By Hermann Stahl

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**Extra info for Abriss einer Theorie der algebraischen Funktionen einer Veraenderlichen**

**Example text**

1A /. We show that the unit property of satisfies the counit property. For f 2 Aı , r 2 K, a 2 A, . r ˝ a/ D . f /. mA . r ˝ a/: The reader should note that we have applied the unit property of line 3 to line 4 above. Thus . Aı ; Aı ; coalgebra. A0 / is a If A is a K-algebra that is finite dimensional as a K vector space, then Aı D A and A is a K-coalgebra. 9). Let S be a finite monoid, jSj D m, and let KS be the monoid ring. The K-algebra KS is finite dimensional over K on the basis f1 D g0 ; g1 ; g2 ; : : : ; gm 1 g.

A is mA W A ! A ˝K A/ . A ˝K A/ . Hence, in the case that A is infinite dimensional, mA may not induce the required comultiplication map A ! A ˝K A . A / 6Â A ˝K A . 2. Let A D KŒx be the K-algebra of polynomials in x. Let mKŒx W KŒx ˝K KŒx ! KŒx be the multiplication map with transpose mKŒx W KŒx ! KŒx ˝K KŒx/ . Let fsn g be the sequence in K given as 1; 0; 0; 1; 1; 1; 0; 0; 0; 0; : : : (that is, one 1, followed by two 0’s, followed by three 1’s, and so on). xj / D ıi;j , i; j 0. f / 62 KŒx ˝K KŒx .

If A is finite dimensional as a K-vector space, then Aı D A . Proof. Exercise. 9). We need three propositions. 6. Let I be an ideal of A and let s W A ! A=I be the canonical surjection of vector spaces. A=I/ ! A=I/ , a 2 A. Then s is an injection. Proof. A=I/ . a//. a C I/, and so f D g. 7. Let f 2 Aı and suppose that f vanishes on the ideal I of A. f / D f . Proof. The element f 2 Aı Â A is a K-module homomorphism f W A ! f /. Let s W A ! A=I denote the canonical surjection. Then by the universal mapping property for kernels there exists a K-module homomorphism f W A=I !