you can conclude that $\langle x\rangle$ is maximal.

However, it is not true in general that if $a$ is irreducible in a domain $D$, then $D/\langle a\rangle$ is a field. For example, $2$ is irreducible in $\mathbb{Z}[\sqrt{-5}]$, but $\langle 2\rangle$ is not a maximal ideal of $\mathbb{Z}[\sqrt{-5}]$. So you need more than just "$f(x) = x$ is irreducible in $R[x]$" to conclude "$R[x]/\langle x\rangle$ is a field."

(For another, simpler example, consider the case of $\langle x \rangle$ in $R[x,y]$. You can still say that "since $R$ is a field, $x$ is irreducible", but it is no longer true that $R[x,y]/\langle x\rangle$ is a field; in fact, $\langle x\rangle$ is not maximal in this ring, since it is properly contained in the proper ideal $\langle x,y\rangle$.)

But perhaps you can show that the homomorphism $R[x]\to R$ given by "evaluation at $0$" is onto, and has kernel $\langle x\rangle$? That will show that $R[x]/\langle x\rangle \cong R$ is a field.

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    True, since $\mathbb{Z}[\sqrt-5]$ is not a UFD2012-04-09
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    @Belgi: True; but you need more than just UFD. $x$ is irreducible in the UFD $\mathbb{R}[x,y]$, but $\langle x\rangle$ is not maximal. In general, is $a$ is irreducible in a domain, then $\langle a\rangle$ is maximal *among principal ideals*. But it's possible that it may be nonmaximal; PID suffices for the implication, which will happen here, but the proffered reasons are insufficient.2012-04-09
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    You take me back to the lectures in ring theory :) again you are right2012-04-09
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    @ArturoMagidin I did show that R[x]/ is isomorphic to R is a field, then I noticed that there is actually a theorem that says if F is a field and monic p(x) in F[x] is irriducible in F[x], then F[x]/ is a field.2012-04-09
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    @Shannon: Then you need to invoke the theorem explicitly; like I said, what you write is not sufficient on its face, since there are situations where "$a$ is irreducible" does not imply $D/\langle a\rangle$ is a field. I would suggest just showing that $R[x]/\langle x\rangle$ is isomorphic to $R$ and be done.2012-04-09
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    @ArturoMagidin Thank you. I will do that. By the way, if R is not a field instead R is an integral domain, then not all polynomials can be written as xf(x)+a, where f(x) in R[x] and a in real number. So R[x]/ is not a field, even not an intergral domain, right?2012-04-09
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    @Shannon: If $R$ is an integral domain that is not a field, it is *still* true that every polynomial can be written uniquely as $g(x) = xf(x) + a$, with $a$ constant; the division algorithm works for any polynomial with leading coefficient at unit, and $x$ is such a polynomial. We always have $R[x]/\langle x\rangle \cong R$, so the quotient would be isomorphic to $R$ and hence a domain, though not a field unless $R$ is a field.2012-04-09
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    @ArturoMagidin: Thanks for your help me understanding all this. Since R is an integral domain, then is just an ideal, not a maximal ideal any more.2012-04-09
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    @Shannon: Correct. Note that the notation $\langle a\rangle$ represents "the smallest ideal of the ring that contains $a$", so it is *necessarily* an ideal, no matter what ring you are working on and what element you are working with. For *any* commutative ring $R$ (domain or not), $\langle x\rangle$ is an ideal of $R[x]$ and $R[x]/\langle x\rangle = 0$. The ideal $\langle x\rangle$ is a prime ideal if and only if $R$ is a domain, and it is a maximal ideal if and only if $R$ is a field.2012-04-09
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    @ArturoMagidin: Got you. So in this case, is a prime ideal. I really learnt a lot from you.2012-04-09
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    @Shannon: Yes; note that I mistyped in the above comment. "...and $R[x]/\langle x\rangle = 0$" should be "...and $R[x]/\langle x\rangle \cong R$."2012-04-09
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    @ArturoMagidin I noticed that.:p2012-04-09
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If you know that $R[x]$ is a PID when $R$ is a field, then any ideal above $\langle x\rangle$ must be generated by some monic polynomial $f$ which divides $x$. Hence $f=x$ or $f=1$ and so the ideal is either $\langle x\rangle$ or $R[x]$, and $\langle x\rangle$ is maximal.

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Hint $\rm\ f\not\in\! (x)\ \Rightarrow\ (x,f) = (x,\: f\ mod\ x) = (x,f(0)) = (1)\:$ by $\rm\:f(0)\neq 0\ \Rightarrow\ f(0)$ unit

Or: $\rm\ mod\ x\!:\:\ x\equiv 0\ \Rightarrow\ f(x)\equiv f(0) =$ unit or $0,\:$ so $\rm\:R[x]\ mod\ x\:$ is a field.