Intuitive understanding of ideal $I = (x+1,x^2+1)$ and the quotient $\Bbb Z[x]/I$

Question

For a principal ideal, I have a good understanding of what it will look like in terms of making all multiples of the ideal $0$ (gluing them with zero). I don't on the other hand have any intuition of what $$\mathbb{Z}[x]/(x+1,x^{2}+1)$$ will look like, I believe it contains $\mathbb{Z}/(x+1)$ and $\mathbb{Z}/(x^{2}+1)$ but apart from that I don't know. I believe it’s $\mathbb{F}_{2}$.

Answer 1

Hint $\!\bmod I=(\color{#c00}{x\!-\!a},f(x),g(x),\ldots)\!:\ \color{#c00}{x\equiv a}\,\Rightarrow\, f(\color{#c00}x)\equiv f(\color{#c00}a),\, g(\color{#c00}x)\equiv g(\color{#c00}a),\,\ldots$

therefore: $\,\ I = (x\!-\!a,f(a),g(a),\ldots),\ $ where we used the Polynomial Congruence Rule.

So, in OP: $\, \ I = (x\!+\!1,\, x^2\!+1) = (x\!+\!1,\,\color{#0a0}2)\ $ since $\,f(x)=x^2+1\,\Rightarrow\,\color{#0a0}{f(-1) = 2}$

So $\,\Bbb Z[x]/I = \Bbb Z[x]/(x\!+\!1,2)\cong \Bbb Z[x]/(x\!+\!1)/((2,x\!+\!1)/(x\!+\!1) \cong \Bbb Z/2\,$ via Third Isom. theorem.

Remark $ $ Above is a sort of ideal form of the basic step in the Euclidean algorithm for the gcd, viz. $$(h,f,g,\ldots) = (h,\, f\bmod h,\, g\bmod h,\ldots)$$

i.e. we can mod out all the other generators by any generator while preserving the ideal. More generally ideals are preserved under any unimodular transformation of the generators, which may be viewed as an ideal form of a "change of basis". The Euclidean algorithm generalizes in various ways, e.g. to Hermite (or Smith) normal forms, and other standard basis algorithms e.g. Grobner bases.

Such standard bases often yield a more "intuitive understanding" of the ideal, being "simpler" in various ways, e.g. they may be in triangular form and/or be a module basis, which makes it clear how to use the basis as effective normal-form rewriting rules (e.g. see here), and may also make it easier to deduce properties of the quotient ring.

Answer 2

You can think of it as imposing several conditions on the element $x$: if $R=\mathbb{Z}[x]/I$ with $I=(x+1,x^2+1)$, then $R$ is generated by some element $x$ which has to satisfy the two relations $x+1=0$ and $x^2+1=0$.

So in this case, we must have $x=-1$ by the first relation, but also $x^2=-1$, so $1=(-1)^2=-1$, which means that indeed $R$ must have characteristic $2$, and $x=1$. In the end, $R$ is the ring of characteristic $2$ generated by an element which is actually $1$, so it is $\mathbb{F}_2$ as you suspected.

Note that $R$ does not contain $\mathbb{Z}[x]/(x+1)$ and $\mathbb{Z}[x]/(x^2+1)$, it is a quotient of each of those rings (for instance, $\mathbb{Z}[x]/(x+1)\simeq \mathbb{Z}$, you you see it cannot be a subring of $R\simeq \mathbb{F}_2$).