$K$ is a finite field and $F$ is the set of all functions of $K$ in $K$. In $F$ we define $(f+g)(a)=f(a)+g(a)$ and $(fg)(a)=f(a)g(a)$.

Show that $F$ is isomorphic a $k[x]/I$ for a some ideal $I$ of $k[x]$ , and find this $I$.

Is correct my solution and $I$ is only this $\ker(\Phi)$? What influences $K$ be field and not just ring.

$\Phi$ is homomorphism , because $(f+g)(a)=f(a)+g(a)$ and $(fg)(a)=f(a)g(a)$.

if $k[x]/I$ is isomorphic a $\Phi$ => $\ker(\Phi)= I = \{p(x) \in k[x] \mid \Phi(p(x))=0\}$ and $\Phi$ is homomorphism

So $k[x]/I$ is isomorphic a $\Phi$ , and $I = \ker(\Phi)$

  • $\begingroup$ Your notation seems a little confused. Either $F$ is the set of functions, of $F$ is a homomorphism. Pick one, and call the other one $g$ or something. $\endgroup$ – Prahlad Vaidyanathan Oct 19 '13 at 12:07

Let $F = \{f : k \to k\}$, then, under the operations above, $F$ is a ring.

  1. Define $\varphi : k[x] \to F$ given by $\varphi(f) = f$, where the $f$ on the RHS is thought of as a function on $k$. This is clearly a homomorphism, so we would be done by the first isomorphism theorem if we could only show that it is surjective.

  2. Now write $F = \{x_0, x_1, \ldots, x_n\}$. Given $g \in F$, write $y_i = g(x_i)$, and consider the interpolating polynomial $$ p(x) = \sum_{i=0}^n y_i \prod_{j\neq i} \frac{x-x_j}{x_i - x_j} $$ Then, $p \in k[x]$ (Note : $p$ makes sense because $k$ is a field), and $$ \varphi(p)(x_i) = y_i = g(x_i) \quad\forall i \in \{0, 1,\ldots, n\} $$ Hence, $\varphi(p) = g$ (ie., as a function on $k$, $p = g$).

This completes the proof.

  • $\begingroup$ hello, because F={f:k→k} has (f+g)(a)=f(a)+g(a) and (fg)(a)=f(a)g(a) , F is ring, right? φ:k[x]→F given by φ(f)=f is ring homomorphism , so why i have show that it is surjective to use first isomorphism theorem ? In part 2 was proofed that φ is surjective? The Ideal I would be ker(φ)? sorry i´m learning... $\endgroup$ – andre Oct 19 '13 at 14:36
  • $\begingroup$ Note that the kernel of $\phi$ is exactly the ideal generated by $x^q - x$ where $q$ is the order of $F$. This is because $x^q = x$ for all $x \in F$ (since the group of units is a group of order $q-1$), so $x^q - x$ is in fact in the kernel. On the other hand, no polynomial of degree strictly less than $q$ can evaluate to 0 at every point unless it's the 0 polynomial, since otherwise it would have too many roots. Now, knowing that $F[x]$ is a PID completes the proof. $\endgroup$ – Daniel Schepler Oct 24 '17 at 1:12

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