Example of prime ideal pulling back to non-prime.

Question

In non-commutative rings, is it possible for a homomorphism of a prime ideal not to pullback to a prime ideal? A question on my homework seems to suggest it is possible. I'm having a difficult time thinking of ideals at all in non-commutative rings.

Specifically I seek $R$, $S$ unital rings $f:R\rightarrow S$ ring homomorphism s.t $P$ is prime in $S$ but $f^{-1}(P)$ is not prime in $R$

Answer 1

For arbitrary unital morphism, this is possible. Let $S$ be the ring of $2\times 2$ matrices with real coefficients; the only ideals are $(0)$ and $S$, with the former prime.

Let $R$ be the subring consisting of matrices of the form $$\left(\begin{array}{cc} * & *\\ 0 & * \end{array}\right)$$ (that is, the $(2,1)$ entry is zero). Let $f\colon R\to S$ be the inclusion. These rings are unital, and the function is unital.

I take the definition of “prime ideal” to be the usual one for noncommutative rings: $P$ is prime if and only if $P$ is not the whole ring, and if $A$ and $B$ are ideals such that $AB\subseteq P$, then either $A\subseteq P$ or $B\subseteq P$.

Among the ideals in $R$ is the ideal $M$ that consists of all matrices of the form $$\left(\begin{array}{cc} 0 & *\\ 0 & 0 \end{array}\right).$$ Note that the product of two such matrices is the zero matrix. So $M^2=(0)$ (since $M^2$ consists of finite sums of products of two elements of $M$).

The pullback of the prime ideal $(0)$ of $S$ is the zero ideal in $R$. But while $(0)$ is prime in $S$, it is not prime in $R$, since $M^2\subseteq (0)$ but $M\not\subseteq (0)$.

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If you require your $f$ to be surjective and unital, the pullback of a prime ideal is a prime ideal. Say $f$ is surjective, and let $N$ be the kernel of $f$.

Say $P\triangleleft S$ is prime; note that $1\notin S$, so $Q=f^{-1}(P)$ does not contain $1$, hence is a proper ideal. Note that $N\subseteq Q$.

Let $A$ and $B$ be ideals of $R$ such that $AB\subseteq Q$. The ideals $A+N$ and $B+N$ contain $N$, and moreover, $$(A+N)(B+N)\subseteq AB + NB + AN + N^2\subseteq AB+N\subseteq Q+N \subseteq Q.$$ Now, by the lattice isomorphism, $A+N$ and $B+N$ correspond to ideals $I$ and $J$ of $S$, and $$IJ = f(A+N)f(B+N)=f((A+N)(B+N))\subseteq f(Q)=P,$$ hence either $I\subseteq P$ or $J\subseteq P$; hence $A\subseteq A+N=f^{-1}(I)\subseteq f^{-1}(P)=Q$ or $B\subseteq B+N=f^{-1}(J)\subseteq f^{-1}(P)=Q$.

Note that the reason this argument fails in the example above is that the image of $M$ is not an ideal, and the smallest ideal of $S$ that contains $M$ is the whole ring. Thus, $f(M^2)=f(M)f(M)$ is no longer a product of ideals contained in $(0)$, so you can’t argue that $f(M)$ must be contained in $0$.

Answer 2

It is not possible. We assume that $f$ is surjective (otherwise the image of an ideal may not be an ideal). If $(a)(b)\subseteq f^{-1}(P)$ then $f((a)(b))=(f(a))(f(b))\subseteq P$, then $(f(a))$ or $(f(b))$ is inside $P$, so $(a)$ or $(b)$ is inside $f^{-1}(P)$.