What does $\mathbb Z[x] / (2,x)$ look like?

Question

So $(2,x)$ is all linear combinations of $2$ and $x$. So the only thing that should not be in $(2,x)$ is odd coefficients and $1$ correct? Then $\mathbb Z[x] / (2,x) = \{p(x) + (2,x) | p(x) \in \mathbb Z[x]\}$. Isn't this simply identity? I am not sure. Any insight is appreciated, thanks.

Answer 1

By definition,

$(2, x) = \{2f(x) + xg(x), \; f(x), g(x) \in \Bbb Z[x] \}; \tag 1$

I claim that

$(2, x) = \{2f(x) + xg(x), \; f(x), g(x) \in \Bbb Z[x] \}$ $= \left \{ p(x) = \displaystyle \sum_0^n p_i x^i \in \Bbb Z[x], \; \; p_0 \in 2\Bbb Z, \; p_i \in \Bbb Z, \; 1 \le i \le n \right \}; \tag 2$

that is, the elements of $(2, x)$ are precisely those polynomials in $\Bbb Z[x]$ having even constant term; for if

$q(x) = \displaystyle \sum_0^m q_i x^i, \; q_0 = 2s \in 2\Bbb Z, \; q_i \in \Bbb Z, \; 1 \le i \le m; \tag 3$

we may write

$q(x) = 2s + x \displaystyle \sum_1^m q_i x^{i - 1}; \tag 4$

taking

$f(x) = s, \; g(x) = \displaystyle \sum_1^m q_i x^{i - 1} \tag 5$

shows that

$q(x) \in (2, x); \tag 6$

going the other way is even easier, since $xg(x)$ has no term of degree $0$; thus the constant term of $2f(x) + xg(x)$ is $2f_0 \in 2\Bbb Z$, i.e. every element of $(2, x)$ has constant term in $2\Bbb Z$, the evens.

Bearing this description of $(2, x)$ in mind, it is easy to find $\Bbb Z[x]/(2, x)$; we recall the elements of this ring are the (additive) cosets of $(2, x)$ in $\Bbb Z[x]$, and that for

$a(x), b(x) \in \Bbb Z[x], \tag 7$

we have

$a(x) + (2, x) = b(x) + (2, x) \Longleftrightarrow a(x) - b(x) \in (2, x) \Longleftrightarrow a_0 - b_0 \in 2\Bbb Z; \tag 8$

that is, the constant terms of $a(x)$ and $b(x)$ differ by an even integer; thus either $a_0$, $b_0$ are both even, or they are both odd; furthermore, all polynomials with even constant term lie in the same coset $(2, x)$, as do all those whose constant term is odd lie in $1 + (2, x)$ ; thus there are precisely two cosets of $(2, x)$, namely $0 + (2, x) = (2, x)$ and $1 + (2, x)$. The quotient ring $\Bbb Z[x]/(2, x)$ therefore is the unique two-element unital ring $\Bbb Z_2$:

$\Bbb Z[x]/(2, x) = \Bbb Z_2, \tag 9$

which is of course a field.

A even more detailed view of the sense in which $\Bbb Z[x]/(2, x)$ "looks like" $\Bbb Z_2$ may be had by considering the homomorphism

$\theta:\Bbb Z[x] \to \Bbb Z_2 \tag{10}$

which maps a polynomial

$p(x) = \displaystyle \sum_0^n p_ix^i \in \Bbb Z[x] \tag{11}$

to $p_0 \mod 2$:

$\theta \left ( \displaystyle \sum_0^n p_ix^i \right) = p_0 \mod 2 \in \Bbb Z_2; \tag{12}$

it is easy to verify that $\theta$ is a homomorphism; indeed it is evaluation at $0$ followed by reduction $\mod 2$; I leave the details to members of my vast audience. In the preceding discussion we have in fact shown that

$\ker \theta = (2, x), \tag{13}$

whence (9). Finally, as a "side benefit" of this analysis we see that $(2, x)$ is a maximal ideal in $\Bbb Z[x]$, since the quotient ring $\Bbb Z[x]/(2, x)$ is the field $\Bbb Z_2$.

Answer 2

There are ring homomorphisms $f:\mathbb Z[x]/(2,x)\to \mathbb Z/2\mathbb Z:f(x)\mapsto f(0),$ and $g:\mathbb Z/2\mathbb Z\to\mathbb Z[x]/(2,x):a\mapsto a.$ It is easily seen that these are inverses of each other, and hence, $\mathbb Z[x]/(2,x)\cong \mathbb Z/2\mathbb Z.$

What to check:

• That these maps are well-defined

• That these maps are ring homomorphisms

• That these maps are inverses of each other