$\mathbb K[s, t]/\langle s-t\rangle\simeq \mathbb K[t]$ as $\mathbb K$-algebras?

Question

How can I show there is an isomorphism of $\mathbb K$-algebras (where $\mathbb K$ is a field): $$\mathbb K[s, t]/\langle s-t\rangle\simeq \mathbb K[t]?$$

Above $\mathbb K[s, t]$ and $\mathbb K[t]$ are the $\mathbb K$-algebras of polynomials in the variables $s$, $t$ and $t$, respectively.

I tried doing it but I don't feel much confortable working with these algebras.

Thanks.

Answer 1

I agree that working with the object on the left is a bit uncomfortable, that is why we are so happy that it is isomorphic to the object on the right! However, before we 'know' (i.e. are allowed to use) that fact, the situation is not completely hopeless either.

I think what confuses you is how to write down homomorphisms whose domain is defined as a quotient of something else instead as an algebra where we can write down all the elements such as $K[s, t]$. The secret to defining morphisms (like the one MooS proposes) on the quotient $K[s, t]/I$ (with $I$ any ideal) is this:

1. Write down the morphism $\phi$ as if the domain is just $K[s, t]$
2. Check that it is well defined on the quotient.

Step 1 is the reason this two step plan is so nice: we can just think of the domain of $\phi$ as if it are just polynomials. Replace any element of the quotient (formally an equivalence class of elements of $K[s, t]$) that you want to feed into $\phi$ by one of its representatives in $K[s, t]$. Of course this only works if step 2 checks out.

Step 2 means: if two elements of $K[s, t]$ are sent to the same element of $K[s, t]/I$ when quotienting $I$ out, does the map $\phi$ behave the same on these elements?

It is easy to see that we need a 'yes' answer to that in order to make sense of $\phi$ in the quotient algebra. Now when are two elements $x$ and $y$ in $K[s, t]$ sent to the same element in the quotient? Exactly when they differ by an element of $I$!

So... $\phi$ is well defined on $K[s, t]/I$ when $\phi(x) = \phi(x + i)$ for every $i \in I$. But $\phi$ is linear so this happens exactly when $\phi(i) = 0$ for every $i \in I$.

In summary: step 2 can be replaced by:

2. Check that $\phi$ sends every element of $I$ to $0$.

Answer 2

I'd forgotten I had seen what @Vicent suggested as a theorem in my algebra course:

Theorem: Let $f:A\longrightarrow B$ be a $R$-algebra morphism and $I$ an ideal of $A$ such that $I\subseteq \textrm{ker}(f)$. There exists a unique $R$-algebra morphism $\overline{f}:A/I\longrightarrow B$ such that $\overline{f}\circ \pi_I=f$ where $\pi_I:A\longrightarrow A/I$ is the canonical projection.

The proof is straightforward.

Briefly, my problem amounts finding a morphism such that $\langle s-t\rangle\subset \textrm{ker}(f)$. At least this will induce a morphism in the quotient (which will turn out being an isomorphism).