关于ℝ×ℂ中极大理想的求解问询
Hey there! Let's work through this problem together—since you already understand maximal ideals in ℤ, we can build on that intuition for ℝ×ℂ.
First, let's recall a key fact about product rings: If you have a ring of the form R × S, every ideal of R × S is of the form I × J, where I is an ideal of R and J is an ideal of S. That's a foundational rule here, so keep it in mind.
Now, for an ideal I × J to be maximal in R × S, the quotient ring (R × S)/(I × J) needs to be a field (since maximal ideals correspond to field quotients). Notice that (R × S)/(I × J) ≅ (R/I) × (S/J). When is a product of rings a field? Only if one of the factors is a field and the other is the trivial ring (i.e., just {0}). That's because a field can't have non-trivial idempotents, but a product of two non-trivial rings does (like (1,0)).
So applying this to ℝ × ℂ:
- First, let's look at
ℝ: it's a field! Fields have exactly two ideals: the trivial ideal{0}and the entire fieldℝ. The only maximal ideal here is{0}, since it's the only proper ideal (maximal ideals have to be proper, meaning they don't equal the whole ring). - Next,
ℂis also a field—same logic applies: its only maximal ideal is{0}.
Now let's map this back to ℝ × ℂ:
- If we take the maximal ideal of
ℝ({0}) and pair it with the entire ringℂ, we get the ideal{0} × ℂ. The quotient ring here is(ℝ × ℂ)/({0} × ℂ) ≅ ℝ, which is a field—so this is a maximal ideal. - If we take the entire ring
ℝand pair it with the maximal ideal ofℂ({0}), we getℝ × {0}. The quotient ring here is(ℝ × ℂ)/(ℝ × {0}) ≅ ℂ, also a field—so this is another maximal ideal.
Are there any other maximal ideals? Let's see: suppose we tried an ideal like I × J where both I and J are proper ideals. But since ℝ and ℂ only have {0} as a proper ideal, that would give us {0} × {0}. The quotient ring here is ℝ × ℂ itself, which is not a field—so that's not maximal.
You mentioned being unsure about uncountable sets like ℝ or ℂ compared to ℤ. The key difference here is that ℝ and ℂ are fields, whereas ℤ is just a commutative ring with unity. For any field (countable or uncountable), the only maximal ideal is the trivial ideal {0}, because fields don't have any non-trivial proper ideals. That's why the logic here is different from ℤ, where maximal ideals correspond to prime integers.
备注：内容来源于stack exchange，提问作者Moll123
