Show that $K[x]/(x^2)$ is algebraic over field $K$

Question

Show that $K[x]/(x^2)$ is algebraic over field $K$.

Answer 1

This question is a little off the well-beaten path, since $x^2$ is not irreducible in $K[x]$, which means that $K[x]/(x^2)$ is not a field; indeed the coset

$\bar x = x + (x^2) \in K[x]/(x^2) \tag 1$

is nilpotent, viz.

$\bar x^2 = x^2 + (x^2) = (x^2) = 0 \in K[x]/(x^2); \tag 2$

nevertheless, even though $K[x]/(x^2)$ is not a field or even an integral domain, we may meaningfully ask if its elements satisfy polynomials in $K[x]$, which is tantamount to them being algebraic over $K$. This program is facilitated by first characterizing the elements of $K[x]/(x^2)$ themselves. Evidently the members of $K[x]$ are all of the form $a \bar x + b$ with $a, b \in K$, since for any $p(x) \in K[x]$ we may write

$p(x) = \displaystyle \sum_0^{\deg p} p_i x^i$ $= p_0 + p_1 x + \displaystyle \sum_2^{\deg p}p_i x^i = p_0 + p_1 x + x^2 \sum_2^{\deg p} p_i x^{i - 2}, \tag 3$

so in passing to (projecting onto) $K[x]/(x^2)$ we obtain

$p(\bar x) = \displaystyle \sum_0^{\deg p} p_i \bar x^i$ $= p_0 + p_1 \bar x + \displaystyle \sum_2^{\deg p} p_i \bar x^i = p_0 + p_1 \bar x + \bar x^2 \sum_2^{\deg p} p_i \bar x^{i - 2}, \tag 4$

which reduces to

$p(\bar x) = p_0 + p_1 \bar x \tag 5$

in the light of (2). This shows that every element of $K[x]/(x^2)$ may be written $a \bar x + b$ as claimed.

We have

$(a \bar x + b)^2 = a^2 \bar x^2 + 2ab\bar x + b^2$ $= 2ab\bar x + b^2 = 2b(a \bar x + b) - b^2, \tag 6$

whence

$(a \bar x + b)^2 - 2b(a \bar x + b) + b^2 = 0; \tag 7$

$a \bar x + b$ thus is a zero of

$x^2 - 2bx + b^2 \in K[x]; \tag 8$

this can also be simply seen in one line by noting that

$(a \bar x + b)^2 - 2b(a \bar x + b) + b^2 = ((a \bar x + b) - b)^2$ $= (a \bar x)^2 = a^2 \bar x^2 = 0 \in K[x]/(x^2); \tag 9$

either way you like to look at it, we have shown that every element of $K[x]/(x^2)$ is algebraic, in fact quadratic, over $K$. Thus we may affirm that $K[x]/(x^2)$ is algebraic over $K$ as well.

Answer 2

For commutative rings that are not necessarily fields, the usual notion is of integral element, instead of algebraic element.

$K[x]/(x^2)$ is a finite-dimensional $K$-algebra. As in the field extension case, it is easy to prove that every finite-dimensional unital commutative $K$-algebra is an integral extension of $K$.

This follows because the standard equivalence holds: $a \in A$ is integral over $K$ iff $K[a]$ is a finite dimensional algebra over $K$.