A rule for Ring and ideal

Question

Let $R$ be a ring, $I\in R$ an ideal and $f\in R[X]$ with $a\in R$. How do I show that in $R/I$:

$$f(a+I)=f(a)+I$$

Try showing that $f(a+i) - f(a) \in I$ for $i\in I$. You will need to use the fact that $I$ is an ideal. It's not hard to argue, but you might need an inductive argument. Here's what it would be for a quadratic function: $f(x) = x^2$, then $f(a+i) = (a+i)^2 = a^2 + ai + ia + i^2 = f(a) + i'$ for some $i'\in I$. I'm sure there's some isomorphism type theorem that argues this well, but I always think of algebra at its basest level arguments for whatever reason. — Cameron Williams, Jan 30 '20 at 13:51
Write $f(a+x)$ in powers of $x$. This is, Taylor expansion at $a$, $f(a+x)=f(a)+f'(a)x+\frac{f''(a)}{2!}x^2+...+\frac{f^{(n)}(a)}{n!}x^n$ for some $n$. Then, if $x\in I$ all the part $f'(a)x+\frac{f''(a)}{2!}x^2+...+\frac{f^{(n)}(a)}{n!}x^n$ belongs to $I$. — OscarRascal, Jan 30 '20 at 13:56
@peterag: it is not stated but I think that we can assume that. — KingDingeling, Jan 30 '20 at 14:29
@OscarRascal e.g., in char $2$, look at the poly $x^2$. Then it is true that $(a+x)^2 = a^2 + x^2$, but your "Taylor series" expansion is problematic - $2! = 0$ in characteristic zero, so you can't divide by it - of course, $f^{(1)}(a)=f^{(2)}(a)$ (formal definition on polys, obviously - no limits ) also is equal to zero. — peter a g, Jan 30 '20 at 14:40
Note that this is the quotient-ring ideal-form of the Polynomial Congruence Rule. It is hekpful to keep in mind the more general congruence analogs when learning such topics (see here for more on the relationship between congruences and ideals for general algebraic structures). — Bill Dubuque, Jan 30 '20 at 17:34

score 2 · Accepted Answer · edited Feb 01 '20 at 15:55

It's a little weird to treat $f$, a formal polynomial with coefficients in $R$, as if it were a function, but let's pretend that formal polynomials implicitly invoke the evaluation map for this problem.

For some $d \geq 0$, there are $r_0, \dots r_d \in R$ such that $$ f = r_0 + \sum_{i=1}^d r_i X^i \text{.} $$ (You don't say that your rings have unit, so we can't assume that $R$ has a $1$ that $(a+I)^0$ can be, so we must split the constant term out of the sum.) Then, implicitly use the evaluation map to treat $f$ as also an element of $\{ R[X] \times R \rightarrow R \}$, in fact, we need the generalization to sets as inputs and outputs, so an element of $\{ R[X] \times \mathscr{P}(R) \rightarrow \mathscr{P}(R) \}$, where $\mathscr{P}$ denotes powerset. Then $$ f(a + I) = r_0 + \sum_{i=1}^d r_i (a+I)^i \text{,} $$ a subset of $R$. Clearly, $(a+I)^1 = a+I = a^1+I$. Suppose, for $n \geq 1$, $(a+I)^n = a^n + I$. Then (being careful about order of multiplication since you have not said your rings are commutative), \begin{align*} (a+I)^{n+1} &= (a+I)(a+I)^n \\ &= (a+I)(a^n + I) \\ &= a(a^n + I) + I(a^n + I) \\ &= a^{n+1} + aI + Ia^n + I^2 \\ &\subseteq a^{n+1} + I \text{,} \end{align*} where in the last step we have used that $I$ is a (since you have not said otherwise, two-sided) ideal to recognize $aI \subseteq I$, $Ia^n \subseteq I$, and $I^2 \subseteq I$. So, \begin{align*} f(a + I) &\subseteq r_0 + \sum_{i=1}^d r_i (a^i+I) \\ &\subseteq r_0 + \sum_{i=1}^d (r_ia^i + I) \\ &= r_0 + \sum_{i=1}^d (r_ia^i) + \sum_{i=1}^d I \\ &= f(a) + \sum_{i=1}^d I \text{.} \end{align*} Since an ideal is closed under addition, $\sum_{i=1}^d I \subseteq I$. Therefore, we have the relation $f(a+I) \subseteq f(a) + I$ among subsets of $R$. Under the usual projection $R \rightarrow R/I$, since the right-hand side is contained in the coset of $I$ containing $f(a)$, we have $$ f(a+I) = f(a) + I$$ in $R/I$.

A rule for Ring and ideal

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