Let $R$ be a commutative unital ring and $A$ an associative (unital) $R$-algebra.
What is an augmented $R$-algebra? A (unital) $R$-algebra $A$, together with a (unital) ring morphism $\varepsilon: A\rightarrow R$? What is a morphism of augmented algebras $(A,\varepsilon)$ and $(A',\varepsilon')$? An algebra morphism $f:A\rightarrow A'$ with $\varepsilon'f=\varepsilon$? What is the associated graded algebra of $I=\ker\epsilon$? The module $\bigoplus_{n\in\mathbb{N}}I^n/I^{n+1}$?
I'm trying to understand the following paragraph from Toric Topology and Invariance of SR rings:
Question: If $I\unlhd K[\mathbf{x}]$ and $J\unlhd K[\mathbf{y}]$ are square-free monomial ideals and $f\!:K[\mathbf{x}]/I\rightarrow K[\mathbf{y}]/J$ an isomorphism of $K$-algebras, how can $f$ be modified to satisfy $\epsilon'f=\epsilon$ and become a graded morphism?
From what I understand, augmentation is $\epsilon\!:k[\Delta]\rightarrow k$ that sends $x_i\!\mapsto\!0$, so $\ker\epsilon=\langle x_1,\ldots,x_n\rangle$.
Thanks to YACP, I was eventually able to figure things out. Mentioning augmentations is here not really necessary.
Let $f\!:K[x]/I\!\rightarrow\!K[y]/J$ be an isomorphism of $K$-algebras. Let $\mathfrak{m}\!=\!\langle x_1,\ldots,x_n\rangle\!\unlhd\!K[x]$ and $\mathfrak{m}'\!=\!\langle y_1,\ldots,y_{n'}\rangle\!\unlhd\!K[y]$. We wish to change $f$ into an isomorphism that satisfies $f(\mathfrak{m})\!\subseteq\!\mathfrak{m}'$, i.e. $f(x_i)\!\in\!\mathfrak{m}'$ for every $i$. If $x_i$ is a zero-divisor in $K[x]/I$, then $f(x_i)\!=\!\alpha\!+\!y_1p_1\!+\!\ldots\!+\!y_{n'}p_{n'}$ is a zero-divisor in $K[y]/J$, so $\alpha q\!+\!y_1p_1q\!+\!\ldots\!+\!y_{n'}p_{n'}q\!\in\!J$ for some $q\!\in\!K[y]\!\setminus\!J$, but since $\alpha q$ and $y_jp_jq$ share no monomials, by a) we have $\alpha q\!\in\!J$, which implies $\alpha\!=\!0$, i.e. $f(x_i)\!\in\!\mathfrak{m}'$. If $x_i$ is a non-zero-divisor in $K[x]/I$, then $\forall\mathbf{a}\!: x^\mathbf{a}\!\in\!K[x]\!\setminus\!I \Rightarrow x_ix^\mathbf{a}\!\in\!K[x]\!\setminus\!I$, hence $\forall\mathbf{a}\!: x_ix^\mathbf{a}\!\in\!I \Rightarrow x^\mathbf{a}\!\in\!I$, meaning that w.l.o.g. every generator of $I$ does not contain $x_i$. Let $x_1,\ldots,x_k$ be all the non-zero-divisors among $x_i$ (possibly after renumbering the variables) and $\alpha_1,\ldots,\alpha_n\!\in\!K$ the constant terms of $f(x_1),\ldots,f(x_n)$. Then the isomorphism $\phi\!:K[x]\!\rightarrow\!K[x]$ that sends $x_i\!\mapsto\!x_i\!-\!\alpha_i$ satisfies $\phi(I)\!=\!J$, because every generator of $I$ is not divisible by $x_1,\ldots,x_k$, and $\alpha_{k+1},\ldots,\alpha_n\!=\!0$, so $\phi|_{\text{generators}}\!=\!id$. Thus $\phi$ induces an isomorphism $K[x]/I\!\rightarrow\!K[x]/I$ with $f\phi(\mathfrak{m})\!\subseteq\!\mathfrak{m}'$.
Every ideal $I$ of an $R$-algebra $A$ induces a $\mathbb{N}$-graded $R$-algebra $Gr_IA\!=\! \frac{A}{I}\!\oplus\! \frac{I}{I^2}\!\oplus\! \frac{I^2}{I^3}\!\oplus\!\ldots$ via $(a_k\!+\!I^{k+1})_k\!+\!(a'_k\!+\!I^{k+1})_k\!=\!(a_k\!+\!a'_k\!+\!I^{k+1})$ and $(a_k\!+\!I^{k+1})_k\!\cdot(a'_k\!+\!I^{k+1})_k\!=\!(a_ka'_k\!+\!I^{k+1})$ and $r(a_k\!+\!I^{k+1})\!=\!(ra_k\!+\!I^{k+1})$. Every $R$-algebra morphism $f\!:A\!\rightarrow\!B$ with $f(I)\!\subseteq\!J$ satisfies $f(I^k)\!\subseteq\!J^k$ and hence induces an $R$-algebra morphism $Gr(f)\!:Gr_IA\!\rightarrow\!Gr_JB$ that sends $(a_k\!+\!I^{k+1})_k\!\mapsto\! (f(a_k)\!+\!J^{k+1})_k$, with $Gr(id_A)\!=\!id_{Gr A}$ and $Gr(gf)\!=\!Gr(g)Gr(f)$. In case $R\!=\!K$ and $A\!=\!K[x]/I$ and $I\!=\!\mathfrak{m}/I$, we have $\frac{\mathfrak{m}^k\!/I}{\mathfrak{m}^{k+1}\!/I}\!=\! \{\text{polynomials in }K[x]/I\text{ of degree }k\}$, so there holds $Gr_\mathfrak{m}K[x]/I \cong K[x]/I$ as $K$-algebras. In this way we obtain an isomorphism of $\mathbb{N}$-graded $K$-algebras $Gr(f\phi)\!: K[x]/I\!\rightarrow\!K[y]/J$. $\blacksquare$