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Consider the ring homomorphism $\phi: A[t]\to A^A$ where $A$ is an integral domain and $A^A$ is the ring of all functions on $A$ with values in $A$. Determine the kernel of $\phi$. Be as explicit as possible.

I reasoned that if $f$ has $n$ distinct roots, that is, if every element in $A$ is a root of $f$, then $f\in\ker\phi$. However, my professor disagreed with me and asked me to be more explicit. What am I missing?

chris
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  • There isn't enough information to answer the question as stated. There is more than one homomorphism from $A[t]$ to $A^A$. For example, the zero map and the obvious injection are two such homomorphisms, but they have very different kernels! –  Mar 15 '12 at 15:50
  • @Jack: Presumably the homomorphism is the injection ("plugging in for $t$"), right? – anon Mar 15 '12 at 15:54
  • @anon - If so, then the kernel is very easy to find. But it's not clear to me that this is the homomorphism to which the OP was referring. –  Mar 15 '12 at 15:58
  • @Jack: Considering OP's line of reasoning, and the fact the professor's diagnosis was to "be more explicit," I think it's a safe bet. – anon Mar 15 '12 at 16:01
  • Your answer $$ f(X) \mapsto 0 \Leftrightarrow f(a)=0 \text{ for all } a \in A $$ is correct (given $t\mapsto \operatorname{id}$), but this is just a reformulation of what the kernel is! Hardly explicit :-) – Blah Mar 15 '12 at 17:27

2 Answers2

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Hint $\ $ I presume you mean that $\rm\:t\:$ maps to the identity function. Suppose $\rm\:f(t)\:$ maps to a zero function, i.e. $\rm\:f(A) = 0.\:$ If $\rm\:A\:$ is infinite then $\rm\:f = 0\:$ since a polynomial $\ne 0$ over a domain has no more roots than its degree. If $\rm\:A = \{a_1,\ldots,a_n\}$ is finite, then by the Factor Theorem

$$\rm f(A)=0\iff\forall\: i:\ t\!-\!a_i\ |\ f(t)\iff lcm(t\!-\!a_i) = (t\!-\!a_1)\cdots(t\!-\!a_n)\ |\ f(t),\:$$

Thus the kernel $\rm\:I\:$ is generated by $\rm\:(t-a_1)\cdots(t-a_n).\:$ But $\rm\:t^n - t\in I\:$ since $\rm\:a^{n-1} = 1\:$ for $\rm\:a\ne 0,\:$ by applying Lagrange's theorem to the multiplicative group of the finite field $\rm\:A.\:$ So both polynomials are equal (else their difference is a polynomial $\ne 0$ of degree $\rm < n$ with $\rm\:n\:$ roots).

Community
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Bill Dubuque
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Your question isn't so clear so I'll try to answer in the more coplete way.

There are two cases to consider:

  1. $A$ is a finite integral domain;
  2. $A$ is an infinite integral domain.

In case 1. $A$ is a field, this can easily be proven observing that for every $a \in A$ because $A$ is an integral domain the map $$x \in A \mapsto a x \in A$$ is a injective endo-function on a finite set, so bijective. In this case clearly we know that for every polinomial $f \in A[x]$ is such that $\varphi(f)=0$ if and only if for every $a \in A$ we have $f(a)=0$ (or more correctely $\phi(f)(a)=0$). This, for the polynomial remainder theorem, imply that for each $a \in A$ have to be $(x-a) \mid f$ and so $\prod_{a \in A}(x-a) \mid f$ (because the polynomial $x-a$ are coprime).

Let's pass to the more interesting case, the case 2: $A$ is infinite. Let $K$ be the field of fractions of $A$, clearly $K$ is infinite. If $f \in A[x]$ is such that $\phi(f)=0$ this implies that for every $a \in A$ we have that $(x-a) \mid f$ as polynomial in $K[x]$, but from this would follow that $f$ should have infinite degree, that's possible if and only if $f=0$ in $K[x]$ and so in $A[x]$ too.

Giorgio Mossa
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