MHB Is this the only right proposition?

[evinda]

Hello! (Wave)

We are given a ring $R$, polynomials $a$ and $b$ in $R[x]$, an integer $n$, and $c \in R$. We consider the (not necessary true) propositions $P_1$, $P_2$ and $P_3$:

$P_1$: If for any $t \in R$ it holds $a(t)=b(t)$ then $a=b$.

$P_2$: If there are $9$ different elements of $R$ then there are three different ideals of $R$.

$P_3$: If $R$ is an integral domain and $nc=0$ in $R$ then $n=0$ or $c=0$.

Let $A=\{ j \in \{1,2,3\} : P_j \text{ always holds }\}$. Then there are five possible answers for $A$:

1. $A=\{3\}$
2. $A=\varnothing$
3. $A=\{2\}$
4. $A=\{1\}$
5. $A=\{1,2\}$

I think that $P_3$ is the only right proposition and thus $A=\{3\}$. Am I right? (Thinking)

 

[I like Serena]

Hello! (Wave)

We are given a ring $R$, polynomials $a$ and $b$ in $R[x]$, an integer $n$, and $c \in R$. We consider the (not necessary true) propositions $P_1$, $P_2$ and $P_3$:

$P_1$: If for any $t \in R$ it holds $a(t)=b(t)$ then $a=b$.

$P_2$: If there are $9$ different elements of $R$ then there are three different ideals of $R$.

$P_3$: If $R$ is an integral domain and $nc=0$ in $R$ then $n=0$ or $c=0$.

Let $A=\{ j \in \{1,2,3\} : P_j \text{ always holds }\}$. Then there are five possible answers for $A$:

1. $A=\{3\}$
2. $A=\varnothing$
3. $A=\{2\}$
4. $A=\{1\}$
5. $A=\{1,2\}$

I think that $P_3$ is the only right proposition and thus $A=\{3\}$. Am I right? (Thinking)

Hey evinda!

$P_3$ is part of the definition of an integral domain, so yes, it is indeed true.
What is your reasoning that $P_1$ and $P_2$ are false? (Wondering)