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Let $\mathbb{F}$ be a field and $R=\mathbb{F}[x]$, the polynomial ring over $\mathbb{F}$. Is the ideal $(x^2-1)$ maximal in $R$? Does the answer depend upon $\mathbb{F}$?

I think of this isomorphism $\mathbb{F}[x]/(x^2-1) \cong \mathbb{F}[i]$ where $\mathbb{F}[i]=\lbrace a+bi:a,b \in \mathbb{F} \rbrace$. Since $\mathbb{F}[i]$ is a field (which I am not quite sure about it), $(x^2-1)$ is maximal.

Can anyone explain to me whether $\mathbb{F}[i]$ is a field or not.

Idonknow
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  • I wonder why the heck the downvote...! Because the OP is wrong assuming the quotient ring is a field? But that's what he's asking! Of course, +1 because this is a good question. – DonAntonio Apr 23 '13 at 16:05
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    Note that $x$ is not a square root of $-1$. It's not clear what you mean by $i$ when you write $\mathbb F[i]$. What if $\mathbb F=\mathbb C$? – Thomas Andrews Apr 23 '13 at 16:05
  • @Landscape , it all depends: sure... of what ? – DonAntonio Apr 23 '13 at 16:15
  • Because it is a well and clearly redacted question about a mathematical issue where the OP has some problems of understanding, and that's precisely, imfho, what this site is for: to help people with doubts about mathematics. Furthermore, the OP did take time to explain his mind and showed some self effort and work. In my book, this is in fact is a very good question. Perhaps you think a good question is one where some self work is shown and it is mistakeless? – DonAntonio Apr 23 '13 at 16:23
  • @DonAntonio: Now I agree with you. Please let me delete the impolite comments. – 23rd Apr 23 '13 at 16:32
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    Oh, don't worry about that, @Landscape : we all are entitled to my...I mean, our ...own opinions. :) It's just that I think it is not fair to downvote sincere, worked questions just "because", even if we consider the question not to be a good one, or boring or trivial. – DonAntonio Apr 23 '13 at 16:35
  • @DonAntonio: I see. Thank you for your responses. – 23rd Apr 23 '13 at 16:47

6 Answers6

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Hint:

$$x^2-1=(x-1)(x+1)\implies x-1+\langle x^2-1\rangle \;\;\text{is a divisor of zero in}\;\;\Bbb F[x]/\langle x^2-1\rangle\ldots$$

DonAntonio
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5

Hint: If so, the quotient structure is a field. But the quotient structure has zero divisors.

We can also solve the problem "from the definition," without quoting the above result. Prove that $(x-1)$ is an ideal that properly contains $(x^2-1)$.

André Nicolas
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If $p(x)$ is irreducible over a field $F$, then the ideal generated by $p(x)$ is maximal. If $p(x)$ reducible, the ideal is not maximal. Furthermore, if $F$ is a commutative ring with unity, $F/J$ is a field if and only if $J$ is a maximal ideal. So to prove $F[x]/(x^2 - 1)$ is not a field, show that $x^2 -1$ is reducible over $F$.

manthanomen
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Whilst $(x^2 - 1)$ is never maximal, the same is not true of $(x^2 + 1)$:

If $\mathbb{F} = \mathbb{C}$ then $(x + i)(x - i) = (x^2 + 1)$ so $\mathbb{C}[x]/(x^2 + 1)$ is not even an integral domain.

However, if $\mathbb{F} = \mathbb{R}$ then the map $f: \mathbb{R}[x] \rightarrow \mathbb{C}$ given by $f(x) = i$, $f(r) = r$ for $r \in \mathbb{R}$ is a surjective ring homomorphism with kernel $(x^2 + 1)$, and so $\mathbb{R}[x]/(x^2 + 1)$ is a field, so $(x^2 + 1)$ is a maximal ideal.

(In general, $(x^2 + 1)$ is maximal in $\mathbb{F}[x]$ iff there does not exist solutions to $x^2 + 1$ in $\mathbb{F}$.)

I'm posting this because the fact that the OP talks about $\mathbb{F}[x]/(x^2-1) \cong \mathbb{F}[i]$ suggests he might have meant $(x^2+1)$ in the first place.

rschwieb
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Christopher
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Hint $\rm\,\ (x^2\!-\!1)\:$ is not maximal since $\rm (1)\supsetneq (x\!-\!1)\supsetneq (x^2\!-\!1)\ $ by $\rm\ 1\mid x\!-\!1\mid x^2\!-\!1\:$ all properly.

Remark $\ $ For principal ideals: $ $ contains $\equiv$ divides, i.e. $\rm\: (a)\supseteq (b)\!\iff\! a\mid b.\:$ Thus, having no proper containing ideal (maximal) is equivalent to having no proper divisor (irreducible).

Math Gems
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All you need to check is whether $\mathbb F[x]/\langle x^2-1\rangle$ is a field. Note that,

$$x+1,x-1\notin \langle x^2-1\rangle$$

$($$\mathbb F$ is an integral domain $\implies$ $\deg f(x)g(x)=\deg f(x)$$+\deg g(x)$$~\forall~f(x),g(x)\in\mathbb F[x]-\{0\}).$

Consequently $(x+1)+\langle x^2-1\rangle,(x-1)+\langle x^2-1\rangle$ are non-zero elements of $\mathbb F[x]/\langle x^2-1\rangle.$

But $((x+1)+\langle x^2-1\rangle)((x-1)+\langle x^2-1\rangle)=(x^2-1)+\langle x^2-1\rangle=\langle x^2-1\rangle$ implies that $\mathbb F[x]$ has divisors of zero.

Thus $\langle x^2-1\rangle$ is not maximal in $R.$

For your second question you should simply note that we've just used the property of an integral domain.

Sugata Adhya
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