Let R, S be commutative rings with unity, f ∶ R → S be a surjective ring homomorphism, Q ⊆ S be a non-zero prime ideal. Which of the following statements are true?
Let R and S be commutative rings with unity, and let f ∶ R → S be a surjective ring homomorphism. Let Q ⊆ S be a non-zero prime ideal. We need to analyze the properties of the preimage ideal f-1(Q) in R.
Prime Ideal Preimage
A fundamental property of ring homomorphisms relates prime ideals. If f ∶ R → S is a ring homomorphism and Q is a prime ideal in S, then the preimage f-1(Q) is a prime ideal in R, provided f-1(Q) is not equal to R. The preimage f-1(Q) = R if and only if 1S ∈ Q. If Q is a proper prime ideal (Q ≠ S), then 1S ∉ Q, which means 1R ∉ f-1(Q), so f-1(Q) is a proper prime ideal in R.
In this question, Q is given as a non-zero prime ideal in S. Since Q is non-zero, there exists an element $q \in Q$ such that $q \ne 0$. Since f is surjective, there exists an element $r \in R$ such that $f(r) = q$. By definition of the preimage, $r \in f^{-1}(Q)$. If $r$ were 0, then $f(0) = 0 = q$, which contradicts $q \ne 0$. Therefore, $r \ne 0$, and $r \in f^{-1}(Q)$. This shows that $f^{-1}(Q)$ contains a non-zero element, so $f^{-1}(Q)$ is a non-zero ideal in R.
If Q is a proper non-zero prime ideal in S, then f-1(Q) is a proper non-zero prime ideal in R. What if Q=S? If Q=S, then S must be a prime ring, which means S is an integral domain. Since f is surjective, $S \cong R/\ker(f)$. If S is an integral domain, then $R/\ker(f)$ is an integral domain, meaning $\ker(f)$ is a prime ideal in R. In this case, $f^{-1}(Q) = f^{-1}(S) = R$. The ring R itself is a non-zero ideal (since S is non-zero and f is surjective, R must be non-zero). The definition of a prime ideal I in a commutative ring R is that if $ab \in I$, then $a \in I$ or $b \in I$. For $I=R$, $ab \in R$ is always true, and $a \in R$ or $b \in R$ is always true. Thus, R satisfies the definition of a prime ideal. So, f-1(Q) is always a non-zero prime ideal in R for a non-zero prime ideal Q in S.
Based on this, Statement 1: f-1(Q) is a non-zero prime ideal in R, is true.
Maximality Link between Preimage and Ideal
A crucial result concerning surjective ring homomorphisms states that if f ∶ R → S is a surjective ring homomorphism, and Q is an ideal in S, then the preimage f-1(Q) is a maximal ideal in R if and only if Q is a maximal ideal in S. This comes from the isomorphism $R/f^{-1}(Q) \cong S/Q$. $f^{-1}(Q)$ is maximal in R if and only if $R/f^{-1}(Q)$ is a field. $S/Q$ is a field if and only if Q is maximal in S.
Therefore, Statements 2, 3, and 4 can be rephrased as asking under which conditions Q must be a maximal ideal in S, given that Q is a non-zero prime ideal in S.
PID Condition for Maximality
Statement 2: f-1(Q) is a maximal ideal in R if R is a PID. This is equivalent to asking: If R is a PID, is every non-zero prime ideal Q in S necessarily maximal in S?
Since R is a PID, it is an integral domain. $S \cong R/\ker(f)$. Let $I = \ker(f)$. $I$ is an ideal in R. If $I = \{0\}$, then $S \cong R$. Since R is a PID, its non-zero prime ideals are maximal. Q is a non-zero prime ideal in S $\cong$ R, so Q is maximal in S. If $I \ne \{0\}$, then $I = (a)$ for some $a \ne 0$ in the PID R. The prime ideals of $S \cong R/(a)$ correspond to the prime ideals of R containing $(a)$. Let $P$ be a prime ideal in R containing $(a)$. If $P=(a)$, then $S \cong R/(a)$ is an integral domain, which means $(a)$ is a prime ideal in R. Since $a \ne 0$, and R is a PID, $(a)$ must be a maximal ideal. In this case $S$ is a field. The only non-zero prime ideal Q in a field S is S itself, which is maximal. If $P \supsetneq (a)$, then $P$ must be a prime ideal in R properly containing $(a)$. In a PID, every proper ideal is contained in a maximal ideal. Since R is a PID, non-zero prime ideals are maximal. If $P$ is a proper prime ideal containing $(a)$ and $P \ne (a)$, $P$ must be maximal in R. Any non-zero proper prime ideal $P$ in R containing $(a)$ must be of the form $(p)$ where $p$ is a prime element dividing $a$. These ideals $(p)$ are maximal in R. The corresponding ideals in $S \cong R/(a)$ are $(p)/(a)$, which are the non-zero proper prime ideals of S. The quotient $S/((p)/(a)) \cong (R/(a))/((p)/(a)) \cong R/(p)$. Since $(p)$ is maximal in R, $R/(p)$ is a field. Thus, $(p)/(a)$ is maximal in S.
In all cases where R is a PID, any non-zero prime ideal Q in S is maximal in S. Therefore, f-1(Q) is maximal in R.
Statement 2 is true.
Finite Ring Condition for Maximality
Statement 3: f-1(Q) is a maximal ideal in R if R is a finite commutative ring with unity.
If R is a finite commutative ring with unity, and f ∶ R → S is a surjective ring homomorphism, then S is also a finite commutative ring with unity (as it is the image of R). A key property of finite commutative rings with unity is that every prime ideal is a maximal ideal. Since Q is a non-zero prime ideal in S, and S is a finite commutative ring with unity, Q must be a maximal ideal in S.
Since Q is a maximal ideal in S, and $f^{-1}(Q)$ is maximal in R if and only if Q is maximal in S, f-1(Q) is a maximal ideal in R.
Statement 3 is true.
Ring with x5 = x Condition for Maximality
Statement 4: f-1(Q) is a maximal ideal in R if x5 = x for all x ∈ R.
If R is a ring such that $x^5=x$ for all $x \in R$, then R is commutative and has unity. The property $x^5=x$ is preserved under surjective homomorphisms. For any $y \in S$, since f is surjective, there exists $x \in R$ such that $f(x)=y$. Then $y^5 = (f(x))^5 = f(x^5) = f(x) = y$. Thus, S also satisfies the property $y^5=y$ for all $y \in S$.
Let I be any prime ideal in a ring T where $t^5=t$ for all $t \in T$. Consider the quotient ring T/I. Since I is prime, T/I is an integral domain. For any element $\bar{t} \in T/I$, we have $\bar{t}^5 = \bar{t}$. Rearranging gives $\bar{t}^5 - \bar{t} = 0$, or $\bar{t}(\bar{t}^4 - 1) = 0$. Since T/I is an integral domain, either $\bar{t} = 0$ or $\bar{t}^4 - 1 = 0$, which means $\bar{t}^4 = 1$. If $\bar{t} \ne 0$, then $\bar{t}^4 = 1$, which implies $\bar{t}$ is a unit in T/I (its inverse is $\bar{t}^3$). Therefore, every non-zero element in T/I is a unit. This means T/I is a field. Since T/I is a field, the ideal I must be maximal in T.
Since Q is a non-zero prime ideal in S, and S is a ring where $y^5=y$ for all $y \in S$, Q must be a maximal ideal in S.
Since Q is a maximal ideal in S, and $f^{-1}(Q)$ is maximal in R if and only if Q is maximal in S, f-1(Q) is a maximal ideal in R.
Statement 4 is true.
Based on the analysis, all four statements are true.
Important Questions from Rings & Ideals
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