Q:

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

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?

A:

$\textit{proof.}$
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.

