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Determine the maximal ideals of $\mathbb{R}^2$.

Well for any real number that is not divisible by another number other than 1 and itself generates a maximal ideal for $\mathbb{R}$. Is that right? Would it be the same as for $\mathbb{R}^2$?

user26857
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sarah
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  • I'm kind of confused by what you're asking. $\mathbb R$ is a field, so it has only one proper ideal (the zero ideal). You can use that to try to find the ideals of $\mathbb R^2$. – Ian Coley Oct 20 '13 at 05:12
  • No, that is not right: there doesn't exist "any real number that is not divisible by another number other than $1$ and itself." Fields have two ideals: $0$ and the whole field. – anon Oct 20 '13 at 05:14
  • I guess OP means the product ring structure where addition and multiplication are defined componentwise. – user43208 Oct 20 '13 at 05:29
  • I wonder how this question was answered and accepted by the OP without even knowing what product is the OP taking for $;\Bbb R^2;$...unless (s)he meant actually the external direct product $;\Bbb R\times\Bbb R;$ of the real field with itself. Mistery... – DonAntonio Oct 20 '13 at 10:55

1 Answers1

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Not quite, since there aren't any primes in $\mathbb{R}$ - every non-zero element has a multiplicative inverse. Any non-trivial ideal in $\mathbb{R}$ is, in fact, $\mathbb{R}$. For if $r \in I$ and $r \ne 0$, then

$$\frac{1}{r} I \subseteq I \implies \frac{1}{r} r \in I \implies 1 \in I$$

And any ideal containing $1$ is the entire ring. So $0$ is actually the unique maximal ideal.

On the other hand, suppose that $(a, b) \in I$ for some ideal $I \subseteq \mathbb{R} \times \mathbb{R}$. Consider some cases:

  • If $a$ and $b$ are both non-zero, show that $0 \ne I = \mathbb{R} \times \mathbb{R}$ by constructing something analogous to $1/r$.

  • If every element of $I$ looks like $(0, b)$, conclude that $I = 0 \times \mathbb{R}$ by thinking about the element $(0, \frac{1}{b})$.

  • Likewise for $(a, 0)$.