I'm asked to prove that the quotient ring $\mathbb{Z}[x]/\langle x, 2\rangle$ is isomorphic to $\mathbb{Z}_2$.

The first thing that came to my mind was to apply the Theorem of Homomorphism of Rings, but to do so I need to construct a homomorphism $\phi: \mathbb{Z} \to \mathbb{Z}_2$, such that $\mathrm{Ker}\phi = \langle x, 2\rangle$ ($\langle x, 2\rangle$ is an ideal of $\mathbb{Z}[x]$). But

What elements are in $\mathbb{Z}[x]/\langle x, 2\rangle$ in the first place?

I know that $\langle x, 2\rangle$ of all the polynomials with even constant term, but I can't figure out how the elements of $\mathbb{Z}[x]/\langle x, 2\rangle$ look like.

Any help would be appreciated.

  • $\begingroup$ $(2,x)$ is not an ideal in $\mathbb{Z}$, because $x$ is an indeterminate, and not an integer. It is an ideal in $\mathbb{Z}[x]$. $\endgroup$ – Dietrich Burde May 18 '17 at 19:11
  • $\begingroup$ @DietrichBurde Ups. I meant in $\mathbb{Z}[x]$. I fixed it. $\endgroup$ – Jazz May 18 '17 at 19:57

To find your morphism, let $f: \mathbb{Z}[X] \to \mathbb{Z}$ with $f(P) = P(0)$. It is easy to check that $f$ is a surjective ring morphism. Moreover, let $\pi$ be the canonical projection $\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}$.

Thus $\pi\circ f $ is a surjective ring morphism. What is its kernel ? The first isomorphism theorem then concludes the proof.


First note that $\langle x \rangle$ is the set of all polynomials with $0$ constant term.

Two elements $p(x), q(x) \in \mathbb{Z}[x]$ are identified in $\mathbb{Z}[x]/\langle x,2\rangle$ if and only if their difference $p(x)-q(x)$ is an element of $\langle x,2 \rangle$.

Observe that we must have that all polynomials with the same constant term are identified in the quotient ring, since their difference would have $0$ constant term, and thus be an element of $\langle x \rangle \subset \langle x , 2 \rangle$. So in fact every polynomial is in the same equivalence class as its constant term! Thus we have at most one equivalence class for each integer.

Now note that $\langle 2 \rangle$ contains all polynomials with even coefficients, and hence also contain all even integers. So if two constant have the same parity they are identified, since then their difference is in $\langle 2 \rangle \subseteq \langle x ,2 \rangle$.

We have observed that polynomials are identified with their constant term, and constant terms are identified if they have the same parity. Thus we have at most two equivalence classes; one for polynomials with even constant term, and another for polynomials with odd constant term.

Now to see that we have at least two equivalence classes, we can show that $1$ is not identified with $0$, which amounts to showing that $1-0 = 1 \not \in \langle x , 2 \rangle$. This should not be too much work if you recall that two polynomials are the same in $\mathbb{Z}[x]$ if and only if they have the same coefficients.


We have $$ \mathbb{Z}[x]/(2,x)\cong (\mathbb{Z}/2)[x]/(x)\cong \mathbb{Z}/2, $$ which says that a polynomial $f(x)\in \mathbb{Z}[x]$ modulo $x$ is just like an integer, and then modulo $2$ we obtain just $\mathbb{Z}/2$ (or first modulo $2$, and then modulo $x$).

Remark: Let $p$ be a prime number and $f(x)\in \mathbb{Z}[x]$ with $gcd(p,f(x))=1$. Then $$ \mathbb{Z}[x]/(p,f(x))\cong (\mathbb{Z}/p)[x]/(f(x)). $$

  • $\begingroup$ How did you come up with the quotient ring of the middle? So $\mathbb{Z}[x]/(2,x)$ is the set of all polynomials modulo $x$ and $2$ ''separately''? $\endgroup$ – Jazz May 18 '17 at 19:08
  • $\begingroup$ Well, basically yes, because $gcd(2,x)=1$ in $\mathbb{Z}[x]$. It is $R[x]/(x)\cong R$ with $R=\mathbb{Z}/2$. $\endgroup$ – Dietrich Burde May 18 '17 at 19:09

Well, the first homomorphism theorem states that $\mathrm{Im} f \cong A/\ker f$ for some morphism $f:A \to B$. In particular, this implies that $(A/J)/(I/J) \cong A/I$. For your isomorphism, note that $k[x]/(x,2) \cong \frac{\mathbb Z[x]/(2)}{(x,2)/(2)}=\mathbb Z_2[x]/(x) =\mathbb Z_2$ where the last isomorphism is induced by the relation $x \mapsto 0$ in the quotient ring.


Let $\pi : R \to R/I$ be the canonical morphism of the quotient.

By far the most common ways to express an element $R/I$ boil down to

  • by $\pi(r)$, where $r$ expresses an element of $R$
  • by $\pi(r)$, where $r$ expresses an element of $R$, but restricted to a carefully chosen selection of one element per congruence class
  • by $\theta(s)$, where $s$ expresses an element of $S$ and $\theta : S \to R/I$ is an isomorphism

If the second bullet point is unclear, a common example is that first introductions to the integers modulo $N$ often express elements of that ring as the integers $\{ 0, 1, 2, \ldots, N-1 \}$.

When $R/I$ is constructed as a ring of cosets, $\pi(r) = r + I$. Sometimes the notation $[r]_I$ is used.


Suppose we have:

$f(x) = a_0 + a_1x +\cdots + a_nx^n$.

Since $a_0 \in \Bbb Z$, we can re-write (via the Euclidean (division) algorithm) this as:

$(2k + r) + x(a_1 + \cdots + a_nx^{n-1}) = r + 2k + xg(x)$, where $r = 0$ or $r = 1$.

Now $2k \in \langle 2,x\rangle$, and $xg(x) \in \langle 2,x\rangle$, so:

$2k + xg(x) \in \langle 2,x\rangle$, and thus:

$f(x) + \langle 2,x\rangle = r + \langle 2,x\rangle$.

This shows that $\Bbb Z[x]/\langle 2,x\rangle$ has (at most) two distinct cosets.

Let's answer your question: what do elements of $\langle 2,x\rangle$ look like?

As we saw above, $\langle 2,x\rangle$ clearly contains the set:

$J = \{2k + xg(x)| k\in \Bbb Z, g(x) \in \Bbb Z[x]\}$.

This is clearly an additive subgroup of $\Bbb Z[x]$. If we take any

$p(x) = b_0 + b_1x +\cdots +b_mx^m = b_0 + xh(x)$ (where of course:

$h(x) = b_1 + b_2x + \cdots + b_mx^{m-1}$), then we have:

$p(x)(2k + xg(x)) = (b_0 + xh(x))(2k + xg(x))$

$= 2b_0k + x(b_0g(x) + 2kh(x) + xg(x)h(x)) \in J$.

So $J$ is an ideal contained in $\langle 2,x\rangle$.

But $J$ contains $x = 2\cdot 0 + x\cdot 1$, and $2 = 2\cdot 1 + x\cdot 0$, that is, $J$ is an ideal containing $2$ and $x$. Since $\langle 2,x\rangle $ is the smallest such ideal, we have $\langle 2,x\rangle \subseteq J$.

We thus have $J = \langle 2,x\rangle$.

Returning to our original discussion, we now show that $\Bbb Z[x]/J$ has exactly two cosets (we already know it has at most two).

For suppose $J = 1+J$. This means $1 - 0 = 1 \in J$, so that $1 = 2k + xg(x)$.

Since $1$ is a constant polynomial, we must have for $1 = 2k + xg(x)$, that $g(x) = 0$ (arguing by degree), leading to $1 = 2k$, for an integer $k$, contradiction.

Thus $f(x) \mapsto f(x) + J$ is an onto ring-homomorphism to $\{J,1+J\}$. The isomorphism with $\Bbb Z_2$ should now be clear.


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