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Exactly like in the title,

here was my thought process:

If $P\times Q$ must be a field, every element (besides $(0, 0)$) must have its inverse. Let's take $(p, q)$ for any $p$ from $P$ and $q$ from $Q$. For it to have an inverse, $p$ and $q$ both must have inverses therefore $P$ and $Q$ are both fields, however we can take $(p, 0)$ or $(0, q)$ which doesn't have an inverse since $0$ doesn't have an inverse (?), therefore both $(p, 0)$ and $(0, q)$ don't have inverses so such field can not exist. Am I wrong?

Thanks for any help.

Red Five
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Michal Pielka
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  • Instead of going about it that way, ask what it means for $(p,0)$ to have an inverse. You've missed a case by assuming that it doesn't have an inverse. – Brian Moehring May 14 '24 at 23:59
  • @BrianMoehring (p, 0)^(-1) = (p^(-1), 0^(-1)) so P MUST be a field, but then again 0 doesn't have an inverse for all I know? – Michal Pielka May 15 '24 at 00:01
  • For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. – Bill Dubuque May 15 '24 at 00:03
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    You either need to use the definition of "has an inverse", or you need to ask when $0$ has an inverse. We never said it's an axiom that $0$ doesn't have an inverse. That is a consequence of another [admittedly extremely common] assumption that we don't have here. – Brian Moehring May 15 '24 at 00:18
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    Your question should be clear without the title. After the title has drawn someone's attention to the question by giving a good description, its purpose is done. The title is not the first sentence of your question, so make sure that the question body does not rely on specific information in the title. – Martin R May 15 '24 at 02:36

1 Answers1

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In short, you are right to look at elements of the form $(p,0)$, but you seem to be jumping to conclusions. Instead, start from the definitions and deduce something about $P\times Q$.

Suppose $(p,0) \in P\times Q$ is nonzero (I'm assuming something about $P$ here... do you see what it is?) and $P\times Q$ is a field.

Then $(p,0)$ has an inverse in $P\times Q$.

Then there is some $(a,b) \in P\times Q$ such that $$(pa,0) = (p,0) \cdot (a,b) = (1_P, 1_Q)$$ By definition, this means $pa = 1_P$ and $0 = 1_Q$.

This means that $Q = \{0\}$ is the zero ring.

From here, it's enough to note that $P \times \{0\} \cong \{0\} \times P \cong P$, so if $P \times \{0\}$ is a field, then $P$ itself is a field.

Then, to summarize, if $P,Q$ are rings and $P\times Q$ is a field, then $$\{P,Q\} = \{\{0\}, k\}$$ for some field $k$ (that is, one is the zero ring and the other is a field)

Note that this doesn't contradict If $R$ and $S$ are rings then $R \times S$ is never a field (or domain) or any other similar question, since it has the additional assumption that both $R,S$ are nonzero rings (which has unfortunately been omitted from that title)

Brian Moehring
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