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I was asked in homework to think about maximal ideals in polynomial rings $\mathbb{R}[x]$ and $\mathbb{C}[x]$. I have realized that: $\forall c\in\mathbb{R},\;I_c : = \{p(x)\in\mathbb{R}[x]\;|\;p(c) = 0\}$ is an ideal (similar for $\mathbb{C}[x]$), now in order to prove it to be maximal, I need to show: $$I_c\subset J\subsetneq A,\;J\text{ is an ideal}\Longrightarrow I_c = J$$ which I have difficulty showing.

Secondly, I don't know how to show that all maximal ideas are in the form of $I_c$. Some help please. Thank you.

mez
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2 Answers2

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Hint $\ $ Polynomial rings over fields enjoy a (Euclidean) division algorithm, hence every ideal is principal, generated by an element of minimal degree (= gcd of all elements). But for principal ideals: contains $\!\iff\!$ 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). Summing up, we have deduced that, in a PID, $\rm\ (f)\:$ is maximal $\rm\!\iff\! f\:$ is irreducible. Thus the problem reduces to determining the irreducible polynomials in $\rm\,\Bbb R[x]\,$ and $\rm\,\Bbb C[x],\,$ which is straightforward.

Math Gems
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  • Thank you very much. I understand now the idea. I was thinking about the field $\mathbb{R}[x]\backslash I$ and $\mathbb{C}[x]\backslash I$ where $I$ is the maximal ideal in the ring. I think they are correspondingly isomorphic to $\mathbb{R}$ and $\mathbb{C}$, am I right? What else can I say about them? – mez Mar 04 '13 at 13:08
  • @mezhang Consider $\rm, \Bbb R[x]/(x^2+1) \cong \Bbb C\ \ $ – Math Gems Mar 04 '13 at 15:35
  • Thanks, I realized that later. – mez Mar 04 '13 at 19:28
  • Also, to prove isomorphism, I construct map $\Phi: \mathbb{C}[x]\backslash I_c\to \mathbb{C}$, $\Phi([p(x)]) = p(c)$. Then it is true that if this homomorphism is bijection then it is isomorphism right? (unlike homeomorphism which require both maps to be continuous) – mez Mar 04 '13 at 20:35
  • Use the first isomorphism theorem: show that the image is $,\Bbb C,$ and the kernel is $\rm,(x-c),$ hence by the theorem $\rm\ Im,\Phi = \Bbb C \cong \Bbb C[x]/(x-c) = \Bbb C[x]/(ker, \Phi).\ \ $ – Math Gems Mar 04 '13 at 20:46
  • I didn't know that isomorphism theorem is for field as well, I thought it is for groups. How do I justify this? – mez Mar 04 '13 at 20:53
  • It's true for rings too (and generalizes to all algebraic structures using congruences). It's a fundamental result that is essential to know. You will find it in every algebra textbook. – Math Gems Mar 04 '13 at 20:59
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Hint: An ideal $I$ is maximal if and only if the quotient $\mathbb{R}[x]/I$ is a field. Try to find some $f:\mathbb{R}[x]\to \mathbb{R}$ with kernel $I_c$.

Sigur
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