Source For H D R 0 ( A ) = K H^0_{dr}(A) = K H D R 0 ​ ( A ) = K For F.g. Smooth K K K -algebra

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Introduction

In the realm of abstract algebra and algebraic geometry, the study of relative differentials and universal derivations plays a crucial role in understanding the properties of $K$-algebras. Given a field $K$ and a $K$-algebra $A$, we are interested in determining the sufficient conditions for the equality $H^0_{dr}(A) = K$, where $H^0_{dr}(A)$ denotes the zeroth de Rham cohomology group of $A$. In this article, we will delve into the world of smooth $K$-algebras and explore the source for this equality.

Background and Notation

Let $K$ be a field and $A$ a $K$-algebra. We denote the relative differentials of $A$ over $K$ by $\Omega_{A/K}$, and the universal derivation by $d \colon A \to \Omega_{A/K}$. The de Rham cohomology groups of $A$ are defined as the cohomology groups of the complex $(\Omega_{A/K}, d)$. Specifically, we are interested in the zeroth de Rham cohomology group, denoted by $H^0_{dr}(A)$.

Smooth $K$-Algebras

A $K$-algebra $A$ is said to be smooth if it is finitely generated and has a regular derivation $d \colon A \to \Omega_{A/K}$. In other words, $A$ is smooth if it has a universal derivation that is regular. This condition is crucial in ensuring that the relative differentials $\Omega_{A/K}$ are a free $A$-module of finite rank.

The Equality $H^0_{dr}(A) = K$

We are now ready to state the main result of this article. Let $A$ be a finitely generated smooth $K$-algebra. Then, we have the following equality:

$$H^0_{dr}(A) = K$$

This result is a direct consequence of the fact that the relative differentials $\Omega_{A/K}$ are a free $A$-module of finite rank. Specifically, we have the following short exact sequence:

$$0 \to K \to \Omega_{A/K} \to \Omega_{A/K}/K \to 0$$

where $K$ is embedded in $\Omega_{A/K}$ via the universal derivation $d \colon A \to \Omega_{A/K}$. Taking the cohomology groups of this sequence, we obtain the following long exact sequence:

$$\cdots \to H^0_{dr}(A) \to H^0_{dr}(\Omega_{A/K}/K) \to H^1_{dr}(A) \to \cdots$$

Since $A$ is smooth, we have $H^1_{dr}(A) = 0$. Therefore, the long exact sequence reduces to:

$$\cdots \to H^0_{dr}(A) \to H^0_{dr}(\Omega_{A/K}/K) \to 0 \to \cdots$$

This implies that $H^0_{dr}(A) = K$, as desired.

Proof of the Result

We now provide a proof of the main result. Let $A$ be a finitely generated smooth $K$-algebra. We need to show that $H^0_{dr}(A) = K$. To do this, we will use the fact that the relative differentials $\Omega_{A/K}$ are a free $A$-module of finite rank.

Step 1: Embedding $K$ in $\Omega_{A/K}$

We start by embedding $K$ in $\Omega_{A/K}$ via the universal derivation $d \colon A \to \Omega_{A/K}$. Specifically, we define a map $\phi \colon K \to \Omega_{A/K}$ by $\phi(k) = d(k)$ for all $k \in K$. This map is clearly a $K$-algebra homomorphism.

Step 2: Showing that $\phi$ is injective

We now show that the map $\phi$ is injective. Suppose that $\phi(k) = 0$ for some $k \in K$. Then, we have $d(k) = 0$, which implies that $k = 0$. Therefore, $\phi$ is injective.

Step 3: Showing that $\phi$ is surjective

We now show that the map $\phi$ is surjective. Let $\omega \in \Omega_{A/K}$. We need to show that there exists $k \in K$ such that $\phi(k) = \omega$. Since $\Omega_{A/K}$ is a free $A$-module of finite rank, we can write $\omega = \sum_{i=1}^n a_i \omega_i$ for some $a_i \in A$ and $\omega_i \in \Omega_{A/K}$. We can now define a map $\psi \colon K \to \Omega_{A/K}$ by $\psi(k) = \sum_{i=1}^n a_i d(k)$ for all $k \in K$. This map is clearly a $K$-algebra homomorphism. Moreover, we have $\psi(k) = \omega$ for all $k \in K$. Therefore, $\phi$ is surjective.

Step 4: Conclusion

We have now shown that the map $\phi$ is both injective and surjective. Therefore, $\phi$ is an isomorphism of $K$-algebras. This implies that $H^0_{dr}(A) = K$, as desired.

Conclusion

In this article, we have shown that the zeroth de Rham cohomology group of a finitely generated smooth $K$-algebra $A$ is equal to $K$. This result is a direct consequence of the fact that the relative differentials $\Omega_{A/K}$ are a free $A$-module of finite rank. We hope that this result will be useful in the study of abstract algebra and algebraic geometry.

References

• [1] Bourbaki, N. (1959). Algebra I. Hermann.
• [2] Cartan, H. (1958). Cours d'algèbre. Hermann.
• [3] Serre, J.-P. (1959). Cohomologie galoisienne. Springer-Verlag.

Appendix

In this appendix we provide some additional information on the de Rham cohomology groups of $K$-algebras. Specifically, we show that the de Rham cohomology groups of a finitely generated smooth $K$-algebra $A$ are isomorphic to the cohomology groups of the complex $(\Omega_{A/K}, d)$.

Step 1: Defining the complex $(\Omega_{A/K}, d)$

We start by defining the complex $(\Omega_{A/K}, d)$. Specifically, we define a differential $d \colon \Omega_{A/K} \to \Omega_{A/K}$ by $d(\omega) = d(\omega)$ for all $\omega \in \Omega_{A/K}$. This differential is clearly a $K$-algebra homomorphism.

Step 2: Showing that the complex $(\Omega_{A/K}, d)$ is exact

We now show that the complex $(\Omega_{A/K}, d)$ is exact. Specifically, we need to show that the kernel of the differential $d$ is equal to the image of the differential $d$. Let $\omega \in \Omega_{A/K}$ be such that $d(\omega) = 0$. Then, we have $d(\omega) = 0$, which implies that $\omega = 0$. Therefore, the kernel of the differential $d$ is equal to the zero ideal. On the other hand, let $\omega \in \Omega_{A/K}$ be such that $\omega = d(\omega')$ for some $\omega' \in \Omega_{A/K}$. Then, we have $d(\omega) = d(d(\omega')) = 0$, which implies that $\omega = 0$. Therefore, the image of the differential $d$ is equal to the zero ideal. This shows that the complex $(\Omega_{A/K}, d)$ is exact.

Step 3: Conclusion

We have now shown that the complex $(\Omega_{A/K}, d)$ is exact. This implies that the de Rham cohomology groups of a finitely generated smooth $K$-algebra $A$ are isomorphic to the cohomology groups of the complex $(\Omega_{A/K}, d)$.

References

• [1] Bourbaki, N. (1959). Algebra I. Hermann.
• [2] Cartan, H. (1958). Cours d'algèbre. Hermann.
• [3] Serre, J.-P. (1959). Cohomologie galoisienne. Springer-Verlag.
  Q&A: Source for $H^0_{dr}(A) = K$ for f.g. smooth $K$-algebra ===========================================================

Introduction

In our previous article, we explored the source for the equality $H^0_{dr}(A) = K$ for finitely generated smooth $K$-algebras. In this article, we will answer some of the most frequently asked questions related to this topic.

Q: What is the significance of the equality $H^0_{dr}(A) = K$?

A: The equality $H^0_{dr}(A) = K$ has significant implications in the study of abstract algebra and algebraic geometry. It provides a connection between the de Rham cohomology groups of a $K$-algebra and the field $K$ itself.

Q: What is the relationship between the relative differentials $\Omega_{A/K}$ and the universal derivation $d \colon A \to \Omega_{A/K}$?

A: The relative differentials $\Omega_{A/K}$ are a free $A$-module of finite rank, and the universal derivation $d \colon A \to \Omega_{A/K}$ is a regular derivation. This means that the relative differentials $\Omega_{A/K}$ are a module over the $K$-algebra $A$, and the universal derivation $d$ is a homomorphism from $A$ to $\Omega_{A/K}$.

Q: How does the smoothness of the $K$-algebra $A$ affect the de Rham cohomology groups?

A: The smoothness of the $K$-algebra $A$ ensures that the relative differentials $\Omega_{A/K}$ are a free $A$-module of finite rank. This, in turn, implies that the de Rham cohomology groups of $A$ are isomorphic to the cohomology groups of the complex $(\Omega_{A/K}, d)$.

Q: Can you provide an example of a finitely generated smooth $K$-algebra $A$ for which $H^0_{dr}(A) = K$?

A: Yes, consider the polynomial ring $A = K[x_1, \ldots, x_n]$. This is a finitely generated smooth $K$-algebra, and we have $H^0_{dr}(A) = K$.

Q: How does the equality $H^0_{dr}(A) = K$ relate to the study of algebraic geometry?

A: The equality $H^0_{dr}(A) = K$ has significant implications in the study of algebraic geometry. It provides a connection between the de Rham cohomology groups of a $K$-algebra and the field $K$ itself, which is essential in the study of algebraic geometry.

Q: Can you provide a proof of the result $H^0_{dr}(A) = K$ for a finitely generated smooth $K$-algebra $A$?

A: Yes, we can provide a proof of the result $H^0_{dr}(A) = K$ for a finitely generated smooth $K$-algebra $A$. The involves showing that the relative differentials $\Omega_{A/K}$ are a free $A$-module of finite rank, and then using this fact to show that the de Rham cohomology groups of $A$ are isomorphic to the cohomology groups of the complex $(\Omega_{A/K}, d)$.

Q: What are some of the applications of the result $H^0_{dr}(A) = K$?

A: The result $H^0_{dr}(A) = K$ has significant applications in the study of abstract algebra and algebraic geometry. It provides a connection between the de Rham cohomology groups of a $K$-algebra and the field $K$ itself, which is essential in the study of algebraic geometry.

Conclusion

In this article, we have answered some of the most frequently asked questions related to the source for the equality $H^0_{dr}(A) = K$ for finitely generated smooth $K$-algebras. We hope that this article has provided a useful resource for those interested in the study of abstract algebra and algebraic geometry.

References

• [1] Bourbaki, N. (1959). Algebra I. Hermann.
• [2] Cartan, H. (1958). Cours d'algèbre. Hermann.
• [3] Serre, J.-P. (1959). Cohomologie galoisienne. Springer-Verlag.

Appendix

In this appendix, we provide some additional information on the de Rham cohomology groups of $K$-algebras. Specifically, we show that the de Rham cohomology groups of a finitely generated smooth $K$-algebra $A$ are isomorphic to the cohomology groups of the complex $(\Omega_{A/K}, d)$.

Step 1: Defining the complex $(\Omega_{A/K}, d)$

We start by defining the complex $(\Omega_{A/K}, d)$. Specifically, we define a differential $d \colon \Omega_{A/K} \to \Omega_{A/K}$ by $d(\omega) = d(\omega)$ for all $\omega \in \Omega_{A/K}$. This differential is clearly a $K$-algebra homomorphism.

Step 2: Showing that the complex $(\Omega_{A/K}, d)$ is exact

We now show that the complex $(\Omega_{A/K}, d)$ is exact. Specifically, we need to show that the kernel of the differential $d$ is equal to the image of the differential $d$. Let $\omega \in \Omega_{A/K}$ be such that $d(\omega) = 0$. Then, we have $d(\omega) = 0$, which implies that $\omega = 0$. Therefore, the kernel of the differential $d$ is equal to the zero ideal. On the other hand, let $\omega \in \Omega_{A/K}$ be such that $\omega = d(\omega')$ for some $\omega' \in \Omega_{A/K}$. Then, we have $d(\omega) = d(d(\omega')) = 0$, which implies that $\omega = 0$. Therefore, the image of the differential $d$ is equal the zero ideal. This shows that the complex $(\Omega_{A/K}, d)$ is exact.

Step 3: Conclusion

We have now shown that the complex $(\Omega_{A/K}, d)$ is exact. This implies that the de Rham cohomology groups of a finitely generated smooth $K$-algebra $A$ are isomorphic to the cohomology groups of the complex $(\Omega_{A/K}, d)$.

References

• [1] Bourbaki, N. (1959). Algebra I. Hermann.
• [2] Cartan, H. (1958). Cours d'algèbre. Hermann.
• [3] Serre, J.-P. (1959). Cohomologie galoisienne. Springer-Verlag.