# $A\subseteq B$ integral domains with surjective multiplication, then the localization by all monic polynomials evaluated at some point is nonzero

Let $A\subseteq B$ be two integral domains such that the multiplication function $m: A \times (B \setminus A) \to B$, $m(x,y)=xy$, is surjective.

Let $S \subset A[x]$ be the set of all monic polynomials. For each $b \in B$, let ${v}_b: A[x] \to B$ be the evaluation homomorphism, given by $v_b(f) = f(b)$.

Prove that there exists $b_0 \in B$ such that $({v}_{b_0}(S))^{-1}B \neq \{0\}$.

We look for some $b_0,b \in B, f \in S$ such that $\frac{b}{f(b_0)} \neq \frac{0}{1}$. Thus we look for some elements $b_0, b \in B$ such that for all $g \in S$, $b\cdot g(b_0)\neq 0$ in $B$. Since $B$ is an integral domain, this is equivalent to requiring that both $b, g(b_0)$ are nonzero. Hence the choice of $b$ doesn't seem to matter, and we only want to find a suitable $b_0 \in B$ that is not the root of any monic polynomial in $A[x]$.

Clearly we must have $b_0 \in B \setminus A$, since otherwise we can take $g = x-b_0 \in S$ and get $b\cdot g(b_0) = 0$ no matter how we choose $b$.

We didn't use our given information that $m$ is surjective and $A$ is an integral domain. The surjectivity implies that there are $x \in A, y \in B \setminus A$, with $xy=1$. Thus $x,y$ are units in $B$. Which seems to be a dead-end.

• What does $v_b(S)^{-1}$ mean? – Daniel McLaury Jan 21 '16 at 22:24
• Image as a function, then taking a localization. – Emolga Jan 21 '16 at 22:25
• If there is no such $b_0$, then $B$ is an integral extension of $A$. Maybe there are some results from commutative ring theory which you can apply, e.g. the lying over theorem. – D_S Jan 21 '16 at 22:40
• @Leullame I suppose that $x\in B$ was a typo, so I've corrected it. – user26857 Jan 22 '16 at 20:37

Let $A\subseteq B$ be a ring extension such that the multiplication function $m: A \times (B \setminus A) \to B$, $m(x,y)=xy$, is surjective. Show that $A\subseteq B$ is not integral.
Let me start form here: The surjectivity implies that there are $x \in A, y \in B \setminus A$, with $xy=1$.
Suppose that $A\subset B$ is an integral extension. Then $y$ is integral over $A$ and write $$y^n+a_1y^{n-1}+\cdots+a_n=0,$$ where $a_i\in A$. Multiply this by $x^n$ and find $1+a_1x+\cdots+a_nx^n=0$, so $x$ is invertible in $A$ hence $y\in A$, a contradiction.