Otdfbt

Adjusted R2 is said to be more unbiased than ordinary R2 as it takes the number of explanatory variables into account.

Can adjusted R2 be used in a model with only an intercept and one independent variable?

Following this, let's say we want to compare two nested linear models, one with 1 independent variable and the other adding an additional variable. I gather that adjusted R2 should be calculated for the bigger model, but should this be compared with the smaller model's ordinary or adjusted R2?

With only one (or only a few) predictor variables, the adjusted R-squared can be used, but it will not be very different from the unadjusted R-square, so it doesn't really matter. The adjustment was invented as a solution to problems caused by variable selection, so if you are not doing variable selection it isn't necessary.

But, if you are using R-square to compare models, you should use the same version in all cases. So maybe just stay with adjusted R-square.

Some additional notes to what has been said so far.

Note that $R^2$ can not decrease if one adds new variable but only increase. So even if you would add random variables $R^2$ can become quite high. See the following example from the R code:

set.seed(10) # make the example reproducible
n <- 100 # sample size
k <- 20 # number of predictors
df <- data.frame(y= rnorm(n), matrix(rnorm(n*(k)), ncol= k)) # generate some *random* data
summary(lm(y ~ ., data= df)) # fit a regression model

# results
# Multiple R-squared: 0.2358
# Adjusted R-squared: 0.0423

$R^2$ is 0.2358% which is way too high if we keep in mind that we used only random variables. On the other hand, the $R^2_adj$ is 0.0423 which is much closer to what we would expect should happen if we use random variables.

This is great but if you use $R^2_adj$ for a few variables, keep in mind that $R^2_adj$ can have negative values. See here:

radj <- rep(NA, ncol(df) - 1) # vector for results
for(i in 2:ncol(df)) # determine radj for every x
radj[i-1] <- summary(lm(y ~ df[ , i], data=df))$adj.r.squared


sum(radj < 0) # number of negative radj
# 11

In this example 11 of 20 predictors have a negative $R^2_adj$. I agree with the suggestion of @kjetil b halvorsen (+1). I just want to point out this property of $R^2_adj$ which you might encounter since you want to use $R^2_adj$ for a few variables and because a negative value might be confusing at first.