Polynomial Ring modulo Ideal is Polynomial ring of cosets of indeterminates.

Question

I wonder if for any arbitrary ideal $I \leq K[X_1,\ldots,X_n]$, the following is true: $$ K[X_1,\ldots,X_n] \text{ mod } I = K[X_1 + I, \ldots, X_n +I].$$

If so, how can one show that?

Answer 1

$\textit{proof.}$ Remember the multi index notation, i.e. $x := (x_1, \ldots, x_n)$ and for $\alpha \in \mathbb{N}^n$ it is $$ x^\alpha = x_1^{\alpha_1}\cdot \ldots\cdot x_n^{\alpha_n}, $$ and $$ | \alpha| := \sum_{i=0}^n \alpha_i .$$ Define $x_i := X_i + I$. Then \begin{align*} & f \in K[x_1, \ldots, x_n]\\ \iff & f = \sum_{|\alpha| \leq n} a_\alpha x^\alpha = \sum_{|\alpha| \leq n} a_\alpha x_1^{\alpha_1}\cdot \ldots \cdot x_n^{\alpha_n}\\ \iff & f = \sum_{|\alpha| \leq n} a_\alpha (X_1 + I)^{\alpha_1}\cdot \ldots \cdot (X_n + I)^{\alpha_n}\\ \iff & f = \sum_{|\alpha| \leq n} a_\alpha (X_1^{\alpha_1} + I)\cdot \ldots \cdot (X_n^{\alpha_n} +I)\\ \iff & f = \sum_{|\alpha| \leq n} a_\alpha X_1^{\alpha_1}\cdot \ldots \cdot X_n^{\alpha_n} + I\\ \iff &f \in K[X_1, \ldots, X_n] \text{ mod } I \end{align*} $(X_i + I)(X_j + I) = (X_iX_j + I)$ by definition of multiplication of equivalence classes.