Find a "simpler description" for $\mathbb{Z}[X]/(X-5,X^2+3)$

Question

The problem asks for a "simpler description" of the ring $\mathbb{Z}[X]/(X-5,X^2+3)$.

I could use the Chinese Remainder Theorem if $\mathbb{Z}$ were replaced by $\mathbb{Q}$, but here the ideals $(X-5)$ and $(X^2+3)$ aren't comaximal.

In general, how should I go about solving a problem like this?

Answer 1

Since $x-5=0$ we can replace every $x$ by a $5$, and since $x^2+3=0$ we can replace every $28$ with $0$.

By composing the two projections from

$$\Bbb Z[x]\to \Bbb Z[x]/(x-5)=B\cong \Bbb Z$$

and the one from $B\to B/I$ where $I$ is the projection of $x^2+3$ in $B$ we get

$$B\to B/28 B$$

we see the image is $\Bbb Z/28\Bbb Z$.

If it helps to visualize, you can verify that $x^2+3=28+10(x-5)+(x-5)^2$, which makes it easier to see the isomorphism since you can rewrite the ideal as $(x-5,28)$ so that you are looking at

$$\Bbb Z[x]/(x-5,28)$$

and can also write the projection as a composition of

$$\Bbb Z[x]\to \left(\Bbb Z/28\Bbb Z\right)[x]\to\left(\Bbb Z/28\Bbb Z\right)[x]/(x-5)\cong\Bbb Z/28\Bbb Z$$

using the isomorphism theorems since it's trivial to see that $(28)\subseteq (x-5,28)$ and that the image of $(x-5, 28)$ in $\left(\Bbb Z/28\Bbb Z\right)[x]$ is $(x-5)$--here we mean this as an ideal of the ring $\left(\Bbb Z/28\Bbb Z\right)[x]$.

Answer 2

The existing answer is much appreciated. I nevertheless am compelled to write an answer which makes more sense in my own head. The quotient $\mathbb{Z}[X]/(X-5,X^2+3)$ is isomorphic to $$\frac{\mathbb{Z}[X]/(X-5)}{(X-5,X^2+3)/(X-5)}$$ The numerator is isomorphic to $\mathbb{Z}$, and a complete set of coset representatives (which are happily closed under addition and multiplication) is the set of all $\overline{f(5)}$ for $f \in \mathbb{Z}[X]$.

For the denominator, one can show that $(X-5,X^2+3) = (28,X-5)$, so $(28, X-5)/(X-5)$ is the set of all products $\overline{f(5)\cdot28}$. The given quotient is the set of all $\overline{f(5)}$ (cyclic, generated by $\overline{1}$), modded out by the set of all $\overline{f(5) \cdot 28}$ (cyclic, generated by $\overline{28}$). So the quotient is just $\cong \mathbb{Z}/28 \mathbb{Z}$.

Answer 3

$ {\rm mod}\ \color{#0a0}{x\!-\!5}\!:\ f(\color{#0a0}x)\equiv f(\color{#0a0}5)\,$ so $\,\color{#c00}{x^2\!+3\equiv 28},\ $ so $\,I = (\color{#0a0}{x\!-\!5},\color{#c00}{x^2\!+3}) = (\color{#0a0}{x\!-\!5},\color{#c00}{28})\ $ a la Euclid.

The natural map $\,h\,:\,\Bbb Z\to \Bbb Z[x]/I\,$ is onto by $\,f(x) \equiv f(5) \pmod I,\, $ by $\,x-5\in I$

$28\in I\Rightarrow\ker h\supseteq 28\Bbb Z.$ $\,n\in\ker h\,\Rightarrow n = (x\!-\!5)f(x)+28g(x)\overset{x\,=\,5}\Rightarrow n = 28 g(5)\,\Rightarrow\,\ker h \subseteq 28\Bbb Z$

Answer 4

In my opinion, the "right" way to think about this calculation is

$$ \mathbf{Z}[x] / (x-5, x^2 + 3) \cong \left( \mathbf{Z}[x] / (x-5) \right) / (x^2 + 3) \cong \mathbf{Z} / (5^2 + 3) = \mathbf{Z} / 28 $$

And the thing to do is to work out what each part of this literally means, and prove whatever lemmas we need to make it true.

In the second ring, there is a canonical way to view $x^2 + 3 \in \mathbf{Z}[x]$ as an element in the quotient $\mathbf{Z}[x] / (x-5)$: specifically as its image under the projection map.

So now that we know what the second ring means, we look at the first isomorphism. Maps from a quotient ring are determined by maps from the parent ring; we intend this map to be the one induced by the identity map on $\mathbf{Z}[x]$.

It's not hard to see the map does give a homomorphism on the quotient, so we need some way to convince ourselves it's an isomorphism. In fact, we can make a general assertion:

Theorem Let $I,J$ be ideals of a ring $R$, and let $\pi : R \to R/I$ be the projection map. Then the identity map on $R$ induces an isomorphism $R / (I+J) \cong (R/I) / \pi(J)$.

One proof amounts to observing that $\pi(I) \cup \pi(J)$ is a generating set for the ideal $(I+J)/I$ of $R/I$, and thus so is $\pi(J)$.

While this is pretty much just the third isomorphism theorem in disguise, I find this form more practical to use; e.g. it's much more natural to use in algebraic calculations like the one at the top of this answer.

The second isomorphism is, of course, induced by $\mathbf{Z}[x] / (x-5) \cong \mathbf{Z}$. Given your comments to the other answer, you were a little skittish about the transport of structure, so let me propose another theorem to prove as an exercise:

Theorem If $f : R \to S$ is an isomorphism, and $I$ is an ideal of $R$, then $f(I)$ is an ideal of $S$, and $f$ induces an isomorphism $R/I \to S/f(I)$.

Generally speaking, all ring-related algebraic structure on a ring $R$ can be transplanted to a ring $S$ by applying an isomorphism $R \to S$ or its inverse. e.g. it gives an isomorphism between lattice of ideals of $R$ and the lattice of ideals of $S$, and we get a bijection between the sets $\hom(R, T)$ and $\hom(S,T)$ for any ring $T$.