Polynomials in Integers with Even Constant Term forms Ideal

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Theorem

Let $\Z \sqbrk X$ be the ring of polynomials in $X$ over $\Z$.

Let $S \subseteq \Z \sqbrk X$ be the set of polynomials over $\Z$ in $X$ which have a constant term which is even.


Then $S$ is an ideal of $\Z \sqbrk X$.


Proof

For example, $X + 2$ is a polynomials over $\Z$ in $X$ with an even constant term.

So $S$ is not empty.


Let $P_1 = \displaystyle \sum_{k \mathop = 0}^n a_k X^k$ and $P_2 = \displaystyle \sum_{k \mathop = 0}^n b_k X^k$ be elements of $S$.

We have:

\(\displaystyle P_1 - P_2\) \(=\) \(\displaystyle \sum_{k \mathop = 0}^n a_k X^k + \sum_{k \mathop = 0}^n b_k X^k\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 0}^{\max \set {m, n} } \paren {a_k - b_k} X^k\)

The constant term of $P_1 - P_2$ is $a_0 - b_0$ which is even.

Thus $P_1 - P_2 \in S$.

Let $P_3 = \sum_{k \mathop = 0}^s c_k X^k \in \Z \sqbrk X$.

Then the constant term of $P_3 \times P_1$ is $c_0 \times a_0$.

As $a_0$ is even, so is $c_0 \times a_0$.

The result follows by Test for Ideal.

$\blacksquare$


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